标签：tmp return OJ int 1200 long 1300 include dp
1206: Card game
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
double dp[55][55][650];
int a[55],b[55];
int check(char ch)
{
if(ch=='A') return 1;
if(ch=='X') return 10;
if(ch=='J') return 11;
if(ch=='Q') return 12;
if(ch=='K') return 13;
return ch-'0';
}
int main()
{
int n,k,t;
scanf("%d",&t);
while(t--)
{
char str[3];
double C=1;
double f1[650],f2[650];
int sum1=0,sum2=0;
scanf("%d%d",&n,&k);
for(int i=1;i<=k;i++)
C=C*(n-i+1)/i;
for(int i=1;i<=n;i++)
{
scanf("%s",str);
a[i]=check(str[0]);
sum1+=a[i];
}
for(int i=1;i<=n;i++)
{
scanf("%s",str);
b[i]=check(str[0]);
sum2+=b[i];
}
memset(dp,0,sizeof(dp));
for(int i=0;i<=n;i++)
dp[i][0][0]=1;
for(int i=1;i<=n;i++)
for(int j=1;j<=i&&j<=k;j++)
for(int d=1;d<=sum1;d++)
{
if(d>=a[i])
dp[i][j][d]=dp[i-1][j][d]+dp[i-1][j-1][d-a[i]];
else
dp[i][j][d]=dp[i-1][j][d];
}
for(int i=1;i<=sum1;i++)
f1[i]=1.0*dp[n][k][i]/C;
memset(dp,0,sizeof(dp));
for(int i=0;i<=n;i++)
dp[i][0][0]=1;
for(int i=1;i<=n;i++)
for(int j=1;j<=i&&j<=k;j++)
for(int d=1;d<=sum2;d++)
{
if(d>=b[i])
dp[i][j][d]=dp[i-1][j][d]+dp[i-1][j-1][d-b[i]];
else
dp[i][j][d]=dp[i-1][j][d];
}
f2[0]=0.0;
for(int i=1;i<=sum1;i++)
{
f2[i]=1.0*dp[n][k][i]/C;
f2[i]+=f2[i-1];
}
double cnt=0;
for(int i=1;i<=sum1;i++)
cnt+=(double)f1[i]*f2[i-1];
printf("%.6lf\n",1.0*cnt);
}
return 0;
}1208: Fibonacci sum
#include<stdio.h>
#include<string.h>
#define MOD 1000000007
#define maxn 25
long long C[25][25];
long long ret[maxn][maxn];
long long init[maxn][maxn];
long long buf[maxn][maxn];
void matrixMul(long long a[][maxn] , long long b[][maxn] , long long n,long long mod)
{
long long i,j,k;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
buf[i][j]=0;
}
}
for(i=0;i<n;i++)
{
for(k=0;k<n;k++)
{
if(a[i][k]==0)
continue;
for(j=0;j<n;j++)
{
if(b[k][j]==0)
continue;
buf[i][j]+=a[i][k]*b[k][j];
if (buf[i][j]>=mod||buf[i][j]<=-mod)
{
buf[i][j]%=mod;
}
}
}
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
a[i][j]=buf[i][j];
}
}
}
void matrixMul(long long n,long long m,long long mod)
{
long long i,j;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
ret[i][j]=(i==j);
}
}
for(;m;m>>=1)
{
if(m&1)
{
matrixMul(ret,init,n,mod);
}
matrixMul(init,init,n,mod);
}
}
void initc()
{
C[0][0]=C[1][0]=C[1][1]=1;
for(int i=2;i<=20;i++)
{
for(int j=0;j<=i&&j<=20;j++)
{
if(j==0)
C[i][j]=1;
else
C[i][j]=(C[i-1][j]+C[i-1][j-1])%MOD;
}
}
}
int main()
{
initc();
int t,n,k,i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&k);
if(n==1)
{
printf("1\n");
continue;
}
else if(n==2)
{
printf("2\n");
continue;
}
memset(init,0,sizeof(init));
for(i=0;i<=k;i++)
{
for(j=0;j<=k-i;j++)
{
init[j][i]=C[k-i][j];
}
}
for(i=0;i<=k;i++)
{
init[i][k+1]=C[k][i];
}
init[k+1][k+1]=1;
matrixMul(k+2,n-2,MOD);
long long ans=0;
for(i=0;i<=k;i++)
{
ans=(ans+ret[i][k+1])%MOD;
}
ans=(ans+2*ret[k+1][k+1])%MOD;
printf("%lld\n",ans);
}
return 0;
}1211: 大整数开平方练习
import java.util.*;
import java.math.*;
public class Main {
public static void main(String [] args)
{
Scanner input = new Scanner(System.in);
while(input.hasNext())
{
String str = input.next();
BigInteger num = new BigInteger(str);
int len = str.length();
int len1 = (len+1)/2;
BigInteger factor = BigInteger.valueOf(20);
BigInteger factor1 = BigInteger.valueOf(10);
BigInteger div = BigInteger.valueOf(0);
BigInteger ans = BigInteger.valueOf(0);
BigInteger left = BigInteger.valueOf(0);
int t = 0 , index = 0;
if(len%2 == 1){
t = str.charAt(0)-'0';
index+=1;
}else{
t = str.charAt(0)-'0';
t = t*10 + str.charAt(1)-'0';
index+=2;
}
int m = (int)Math.sqrt(t*1.0+0.1);
ans = BigInteger.valueOf(m);
div = BigInteger.valueOf(m).multiply(factor);
left = BigInteger.valueOf(t-m*m);
// System.out.println(ans);
BigInteger tmp = BigInteger.valueOf(0);
for(int i=1 ; i<len1 ; i++ , index+=2){
left = left.multiply(BigInteger.valueOf(100));
int val = str.charAt(index)-'0';
val = val*10+str.charAt(index+1)-'0';
left = left.add(BigInteger.valueOf(val));
// System.out.println("div: "+i+" "+div+" "+left);
int j=9;
for(; j>=0 ; j--){
tmp = div.add(BigInteger.valueOf(j));
if(left.compareTo(tmp.multiply(BigInteger.valueOf(j)))>=0) break;
}
left = left.subtract(tmp.multiply(BigInteger.valueOf(j)));
ans = ans.multiply(factor1).add(BigInteger.valueOf(j));
div = ans.multiply(factor);
}
System.out.println(ans);
}
}
}#include <iostream>
#define INF 1e9
using namespace std;
const int maxn = 105;
int n,m;
int dp[maxn][maxn];
void init()
{
for(int i = 0;i < n;++i)
for(int j = 0;j < n;++j)
if(i == j) dp[i][j] = 0;
else dp[i][j] = INF;
}
bool floyd()
{
for(int k = 0;k < n;++k)
for(int i = 0;i < n;++i)
for(int j = 0;j < n;++j)
dp[i][j] = min(dp[i][k] + dp[k][j],dp[i][j]);
for(int i = 0;i < n;++i){
dp[i][n] = 0;
for(int j = 0;j < n;++j){
if(dp[i][n] >= INF) return false;
dp[i][n] = max(dp[i][j],dp[i][n]);
}
}
return true;
}
int main(void)
{
int a,b,c;
while(cin >> n >> m){
init();
while(m--){
cin >> a >> b >> c;
dp[a][b] = dp[b][a] = c;
}
bool flag = floyd();
int res = INF;
for(int i = 0;i < n;++i){
res = min(res,dp[i][n]);
}
if(flag) cout << res << endl;
else cout << "Can not" << endl;
}
return 0;
}
1226: ACM小组的内战
#include<cstdio>
#include<cstring>
#include<cstdlib>
#define SIZE 300
char s1[SIZE], s2[SIZE], s3[SIZE], s4[SIZE], a[SIZE], b[SIZE], c[SIZE], d[SIZE];
bool f1[100], f2[100];
int main()
{
while( scanf( "%s%s%s%s", s1, s2, s3, s4) == 4)
{
memset( f1, false, sizeof f1);
memset( f2, false, sizeof f2);
int len = strlen( s1);
for( int i = 0; i < len; i ++)
{
if( !f1[ s1[i] - 48 ] && !f2[ s2[i] - 48] ) {
a[ s1[i] - 48 ] = s2[i];
f1[ s1[i] - 48 ] = true;
b[ s2[i] - 48 ] = s1[i];
f2[ s2[i] - 48 ] = true;
}
if( f1[ s1[i] - 48 ] && a[ s1[i] - 48 ] != s2[i])
f1[ s1[i] - 48 ] = false;
if( f2[ s2[i] - 48 ] && b[ s2[i] - 48 ] != s1[i])
f2[ s2[i] - 48 ] = false;
}
bool flag = true;
len = strlen(s3);
for( int i = 0; i < len; i ++)
{
if( !f1[ s3[i] - 48 ]) {
flag = false;
break;
}
else
c[i] = a[ s3[i] - 48 ];
}
if( flag)
for( int i = 0; i < len; i ++)
printf( "%c", c[i]);
else
printf( "Wrong");
printf( "\n");
flag = true;
len = strlen(s4);
for( int i = 0; i < len; i ++)
{
if( !f2[ s4[i] - 48 ]) {
flag = false;
break;
}
else
d[i] = b[ s4[i] - 48];
}
if( flag)
for( int i = 0; i < len; i ++)
printf( "%c", d[i]);
else
printf( "Wrong");
printf( "\n");
}
return 0;
}1228: ACM小组的数字游戏
#include<iostream>
using namespace std;
int a[17], i, j, pos;
bool isOk(int positon)
{
if ((a[positon] < 1) || (a[positon]>16))
return false;
for (int i = 0; i < positon; i++)
if (a[i] == a[positon])
return false;
if ((positon != pos) && a[positon] == 1)
return false;
if ((positon == pos) && a[positon] != 1)
return false;
return true;
}
int main()
{
cin >> i >> j;
pos = (i - 1) * 4 + j;
for (a[1] = 1; a[1] <= 16; a[1]++)
if (isOk(1))
for (a[2] = 1; a[2] <= 16; a[2]++)
if (isOk(2))
for (a[3] = 1; a[3] <= 16;a[3]++)
if (isOk(3))
{
a[4]= 34 - a[1] - a[2] - a[3];
if (isOk(4))
for (a[5] = 1; a[5] <= 16;a[5]++)
if (isOk(5))
{
a[6] = 34 - a[1] - a[2] - a[5];
if (isOk(6))
for (a[7] = 1; a[7] <= 16;a[7]++ )
if (isOk(7))
{
a[8] = 34 - a[5] - a[6] - a[7];
if ((a[3] + a[4] + a[7] + a[8] == 34) && isOk(8))
for (a[9] = 1; a[9] <= 16; a[9]++)
if (isOk(9))
for (a[10] = 1; a[10] <= 16;a[10]++)
if (isOk(10))
{
a[11] = 34 - a[6] - a[7] - a[10];
if (isOk(11))
{
a[12] = 34 - a[9] - a[10] - a[11];
if (isOk(12))
{
a[13] = 34 - a[1] - a[5] - a[9];
if (isOk(13) && (a[4] + a[7] + a[10] + a[13] == 34))
{
a[14] = 34 - a[2] - a[6] - a[10];
{
if (isOk(14) && a[9] + a[10] + a[13] + a[14] == 34)
{
a[15] = 34 - a[3] - a[7] - a[11];
if (isOk(15))
{
a[16] = 34 - a[13] - a[14] - a[15];
if (isOk(16) && (a[11] + a[12] + a[15] + a[16] == 34) && (a[1] + a[4] + a[13] + a[16] == 34) && (a[1] + a[6] + a[11] + a[16] == 34))
{
for (int i = 1; i <=16; i++)
{
if (i % 4 != 0)
cout << a[i] << " ";
else
cout <<a[i]<< endl;
}
cout << endl;
}
}
}
}
}
}
}
}
}
}
}
return 0;
} 1230: 平面上的点
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define ll long long
void exgcd(ll a,ll b,ll &x,ll &y)
{
if(b==0)
{
x=1; y=0;
return ;
}
else
{
exgcd(b,a%b,y,x);
y=y-x*(a/b);
}
}
ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}
ll cal(ll xx1,ll xx2,ll yy1,ll yy2,ll a,ll b,ll m)
{
///一定要保证a,b,m为非负
if(m<0)
{
m=-m; a=-a; b=-b;
}
if(a<0)
{
a=-a; ll tmp=xx1;
xx1=-xx2;
xx2=-tmp;
}
if(b<0)
{
b=-b;
ll tmp=yy1;
yy1=-yy2;
yy2=-tmp;
}
ll x0,y0,d1,t1,t2,t3,t4;
ll x1,y1;
d1=gcd(a,b);
if(m%d1!=0) return 0;
a/=d1; b/=d1; m/=d1;
exgcd(a,b,x0,y0);
x0*=m,y0*=m;
///因为是关于t的单调函数,所以对于下边界要向上取整
///对于上边界要向下取整
t1=ceil(double(xx1-x0)/b); t2=floor(double(xx2-x0)/b);
t3=ceil(double(y0-yy2)/a); t4=floor(double(y0-yy1)/a);
t1=max(t1,t3); t2=min(t2,t4);
if(t2<t1) return 0;
return t2-t1+1;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("test.txt","w",stdout);
ll a,b,c,d;
ll x1,x2,y1,y2,z1,z2;
while(scanf("%lld%lld%lld%lld",&a,&b,&c,&d)!=EOF)
{
scanf("%lld%lld%lld%lld%lld%lld",&x1,&x2,&y1,&y2,&z1,&z2);
if(a==0&&b==0)
{
ll cnt=0;
for(int z=z1;z<=z2;z++)
{
if(c*z+d==0)
cnt++;
}
cnt*=(x2-x1+1)*(y2-y1+1);
printf("%lld\n",cnt);
continue;
}
if(a==0)
{
ll cnt=0;
for(int z=z1;z<=z2;z++)
{
ll tmp=-1*(d+c*z);
if(tmp%b==0&&tmp/b>=y1&&tmp/b<=y2)
cnt++;
}
cnt*=(x2-x1+1);
printf("%lld\n",cnt);
continue;
}
if(b==0)
{
ll cnt=0;
for(int z=z1;z<=z2;z++)
{
ll tmp=-1*(c*z+d);
if(tmp%a==0&&tmp/a>=x1&&tmp/a<=x2)
cnt++;
}
cnt*=(y2-y1+1);
printf("%lld\n",cnt);
continue;
}
ll ans=0;
for(int z=z1;z<=z2;z++)
{
ll m=-1*(d+c*z);
ans+=(cal(x1,x2,y1,y2,a,b,m));
}
printf("%lld\n",ans);
}
return 0;
}1232: 懒汉的旅行
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <sstream>
#include <algorithm>
//#include <map>
#include <set>
#include <cmath>
#include <bitset>
#define INF 0x3f3f3f3f
using namespace std;
struct edge{
int v;int len;
};
struct status{
int step,cost,u,oil;
bool operator < (const status &b) const
{
if(step != b.step) return step > b.step;
else if(cost != b.cost) return cost > b.cost;
else if(oil != b.oil) return oil < b.oil;
}
}mark[105];
vector<edge> vec[105];
int N,M,L,Y;
int start,finish;
int bfs()
{
priority_queue<status> q;
q.push((status){0,0,start,L}) ;
while(!q.empty())
{
status now = q.top();
q.pop();
bool f = false;
if(now.cost<mark[now.u].cost) f = true,mark[now.u]=now;
if(now.step<mark[now.u].step) f = true, mark[now.u]=now;
if(now.oil>mark[now.u].oil) f = true, mark[now.u]=now;
if(!f)continue;
if(now.u == finish)
return now.cost;
for(auto &Next : vec[now.u])
{
status tmp = now;
tmp.u = Next.v;
if(Next.len > L)
continue;
if(tmp.oil >= Next.len)
q.push((status){now.step,now.cost,Next.v,now.oil-Next.len});
q.push((status){now.step+1,now.cost+Y*(L-now.oil),Next.v,L-Next.len});
}
}
return -1;
}
int main()
{
while(scanf("%d%d%d%d",&N,&M,&L,&Y) != EOF)
{
for(int i = 0 ; i <= N ; i++)
mark[i]=(status){INF,INF,i,-1},vec[i].clear();
scanf("%d%d",&start,&finish);
for(int i = 0 ; i < M ; i++)
{
int u,v,l;
scanf("%d%d%d",&u,&v,&l);
vec[u].push_back((edge){v,l});
vec[v].push_back((edge){u,l});
}
printf("%d\n",bfs());
}
return 0;
}1233: 病毒的复制
# include <stdio.h>
# define MAX 1000003
typedef struct Matrix
{
long long a,b;
long long c,d;
} matrix;
long long x[2], t;
int solve(matrix mat)
{
int tot;
matrix cur, cross, tmp;
cur = mat;
if (t & 0x1) cross = cur;
else
{
cross.a = cross.d = 1;
cross.b = cross.c = 0;
}
while (t>0)
{
t >>= 1;
tmp = cur;
cur.a = (tmp.a*tmp.a+tmp.b*tmp.c) % MAX;
cur.b = (tmp.b*(tmp.a+tmp.d)) % MAX;
cur.c = (tmp.c*(tmp.a+tmp.d)) % MAX;
cur.d = (tmp.d*tmp.d+tmp.b*tmp.c) % MAX;
tmp = cross;
if (t & 0x1)
{
cross.a = (tmp.a*cur.a+tmp.b*cur.c) % MAX;
cross.b = (tmp.a*cur.b+tmp.b*cur.d) % MAX;
cross.c = (tmp.c*cur.a+tmp.d*cur.c) % MAX;
cross.d = (tmp.c*cur.b+tmp.d*cur.d) % MAX;
}
}
tot = (int)((cross.a*x[0]+cross.b*x[1]+cross.c*x[0]+cross.d*x[1])%MAX);
return tot;
}
int main()
{
int tot;
matrix mat;
while (~scanf("%Illd%lld%lld%lld%lld%lld%lld", &x[0],&x[1],&mat.a,&mat.c,&mat.b,&mat.d,&t))
{
tot = solve(mat);
printf("%d\n", tot);
}
return 0;
}#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
int const maxn=31000;
long long Min[maxn*4],Max[maxn*4];
void Build(int rt,int l,int r)
{
if(l>=r)
{
Min[rt]=0;
return ;
}
int m=(l+r)/2;
Build(rt*2,l,m);
Build(rt*2+1,m+1,r);
Min[rt]=min(Min[2*rt],Min[2*rt+1]);
}
void update(int rt,int l,int r,int x,int num)
{
if(l>=r)
{
Min[rt]=num;
return ;
}
int m=(l+r)/2;
if(m>=x)
update(rt*2,l,m,x,num);
if(m<x)
update(rt*2+1,m+1,r,x,num);
Min[rt]=min(Min[rt*2],Min[2*rt+1]);
}
long long query(int rt,int l,int r,int ql,int qr)
{
long long ans=0x7fffffff;
if(ql<=l&&qr>=r)
{
return Min[rt];
}
int m=(l+r)/2;
if(m>=ql)
ans=min(ans,query(rt*2,l,m,ql,qr));
if(m<qr)
ans=min(ans,query(rt*2+1,m+1,r,ql,qr));
return ans;
}
int main()
{
//freopen("test0.in","r",stdin);
//freopen("out.txt","w",stdout);
int n,t,p;
while(scanf("%d%d%d",&n,&t,&p)!=EOF)
{
memset(Min,0,sizeof(Min));
while(p--)
{
int x,y;
char str[10];
scanf("%d%s",&x,str);
if(str[0]=='p')
{
int l=1,r=n;
int tmp;
while(l<=r)
{
int m=(l+r)/2;
if(query(1,1,n,l,m)<=x)
r=m-1,tmp=m;
else
l=m+1;
}
update(1,1,n,tmp,x+t*60);
printf("%d\n",tmp);
}
else
{
scanf("%d",&y);
int k=query(1,1,n,y,y);
//cout<<k<<endl;
if(k<=x)
printf("Cannot\n");
else
{
printf("Fire!\n");
update(1,1,n,y,x+t*60);
}
}
}
}
return 0;
}#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
char st1[20001],st2[12000001];
int nextt[1000001];
bool equ(int i,char a[],int j,char b[])
{
if(a[i]==b[j]&&a[i+1]==b[j+1]) return true;
return false;
}
void getnext()
{
int i,j,len=strlen(st1);
nextt[0]=-2;i=0;j=-2;
while(i<len)
{
if(j==-2||equ(i,st1,j,st1))
{
i+=2;j+=2;//++i;++j;
if(!equ(i,st1,j,st1)) nextt[i>>1]=j;
else nextt[i>>1]=nextt[j>>1];
}
else j=nextt[j>>1];
}
}
void kmp()
{
int i=0,j=0;
getnext();
int l1=strlen(st1),l2=strlen(st2);
int ans=0;
while(i<l2)
{
if(j==-2||equ(j,st1,i,st2))
i+=2,j+=2;
else j=nextt[ j>>1 ];
if(j==l1) ++ans,j=nextt[j>>1];
}
cout<<ans<<endl;
}
int main()
{
while(scanf("%s",st1)!=EOF){
scanf("%s",st2);
kmp();
}
return 0;
}1241: 数字序列
#include <cstdio>
#include <cstring>
int vis[10010],num[10010],r[10010],fa[10010];//vis表示该头目节点是否访问过 num[i]表示i作为头目节点所表示的数，但不是确切的数，相对于集体来讲的 r[i]表示i节点与头目节点的相对值 fa[i]是i的父节点
int find(int x) //压缩路径，并且使得r[i]等于i相对于根节点的值的大小
{
if(x==fa[x])
{
return x;
}
int tmp=fa[x];
fa[x]=find(fa[x]);
r[x]=r[x]+r[tmp];
return fa[x];
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
char op[2];
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
for(int i=1;i<=n;i++)
{
fa[i]=i;
r[i]=0;
}
int flag=0;
for(int i=0;i<m;i++)
{
scanf("%s",op);
if(op[0]=='B')
{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
int fx=find(a);
int fy=find(b);
if(fx!=fy)//两集体合并
{
fa[fy]=fx;
r[fy]=r[a]-r[b]-c;//更新b的头目节点相对于a的头目节点的大小
if(vis[fy])//如果b的集体已经有确切的值了
{
int tmp=num[fy]-r[fy];
if(vis[fx])//如果a集体也已经有确切的值了
{
if(num[fx]!=tmp)//
{
flag=1;
}
}
else
{
num[fx]=tmp;
vis[fx]=1;
}
}
}
else//ab是一个集体中的
{
if(r[a]-r[b]!=c)
{
flag=1;
}
}
}
else
{
int a,b;
scanf("%d%d",&a,&b);
int fx=find(a);
int tmp=b-r[a];
if(vis[fx])
{
if(num[fx]!=tmp)
{
flag=1;
}
}
else
{
num[fx]=tmp;
vis[fx]=1;
}
}
}
if(flag)
{
printf("RE\n");
}
else
{
for(int i=1;i<=n;i++)
{
if(find(i)==i&&vis[i]==0)
{
flag=1;
break;
}
}
if(flag)
{
printf("WA\n");
}
else
{
printf("AC\n");
}
}
}
return 0;
} 1244: 琼脂糖双向免疫扩散试验
#include<stdio.h>
#include<math.h>
#define EPS 1e-5
int main()
{
double rg,rb,cg,cb,d,t,s,res;
while(scanf("%lf%lf%lf%lf%lf%lf%lf",&rb,&rg,&cb,&cg,&d,&t,&s)!=EOF)
{
///directly calculate
res=s*rg*cg/rb/cb+1.0;
res=d/res;
///判断条件上WA了几次，精度没必要用EPS的
if(rb-res>=0.0000||d-rg-res<=0.0000||(rb*cb-res*t<EPS))
printf("Can not\n");
else
printf("%.4lf\n",res);
}
return 0;
}1245: 信使核糖核酸转录后剪接
#include<stdio.h>
#include<string.h>
#define N 1000010
char t(char c)
{
if(c=='A')
return 'U';
if(c=='G')
return 'C';
if(c=='C')
return 'G';
if(c=='T')
return 'A';
}
char dna[N],mrna[N];
int main()
{
int i,j,la,lb,f;
while(scanf("%s%s",dna,mrna)!=EOF)
{
la=strlen(dna);
lb=strlen(mrna);
f=1;
for(i=0;i<la;++i)
dna[i]=t(dna[i]);
if(la<lb)
f=0;
else
for(i=la-1,j=0;j<lb;--i,++j)
{
for(;i>=0&&(dna[i]!=mrna[j]);--i);
if(i==-1)
{
f=0;//表示匹配失败
break;
}
}
if(f)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}1246: 十二指肠钩口线虫
#include<stdio.h>
int main()
{
int s,v,t,m,e;
while(scanf("%d%d%d%d%d",&s,&v,&t,&m,&e)!=EOF)
{
int res=e*s;
int n=e/t;///完成吸血的个数
int f=(e-m)/t;///完成渗血的个数
if(f<0)
f=0;
res+=f*m*v;
for(int i=f+1;i<=n;++i)
res+=(e-i*t)*v;
printf("%d\n",res);
}
return 0;
}#include<stdio.h>
#include<string.h>
#define N 10001
int main()
{
int n,t1,t2,l,k,u,i,j;
int pre[N],tim[N],hasNext[N],allTime;
while(scanf("%d%d%d",&n,&t1,&t2)!=EOF)
{
memset(hasNext,0,(n+1)*sizeof(int));
for(i=1;i<=n;++i)
{
scanf("%d%d%d",&u,&l,&k);
pre[i]=u;
hasNext[u]=1;
if(k)
{
int a,b;
scanf("%d",&a);
tim[i]=a;
for(j=1;j<k;++j)
{
scanf("%d%d",&a,&b);
tim[i]+=b-a;
}
scanf("%d",&b);
tim[i]+=l-b;
tim[i]=tim[i]*t1+t2;
}
else
tim[i]=l*t1+t2;
}
int minn=1<<30,minInd,maxx=0,maxInd;
for(i=1;i<=n;++i)
{
if(!hasNext[i])
{
allTime=0;
j=i;
while(j!=0)
{
allTime+=tim[j];
j=pre[j];
}
if(allTime<minn)
{
minn=allTime;
minInd=i;
}
if(allTime>maxx)
{
maxx=allTime;
maxInd=i;
}
}
}
printf("%d %d\n",minInd,maxInd);
}
return 0;
}
1248: 非变性聚丙烯酰胺凝胶电泳
#include<stdio.h>
#include<string.h>
#define N 2001
int mass[N];
bool dp[N];
int main()
{
int n,m,i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
m-=18;
for(i=0;i<n;++i)
{
scanf("%d",&mass[i]);
mass[i]-=18;
}
memset(dp,0,sizeof(dp));
dp[0]=1;
for(i=0;i<n;++i)
{
for(j=(m<<1);j>=mass[i];--j)
dp[j]=dp[j]||dp[j-mass[i]];
}
for(i=0;;i++)
if(dp[m-i]||dp[m+i])
{
if(dp[m-i])
j=m+18-i;
else if(dp[m+i])
j=m+18+i;
break;
}
printf("%d\n",j);
}
return 0;
}1249: 竞争性酶抑制剂和同工酶
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
#define N 155
#define INF 1<<30
int n,m,f[N][N];
int s,e,c,pre[N];
int bfs()
{
queue< int > q;
memset(pre,-1,sizeof(pre));
q.push(s);
pre[s]=0;
int k,i,inc=INF;
while(!q.empty())
{
k=q.front();
q.pop();
for(i=1;i<=n;++i)
if(i!=s&&pre[i]==-1&&f[k][i])
{
pre[i]=k;
q.push(i);
inc=min(inc,f[k][i]);
if(i==e)
return inc;
}
}
return 0;
}
int flow()
{
int res=0;
while(1)
{
int inc=bfs();
if(inc==0)
break;
res+=inc;
int i=e,j;
for(;i!=s;)
{
j=pre[i];
f[j][i]-=inc;
f[i][j]+=inc;
i=j;
}
}
return res;
}
int main()
{
while(scanf("%d%d",&n,&m)==2)
{
memset(f,0,sizeof(f));
while(m--)
{
scanf("%d%d%d",&s,&e,&c);
f[s][e]+=c;
}
scanf("%d%d",&s,&e);
for(int i=1;i<=n;++i)
f[e][i]=f[i][s]=0;
printf("%d\n",flow());
}
return 0;
}#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=5005,maxm=20005,INF=1<<30;
int n,m,e;
// v存储对应有向边的终点，nxt存储当前边起点相连的下一条边的编号
// first存储对应起点的第一条边的编号，dis存储当前状态下的从起点
// 开始到当前点的最小权重和（一直在更新），inq储存是否在队列
// 中的标记 ，w储存对应边的权重
int v[maxm],nxt[maxm],w[maxm],first[maxn],dis[maxn],inq[maxn],fa[maxn];
int find(int a){return fa[a]==a?a:fa[a]=find(fa[a]);}
bool merge(int a, int b){
int faa=find(a),fab=find(b);
if(faa==fab)return false;
fa[fab]=faa;
return true;
}
void add(int a, int b, int c){
v[e]=b;
nxt[e]=first[a];
w[e]=c;
first[a]=e++;
}
void spfa(int source){
queue<int> q;
dis[source]=0;
inq[source]=1;
q.push(source);
while(!q.empty()){
int tmp=q.front();
q.pop();
inq[tmp]=0;
for(int i = first[tmp]; i != -1; i=nxt[i]){
if(dis[v[i]]>dis[tmp]+w[i]){
dis[v[i]]=dis[tmp]+w[i];
if(!inq[v[i]]){
q.push(v[i]);
inq[v[i]]=1;
}
}
}
}
}
int main(){
while(~scanf("%d%d",&n,&m)){
e=0;
memset(first,-1,sizeof(first));
memset(inq,0,sizeof(inq));
for(int i = 0; i < maxn; i++)dis[i]=INF;
for(int i = 0; i <= n; i++)fa[i]=i;
for(int i = 0; i < m; i++){
int a,b;
scanf("%d%d",&a,&b);
add(a,b,0);
add(b,a,1);
merge(a,b);
}
if(find(1)==find(n))
spfa(1);
if(dis[n]==INF)
printf("-1\n");
else
printf("%d\n",dis[n]);
}
return 0;
}1258: 维护序列
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std ;
#define MAXN 10005
#define M(x , y) (x+y)>>1
#define L(x) x<<1
#define R(x) x<<1|1
struct Tree{
int l ;
int r ;
int sum ;
int lazy;
}tree[MAXN * 4] ;
int n ;
int m ;
inline void pushup(int c){
int l = L(c) ;
int r = R(c) ;
tree[c].sum = tree[l].sum ^ tree[r].sum ;
if(tree[l].lazy && tree[r].lazy)
tree[c].lazy = 1 ;
}
void build(int l ,int r , int c){
tree[c].l = l ;
tree[c].r = r ;
tree[c].sum = 0 ;
tree[c].lazy = 0 ;
if(l == r)
return ;
build( l , M(l, r) , L(c)) ;
build( (M(l , r)) + 1 , r , R(c)) ;
}
void insert(int v , int l , int c){
if(tree[c].l == l && tree[c].r==l){
tree[c].sum = v ;
if(!v)
tree[c].lazy = 1 ;
return ;
}
int m = M(tree[c].l , tree[c].r) ;
if(l <= m){
insert(v , l , L(c)) ;
}
else{
insert(v , l , R(c)) ;
}
pushup(c) ;
}
void update(int l , int r , int c){
if(tree[c].lazy)
return ;
if(l == r && tree[c].l==tree[c].r && l==tree[c].l){
if(tree[c].sum != 0){
int k = tree[c].sum & (-tree[c].sum) ;
tree[c].sum = tree[c].sum^k ;
if(!tree[c].sum){
tree[c].lazy = 1 ;
}
}
return ;
}
int m = M(tree[c].l , tree[c].r) ;
if(l==tree[c].l && r == tree[c].r){
update(l , m , L(c) ) ;
update(m + 1 , r , R(c)) ;
}
else if(m >= r){
update(l , r , L(c)) ;
}
else if(m < l){
update(l , r , R(c)) ;
}
else {
update(l , m , L(c)) ;
update(m+1 , r , R(c)) ;
}
pushup(c) ;
}
void query(int &ans , int l , int r , int c){
if(tree[c].lazy){
ans ^= 0 ;
return ;
}
if(tree[c].l == l && tree[c].r == r){
ans ^= tree[c].sum ;
return ;
}
int m = M(tree[c].l , tree[c].r) ;
if(r <= m){
query(ans , l , r , L(c)) ;
}
else if(l > m){
query(ans , l , r , R(c)) ;
}
else{
query(ans , l , m , L(c)) ;
query(ans , m + 1 , r , R(c)) ;
}
}
int main(){
while(scanf("%d%d" , &n , &m) != EOF){
int x ;
int y ;
int k ;
build(1 , n , 1) ;
for(int i = 1 ; i <= n ; i ++){
scanf("%d" , &x) ;
insert(x , i , 1 ) ;
}
for(int i = 1 ; i <= m ; i ++){
scanf("%d%d%d" , &k , &x , &y) ;
if(k == 1){
update(x , y , 1) ;
}
else{
int ans = 0 ;
query(ans , x , y , 1) ;
printf("%d\n" , ans) ;
}
}
}
return 0 ;
}
1259: 跳跳
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
struct stu
{
int i,j;
};
int a[101][101],m;
char s[101][101];
queue<struct stu> q;
void xtd(int i,int j)
{
struct stu n;
int i1,j1;
if(s[i][j]>='2'&&s[i][j]<='9')
for(i1=0;i1<m;i1++)
for(j1=0;j1<m;j1++)
if(s[i1][j1]==s[i][j]&&!a[i1][j1]){
n.i=i1;
n.j=j1;
q.push(n);
a[i1][j1]=a[i][j];
}
}
int bfs(struct stu n)
{
int i,j,flag=0;
q.push(n);
a[n.i][n.j]=1;
while(!q.empty()){
n=q.front();
i=n.i;
j=n.j;
q.pop();
if(s[i][j]=='E'){
flag=1;
break;
}
if(s[i][j]!='1'){
if(i-1>=0&&s[i-1][j]!='1'&&!a[i-1][j]){
n.i=i-1;
n.j=j;
q.push(n);
a[i-1][j]=a[i][j]+1;
xtd(i-1,j);
}
if(i+1<m&&s[i+1][j]!='1'&&!a[i+1][j]){
n.i=i+1;
n.j=j;
q.push(n);
a[i+1][j]=a[i][j]+1;
xtd(i+1,j);
}
if(j-1>=0&&s[i][j-1]!='1'&&!a[i][j-1]){
n.i=i;
n.j=j-1;
q.push(n);
a[i][j-1]=a[i][j]+1;
xtd(i,j-1);
}
if(j+1<m&&s[i][j+1]!='1'&&!a[i][j+1]){
n.i=i;
n.j=j+1;
q.push(n);
a[i][j+1]=a[i][j]+1;
xtd(i,j+1);
}
}
}
return flag;
}
int main()
{
int i,j,t;
struct stu n,k;
while(scanf("%d",&m)!=EOF){
getchar();
for(i=0;i<m;i++){
for(j=0;j<m;j++){
scanf("%c",&s[i][j]);
if(s[i][j]=='S'){
n.i=i;
n.j=j;
}
else if(s[i][j]=='E'){
k.i=i;
k.j=j;
}
}
getchar();
}
memset(a,0,sizeof(a));
while(!q.empty())
q.pop();
t=bfs(n);
if(t)
printf("%d\n",a[k.i][k.j]-1);
else
printf("Oh No!\n");
}
return 0;
}1262: 安全密码
#include <cstdio>
#include <string.h>
#include <iostream>
using namespace std;
int main()
{
#ifdef LOCAL
freopen("data.in","r",stdin);
freopen("data.out","w",stdout);
#endif
char s[55];
int a1,a2,a3,a4;
while(scanf("%s",s)!=EOF)
{
a1=0;a2=0;a3=0;a4=0;
int l=strlen(s);
for(int i=0;i<l;i++)
{
if(l<8)
break;
if(s[i]>=65&&s[i]<=90)
a1++;
else if(s[i]>=97&&s[i]<=122)
a2++;
else if(s[i]>=48&&s[i]<=57)
a3++;
else
a4++;
}
if(a1*a2*a3||a1*a2*a4||a1*a3*a4||a2*a3*a4)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}1264: 惠民工程
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int pre[1005];
struct node
{
int p1;
int p2;//p1,p2分别对应的两个居民点
int v;//铺设管道所需的费用
} way[1005];
int m,n;
void init()
{
memset(pre,0,sizeof(pre));
memset(way,0,sizeof(way));
for(int i=1;i<=m;i++)//初始化，每一个点当做是一个父节点
pre[i]=i;
}
int cmp(node a,node b)
{
return a.v<b.v;
}
int find(int x)//查找父节点
{
int ret=x;
while(pre[ret]!=ret) //如果查找到的父节点不是它本身，就直接把它当做一个新的父节点
ret=pre[ret];
int t=x,r;//下面是路径压缩
while(t!=ret)
{
r=pre[t]; //在改变上一级额的父节点之前 用临时变量将它记录起来
pre[t]=ret;//把上级节点改为父节点
t=r;
}
return ret;
}
int join(int x,int y) //判断两个点是否连通
{
int fx=find(x);
int fy=find(y);
if(fx!=fy)//如果已经联通就不用管，否则将它并入之前的连通分支中
{
pre[fx]=fy;
return 1;
}
return 0;
}
int main()
{
while(~scanf("%d%d",&m,&n))
{
init();
for(int i=0;i<n;i++)
scanf("%d%d%d",&way[i].p1,&way[i].p2,&way[i].v);
sort(way,way+n,cmp); //排序，目的是为了优先考虑成本低的将它并入连通分支
int ans=0;
for(int i=0;i<n;i++)
if(join(way[i].p1,way[i].p2))//如果已经联通就只需计算管道费用就行
ans+=way[i].v;
int count=0;
for(int i=1;i<=m;i++)
if(pre[i]==i)//判断是否都连通
count++;
if(count>1)//如果不连通
printf("?\n");
else
printf("%d\n",ans);
}
return 0;
}
1268: Pingpang Balls
#include<stdio.h>
int abs(int a, int b)
{
if(a >= b)
return a - b;
else
return b -a;
}
int main()
{
int N, S, D1, D2, D3, T, W[25], i, flag1, flag2, flag3;
scanf("%d", &T);
while(T --)
{
i = flag1 = flag2 = flag3 = 0;
scanf("%d%d%d%d%d", &N, &S, &D1, &D2, &D3);
for(i = 0; i < N; i ++)
scanf("%d", &W[i]);
for(i = 0; i < N; i ++)
{
if(abs(W[i], S) <= D1)
flag3 ++;
else if(abs(W[i], S) > D1 && abs(W[i], S) <= D2)
flag2 ++;
else if(abs(W[i], S) > D2 && abs(W[i], S) <= D3)
flag1 ++;
}
printf("%d %d %d\n", flag3, flag2, flag1);
}
return 0;
}#include<stdio.h>
#include<algorithm>
using namespace std;
char mp[20][25];
int aa[25]= {0,1,2,0,0,0};
int pp[25]= {0,1,2,2,1,0};
int mm[25]= {0,2,2,0,0,0};
int num[15][9]= {{0,2,2,2,2,2},{0,1,2,2,2,2},{0,1,1,2,2,2},{0,1,1,1,2,2},{0,1,1,1,1,2},{0,1,1,1,1,1},{0,2,1,1,1,1},{0,2,2,1,1,1},{0,2,2,2,1,1},{0,2,2,2,2,1},};
void dian(int n,char c)
{
for(int i=1; i<=5; i++)
mp[i][n]=c;
}
void doit(int biao,char c)
{
int *p;
if(c=='A')
p=aa;
else if(c=='M')
p=mm;
else if(c=='P')
p=pp;
else
p=num[c-'0'];
for(int i=1; i<=5; i++)
{
if(p[i]==1)
{
mp[i][biao]='#';
mp[i][biao+1]='.';
}
if(p[i]==2)
{
mp[i][biao]='#';
mp[i][biao+1]='#';
}
if(p[i]==0)
{
mp[i][biao]='.';
mp[i][biao+1]='.';
}
}
}
int main()
{
char s[200];
int t;
scanf("%d",&t);
getchar();
while(t--)
{
gets(s);
int biao=1;
for(int i=0; s[i]!=0; i++)
{
if(s[i]!=' ')
{
dian(biao++,'.');
doit(biao,s[i]);
biao+=2;
}
}
for(int i=1; i<=5; i++)
{
for(int j=1; j<=18; j++)
{
printf("%c",mp[i][j]);
}
puts("");
}
puts("");
}
return 0;
}1272: Karma
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;
const double eps = 1e-8;
const double PI = acos(-1.0);
int sgn(double x)
{
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
struct Point
{
double x,y;
Point() {}
Point(double _x,double _y)
{
x = _x;
y = _y;
}
Point operator -(const Point &b)const
{
return Point(x - b.x,y - b.y);
}
//叉积
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
//点积
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
//绕原点旋转角度B（弧度值），后x,y的变化
void transXY(double B)
{
double tx = x,ty = y;
x = tx*cos(B) - ty*sin(B);
y = tx*sin(B) + ty*cos(B);
}
};
struct Point px[1010],dbx[10];
struct Line
{
Point s,e;
Line() {}
Line(Point _s,Point _e)
{
s = _s;
e = _e;
}
//两直线相交求交点
//第一个值为0表示直线重合，为1表示平行，为0表示相交,为2是相交
//只有第一个值为2时，交点才有意义
pair<int,Point> operator &(const Line &b)const
{
Point res = s;
if(sgn((s-e)^(b.s-b.e)) == 0)
{
if(sgn((s-b.e)^(b.s-b.e)) == 0)
return make_pair(0,res);//重合
else return make_pair(1,res);//平行
}
double t = ((s-b.s)^(b.s-b.e))/((s-e)^(b.s-b.e));
res.x += (e.x-s.x)*t;
res.y += (e.y-s.y)*t;
return make_pair(2,res);
}
};
double dist(Point a,Point b)
{
return sqrt(1.0*(a.x-b.x)*(a.x-b.x)+((a.y-b.y))*((a.y-b.y)));
}
double CalcArea(Point a,Point b,Point c,int n)
{
Point p[3];
p[0]=a;
p[1]=b;
p[2]=c;
double res = 0;
for(int i = 0; i < n; i++)
res += (p[i]^p[(i+1)%n])/2;
return fabs(res);
}
bool OnSeg(Point P,Line L)
{
return
sgn((L.s-P)^(L.e-P)) == 0 &&
sgn((P.x - L.s.x) * (P.x - L.e.x)) <= 0 &&
sgn((P.y - L.s.y) * (P.y - L.e.y)) <= 0;
}
bool ok(Point a,Point b,Point c,Point p)//P点是否在三角形abc内或者在三角形的边上
{
Line ab=Line(a,b);
Line ac=Line(a,c);
Line bc=Line(b,c);
pair<int,Point> j;
j=ab&(ac);
if(j.first!=2)
{
if(!OnSeg(p,ab) && !OnSeg(p,bc) && !OnSeg(p,ac))
return 0;
else
return 1;
}
double spab=CalcArea(p,a,b,3);
double spac=CalcArea(p,a,c,3);
double spbc=CalcArea(p,c,b,3);
double sabc=CalcArea(c,a,b,3);
if(fabs(sabc-(spab+spbc+spac))<=eps)
return 1;
return 0;
}
bool vis[1010];
int main()
{
double x,y;
int t,n,m,i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
scanf("%lf%lf%lf%lf",&dbx[2].x,&dbx[2].y,&dbx[0].x,&dbx[0].y);
for(i=0; i<n; i++)
scanf("%lf%lf",&px[i].x,&px[i].y);
int cont=0;
memset(vis,false,sizeof vis);
for(i=1; i<=m; i++)
{
scanf("%lf%lf",&x,&y);
dbx[i%2].x=x;
dbx[i%2].y=y;
for(j=0; j<n; j++)
{
if(vis[j])
continue;
if(ok(dbx[0],dbx[1],dbx[2],px[j]))
{
cont++;
vis[j]=true;
}
}
}
printf("%d\n",cont);
}
return 0;
}1282: Sphenic Number
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
long long ans[1500005];
long long p[200005], a[200005], tt = 0;
void prime()
{
int i, n=130000, j;
a[1] = 0;
for(i = 2; i <= n; i++)
a[i] = 1;
for(i = 2; i <= n; i++)
if(a[i])
{
p[tt++] = i;
for(j = i; j <= n;j += i)
a[j] = 0;
}
}
int main()
{
prime();
int T, n, t = 0;
for(int i = 0; i < tt; i++)
{
for(int j = i + 1; j < tt; j++)
{
for(int k = j + 1; k < tt; k++)
{
if(p[i] * p[j] * p[k]>539251)
break;
ans[t++] = p[i] * p[j] * p[k];
}
}
}
sort(ans, ans + t);
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
printf("%lld\n", ans[n - 1]);
}
return 0;
}1294: Coins
#include <cstdio>
#include <map>
#include <cstring>
#define LL long long int
using namespace std;
int main()
{
// freopen("in.cpp","r",stdin);
int len,k;
scanf("%d%d",&len,&k);//长度和K值
char a;
int max = 0;//用来记录最后的结果，初始状态为0
map<LL,int> m;
LL sum = 0;
m[0] = 0;
getchar();
for(int i=1; i<=len; ++i)
{
a = getchar();
if(a == 'O')
++sum;//O，加1
else sum -= k;//R,加-K
map<LL,int>::iterator it;
it = m.find(sum);
if(it != m.end())//出现重复
{
if(max < i-(*it).second)//出现更大长度
max = i-(*it).second;//更新MAX
}
else
m[sum] = i;//未出现重复，保存
}
printf("%d\n",max);
return 0;
}1299: Number Transformation II
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#define pi acos(-1)
#define ll long long
#define oo 2139062143
#define MAXN 200005
using namespace std;
struct node
{
int next;
}h[MAXN];
int n,m,dp[MAXN],p[3000000][2];
int main()
{
int x,i,k;
m=0;
memset(h,0,sizeof(h));
memset(p,0,sizeof(p));
for (i=1;i<=MAXN;i++)
for (x=i*2;x<=MAXN;x+=i)
{
p[++m][0]=i;
p[m][1]=h[x].next;
h[x].next=m;
}
while (~scanf("%d%d",&n,&m))
{
memset(dp,0x7f,sizeof(dp));
dp[n]=0;
for (x=n;x<=m;x++)
{
k=h[x].next;
while (k)
{
i=p[k][0];
if (x+i<=m)
if (dp[x+i]==-1 || dp[x+i]>dp[x]+1)
dp[x+i]=dp[x]+1;
k=p[k][1];
}
}
if (dp[m]==oo) printf("sorry, can not transform\n");
else printf("%d\n",dp[m]);
}
return 0;
}
标签：tmp,return,OJ,int,1200,long,1300,include,dp 来源： https://blog.csdn.net/nameofcsdn/article/details/111088226
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