标签：Atcoder const Beginner int void cin push ver 257
A - A to Z String 2
代码：
void solve(int Case) {
int n;
string s = " ";
cin >> n;
int k;
cin >> k;
for (int i = 'A'; i <= 'Z'; i++) {
for (int j = 1; j <= n; j++) {
s.push_back(i);
}
}
cout << s[k] << nline;
}
B - 1D Pawn
代码：
const int N=10010;
int a[N];
int p[N];
void solve(int Case) {
int n;
cin >> n;
int m, q;
cin >> m >> q;
for (int i = 1; i <= m; i++) {
int x;
cin >> x;
p[x] = 1;
}
for (; q--;) {
int l;
cin >> l;
for (int i = 1; i <= n; i++) {
l -= p[i];
if (!l) {
if (i != n and !p[i + 1]) {
p[i] = 0, p[i + 1] = 1;
break;
}
}
}
}
for (int i = 1; i <= n; i++) {
if (p[i]) cout << i << ' ';
}
}
C - Robot Takahashi ]
思路：
前缀和
代码：
const int N = 200010;
int sum[N];
string str;
struct T {
int w, s;
bool operator<(const T &t) const {
return w < t.w;
}
} a[N];
vector<int> v;
int find(int x) {
return lower_bound(all(v), x) - v.begin();
}
void solve(int Case) {
int n;
cin >> n;
cin >> str;
str = " " + str;
for (int i = 1; i <= n; i++) {
auto &[w, s] = a[i];
cin >> w;
s = (str[i] == '1');
}
sort(a + 1, a + 1 + n);
for (int i = 1; i <= n; i++) {
v.push_back(a[i].w);
sum[i] = sum[i - 1] + a[i].s;
}
int res = 0;
for (int i = 1; i <= n; i++) {
int w = a[i].w;
int l = find(w);
res = max(res, l - sum[l] + (sum[n] - sum[l]));
}
res = max({res, sum[n],n-sum[n]});
cout << res << nline;
}
D - Jumping Takahashi 2
思路：
暴力二分bfs，预处理边权跑floyd也可以
代码：
const int N = 220;
using PII = pair<int, int> ;
PII a[N];
int s[N];
bool vis[N];
int n;
bool check(int mid) {
for (int i = 1; i <= n; i++) {
queue<int> q;
q.push(i);
for (int i = 1; i <= n; i++) vis[i] = 0;
vis[i] = 1;
while (q.size()) {
auto t = q.front();
q.pop();
for (int i = 1; i <= n; i++) {
if (vis[i]) continue;
auto [x, y] = a[t];
auto [x1, y1] = a[i];
int d = abs(x - x1) + abs(y - y1);
if (mid >= (d + s[t] - 1) / s[t]) {
vis[i] = true;
q.push(i);
}
}
}
int cnt = 0;
for (int i = 1; i <= n; i++) cnt += (!vis[i]);
if (!cnt) return true;
}
return false;
}
void solve(int Case) {
cin >> n;
for (int i = 1; i <= n; i++) {
auto &[x, y] = a[i];
cin >> x >> y >> s[i];
}
int l = 1, r = 2e18;
while (l < r) {
int mid = l + r >> 1;
if (check(mid)) r = mid;
else l = mid + 1;
}
cout << r << nline;
}
E - Addition and Multiplication 2
思路：
在选择最多个数的情况下选择最大，跑一遍完全背包，然后倒推出方案
代码：
const int N = 1000010;
int f[10][N];
int c[N];
void solve(int Case) {
int n;
cin >> n;
for (int i = 1; i <= 9; i++) cin >> c[i];
for (int i = 1; i <= 9; i++) {
for (int j = 0; j <= n; j++) {
if (j >= c[i])
f[i][j] = max(f[i - 1][j], f[i][j - c[i]] + 1);
else f[i][j]=f[i-1][j];
}
}
string s;
int i = 9;
int j = n;
while (i >= 1) {
while (j >= c[i] and f[i][j - c[i]] + 1 >= f[i][j]) {
s.push_back(i + '0');
j -= c[i];
}
i--;
}
cout << s << nline;
}
F - Teleporter Setting
思路：
题目暗示很明显，建立虚拟原点，第i次询问结果是min(d[1][n],d[1][0]+d[i][n],d[1][i]+d[0][n]);
代码：
const int N = 600010;
int d[3][N];
using PII = pair<int, int>;
vector<PII> h[N];
bool vis[N];
int n;
void dijkstra(int s, int d[]) {
for (int i = 0; i <= n; i++) d[i] = 0x3f3f3f3f;
for (int i = 0; i <= n; i++) vis[i] = 0;
priority_queue<PII, vector<PII>, greater<PII>> q;
d[s] = 0;
q.push({d[s], s});
while (q.size()) {
auto [w, t] = q.top();
q.pop();
if (vis[t]) continue;
vis[t] = 1;
for (auto [ver, nw] : h[t]) {
if (d[ver] > nw + w) {
d[ver] = nw + w;
q.push({d[ver], ver});
}
}
}
}
void solve(int Case) {
int m;
cin >> n >> m;
for (; m--;) {
int u, v;
cin >> u >> v;
h[u].push_back({v, 1});
h[v].push_back({u, 1});
}
dijkstra(1, d[1]);
dijkstra(0, d[0]);
dijkstra(n, d[2]);
for (int i = 1; i <= n; i++) {
int res = min({d[1][n], d[1][i] + d[0][n], d[1][0] + d[2][i]});
if (res >= 0x3f3f3f3f) res = -1;
cout << res << ' ';
}
}
标签：Atcoder,const,Beginner,int,void,cin,push,ver,257 来源： https://www.cnblogs.com/koto-k/p/16413021.html
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