拓扑排序
发糖果
e[++cnt].next = head[u];
e[cnt].to = v;
head[u] = cnt;
打成
e[cnt++].next = head[u];
e[cnt].to = v;
head[u] = cnt;
数论
洛谷P1082
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
int exgcd(int a, int b, int &x, int &y)
{
if(b == 0) {
x = 1, y = 0;
return a;
}
int ret = exgcd(b, a % b, x, y);
int temp = x;
x = y;
y = temp - a / b * y;
return ret;
}
int main()
{
int a, b, x, y;
cin >> a >> b;
int ret = exgcd(a, b, x, y);
x = (x % b + b) % b;
cout << x;
return 0;
}