有理数取余

dalao原文  https://www.luogu.org/blog/cicos/solution-p2613

#include <bits/stdc++.h>
using namespace std;

const int MOD = 19260817;//MOD是题解中的"p"
int x, y;

inline int getint(){
    int res = 0, ch = getchar();
    while(!isdigit(ch) and ch != EOF)
        ch = getchar();
    while(isdigit(ch)){
        res = (res << 3) + (res << 1) + (ch - '0');
        res %= MOD;//直接对MOD取余
        ch = getchar();
    }
    return res;
}
void exgcd(int a, int b){
    if(b == 0){
        x = 1;
        y = 0;
        return;
    }
    exgcd(b, a % b);
    int Last_x = x;
    x = y;
    y = Last_x - a / b * y;
}
int main(){
    int a, b;
    a = getint();
    b = getint();
    if(b == 0){
        puts("Angry!");
        return 0;
    }
    exgcd(b, MOD);
    x = (x % MOD + MOD) % MOD;
    printf("%lld
", a * (long long)(x) % MOD);//小心相乘后爆int
}
原文地址:https://www.cnblogs.com/aprincess/p/11632021.html