poj 1006 剩余定理

无脑套板子….
好像不需要理解什么意思的样子….
具体就是知道余数（除数积/当前）（除数积/当前 mod当前的逆元）求和….
就可以了…
板子有学长写好的…

#include<iostream>#include<algorithm>#include<cmath>#include<string>#include<cstdio>using namespace std;long long a[3],m[3],n;void ex_gcd(int a, int b, int &x, int &y) //板子 {    if (!b)    {        x = 1; y = 0;        return;    }    ex_gcd(b, a%b, y, x);     y -= (a / b) * x;}long long CRT()//板子 {    long long M = 1;    long long ans = 0;    for (int i = 0; i < n; i++)M *= m[i];    for (int i = 0; i < n; i++)     {        int x, y;        long long t = M / m[i];        ex_gcd(t, m[i], x, y);        ans += a[i] * t * x;        ans %= M;    }    ans = (ans + M) % M;    return ans;}int main(){    int q,w,e,r;    n=3;    m[0]=23,m[1]=28,m[2]=33;    int u=0;    while(cin>>q>>w>>e>>r)    {        if(q==-1)break;        a[0]=q,a[1]=w,a[2]=e;        long long i=CRT()-r;        if(i<=0)i+=21252;        printf("Case %d: the next triple peak occurs in %lld days.\n",++u,i);    }    return 0;}