ICode9
精准搜索请尝试： 精确搜索
首页
编程语言>
数据库
系统相关
互联网
其他分享
Python
Java
JavaScript
android
PHP
首页 > 其他分享> 文章详细

上海理工大学联想杯 4题滚粗记

2021-06-14 15:03:45  阅读：325  来源： 互联网

标签：tmp min rep flow 题滚 上海理工大学 粗记 ll define


果然是被生活压垮的人，即将退役和告别大学生活了

早上放下了GRE来看题，写了两个小时就又不得不去学gre了，希望330+

比赛开始20分钟之后才到自习室，然后这个时候排名已经1100开外了

第一题，一看，问最长的等差数列长度，给了公差，所以拿map记录一下就好了

#include <bits/stdc++.h>
using namespace std;
#define limit (2000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x, char c = 't') {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    if(!isalpha(c))*O++ = c;
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int kase;
int n, m,k;
int a[limit];
void solve(){
    cin>>n>>k;
    map<int, int>mp;
    rep(i,1,n){
        cin>>a[i];
    }
    sort(a + 1, a + 1 + n);
    ll ans = 1;
    rep(i,1,n){
        if(!mp.count(a[i])){
            mp[a[i]] = mp[a[i] - k] + 1;
            ans = max(ans, 1ll * mp[a[i]]);
        }
    }
    cout<<ans<<endl;

}
int32_t main() {
#ifdef LOCAL
    FOPEN;
    //FOUT;
#endif
    //FASTIO

    //kase = read();
    //while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}

然后看了下K，问最大匹配，数据还小得一批，笑了，暴力加边，dinic水了过去

#include <bits/stdc++.h>
using namespace std;
#define limit (200000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x, char c = 't') {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    if(!isalpha(c))*O++ = c;
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int kase;
int n,m,vs,ve,p;
int layer[limit],head[limit], cnt;
struct node{
    int to ,next;
    ll flow, w;
}edge[limit];
ll max_flow;
void add_one(int u , int v, ll flow = 0){
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    edge[cnt].flow = flow;
    edge[cnt].w = 0;
    head[u] = cnt++;
}
inline void add(int u, int v, ll flow){
    add_one(u,v,flow);
    add_one(v, u,0);
}
inline void init(bool flag = true){
    if(flag){
        memset(head, -1, sizeof(head));
        cnt = 0;
    }else{
        memset(layer, -1, sizeof(layer));
    }
}
inline bool bfs(){
    init(false);
    queue<int>q;
    layer[vs] = 0;//从第0层开始
    q.push(vs);
    while (q.size()){
        int u = q.front();
        q.pop();
        traverse(u){
            int v = edge[i].to,flow = edge[i].flow;
            if(layer[v] == -1 && flow > 0){
                layer[v] = layer[u] + 1;//迭代加深
                q.push(v);
            }
        }
    }
    return ~layer[ve];
}
ll dfs(int u, ll flow){
    if(u == ve)return flow;
    ll rev_flow = 0,min_flow;
    traverse(u){
        int v =edge[i].to;
        ll t_flow = edge[i].flow;
        if(layer[v] == layer[u] + 1 && t_flow > 0){
            min_flow = dfs(v, min(flow, t_flow));
            flow -= min_flow;
            edge[i].flow -= min_flow;
            rev_flow += min_flow;
            edge[i^1].flow += min_flow;
            if(!flow)break;
        }
    }
    if(!rev_flow)layer[u] = -1;
    return rev_flow;
}
void dinic(){
    while (bfs()){
        max_flow += dfs(vs,inf);
    }
}
int a[limit], b[limit];
void solve(){
    init();
    n = read();
    vs = 90001, ve = vs + 1;
    rep(i,1,n){
        a[i] = read();
        add(vs, i, 1);
    }
    rep(i,1,n){
        b[i] = read();
        add(i + n, ve, 1);
    }
    rep(i,1,n){
        rep(j,1,n){
            if(__gcd(a[i] , b[j]) != 1){
                add(i, j + n, INF);
            }

        }
    }
    dinic();
    write(max_flow);

}
int32_t main() {
#ifdef LOCAL
    FOPEN;
    //FOUT;
#endif
    //FASTIO

    //kase = read();
    //while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}

然后C题，问匹配，好像列出来之后只有6种情况能对答案产生贡献，那么就手动模拟一下就好了

#include <bits/stdc++.h>
using namespace std;
#define limit (2000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x, char c = 't') {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    if(!isalpha(c))*O++ = c;
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int kase;
int n,m,k;
int ca,cat,c,at,t,a;
void solve(){
    cin>>n;
    rep(i,1,n) {
        string str;
        cin >> str;
        if (str == "ca")++ca;
        if (str == "cat")++cat;
        if (str == "a")++a;
        if (str == "at")++at;
        if (str == "t")++t;
        if (str == "c")++c;
    }
    // ca + t
    ll ans = min(ca, t), tmp;
    t -= ans, ca -= ans;
    // c + at
    tmp = min(c, at); ans += tmp;
    c -= tmp, at -= tmp;
    // cat
    ans += cat, cat = 0;
    // c + a + t
    tmp = min({c , a , t}); ans += tmp;
    c -= tmp, a -= tmp, t -= tmp;
    cout<<ans<<endl;

}
int32_t main() {
#ifdef LOCAL
    FOPEN;
    //FOUT;
#endif
    FASTIO

    //kase = read();
    //while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}

然后看了看B，发现题意太过复杂，虽然感觉不难，但是念及自己可怜的分数还是算了，开了J

然后发现J的范围是1e7，很典型的素数筛题，想到根号分解数字，找到最小质因数，然后log计算贡献，再从增量里把这些剪掉就行，暴力加答案。

然后好家伙，tle了，我跟队友说不要写了，这题有坑，然后队友过了一会儿告诉我，tle个锤子，他过了。

然后我发现，好像__int128没关上，那没事了，果断提交并且AC，话说我这个__int128为了谨慎起见还开了fread，没想到这么毒瘤

#include <bits/stdc++.h>
using namespace std;
#define limit (20000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
char buf[1<<23],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
inline ll read(){
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if(x/ 10) print(x / 10);
    *O++=x % 10+'0';
}
void write(ll x, char c = 't') {
    if(x < 0)putchar('-'),x = -x;
    print(x);
    if(!isalpha(c))*O++ = c;
    fwrite(obuf,O-obuf,1,stdout);
    O = obuf;
}
int kase;
ll n,m, k;
ll prime[limit], num[limit],tot;
int a[limit], b[limit];
void get_prime( const int &N=2e7){
    memset(num, 1, sizeof(num));
    tot = 0;
    rep(i ,2,N){
        if(num[i])prime[++tot] = i;
        for(int j = 1 ; j <= tot && i * prime[j] <= N ; ++j){
            num[i * prime[j]] = 0;
            if(i % prime[j] == 0)break;//线性筛
        }
    }
}
void process(ll x){
    if(num[x]){
        a[x] = x;
        b[x] = 1;
        return;
    }
    for(ll i = 2 ; i * i <= x; ++i){
        if(num[i] and x % i == 0){
            a[x] = i;
            ll tmp = x;
            while(tmp % i == 0){
                b[x]++;
                tmp /= i;
            }
            return;
        }
    }


}
void solve(){
    get_prime();
    n = read(),m = read(), k= read();
    rep(i,1 + k,n + k){
        process(i);
    }
    ll ans = 0;
    rep(i,1 + k,n + k){
        ll ti = b[i] - (b[i] / m);
        ll tmp = i;
        rep(j,1,ti){
            tmp /= a[i];
        }
        ans += i - tmp;
    }
    write(ans);
}
int32_t main() {
#ifdef LOCAL
    FOPEN;
    //FOUT;
#endif
    FASTIO

    //kase = read();
    //while (kase--)
    solve();
    cerr << "Time elapsed: " << 1.0*clock()/CLOCKS_PER_SEC << "s\n";
    return 0;
}

然后看了看I题，好像三分写假了，那就撤退好了，今天还有新的任务呢。

感觉之后还是多一些训练，希望有生之年能够成全我一次吧，今年ICPC加油！

标签：tmp,min,rep,flow,题滚,上海理工大学,粗记,ll,define
来源： https://www.cnblogs.com/tiany7/p/14882437.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。

关于我们 | 联系我们 | 留言反馈

专注分享技术，共同学习，共同进步。侵权联系[81616952@qq.com]

Copyright (C)ICode9.com, All Rights Reserved.

ICode9版权所有