标签:次数 long ring 不同的 class begin ++ include return
#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
typedef long long LL;
int main(){
int a, b, c;
cin >> a >> b >> c;
cout << min(c, b / a) << endl;
return 0;
}
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
typedef long long LL;
int res[N];
int main() {
int a, b, c;
cin >> a >> b >> c;
int num = 0;
for (int i = 1; i <= min(a, b); i++) {
if ((a % i == 0) && (b % i == 0)) {
res[num++] = i;
}
}
cout << res[num - c] << endl;
return 0;
}
给出一个长度为n的字符串,由0和1组成,每次可以将相邻的不同的数字删掉(删掉后左右两边变成相邻),问最多可以删掉多少
直接输出出现次数较少的字符的两倍即可
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
typedef long long LL;
stack<int> st;
int main() {
string s;
cin >> s;
int num1=0,num0=0;
for (int i = 0; i < s.size(); i++) {
if (s[i] == ‘0‘) num0++;
else
num1++;
}
cout << 2*min(num1,num0) << endl;
return 0;
}
给出n个点,m条边,问依次删去每个边后,有多少个点对之间不能互达
并查集倒着维护即可,每次删边相当于将两个集合合并起来
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
typedef long long LL;
int n, m, f[N];
LL d[N];
int findf(int x) { return f[x] == x ? x : f[x] = findf(f[x]); }
struct node {
int x, y;
} q[N];
LL ress[N];
LL res = 0;
void Union(int x, int y) {
int fx = findf(x), fy = findf(y);
if (fx != fy) {
res -= d[fx] * d[fy];
f[fx] = fy;
d[fy] += d[fx];
}
}
int main() {
cin >> n >> m;
for (int i = 1; i <= n; i++) f[i] = i, d[i] = 1;
for (int i = 0; i < m; i++) {
cin >> q[i].x >> q[i].y;
}
for (int i = 1; i < n; i++) {
res += (LL)i;
}
for (int i = m - 1; i >= 0; i--) {
ress[i] = res;
int x = q[i].x, y = q[i].y;
Union(x, y);
}
for (int i = 0; i < m; i++) cout << ress[i] << endl;
return 0;
}
标签:次数 long ring 不同的 class begin ++ include return
原文地址:https://www.cnblogs.com/dyhaohaoxuexi/p/14400173.html