POJ_1006_中国剩余

http://poj.org/problem?id=1006

中国剩余定理用来解求模方程组，用到了逆元。

这题三个数互质，直接用扩展欧几里德可得逆元。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

int a[],m[],n,d;

void e_gcd(int a,int b,int &x,int &y)
{
    if(!b)
    {
        x = ;
        y = ;
        return;
    }
    e_gcd(b,a%b,x,y);
    int temp = x;
    x = y;
    y = temp-a/b*y;
}

int CRT()
{
    int M = ,ans = ;
    for(int i = ;i <= n;i++)   M *= m[i];
    for(int i = ;i <= n;i++)
    {
        int x,y,Mi = M/m[i];
        e_gcd(Mi,m[i],x,y);
        ans = (ans+Mi*x*a[i])%M;
    }
    if(ans < ) ans += M;
    return ans;
}

int main()
{
    n = ;
    m[] = ;
    m[] = ;
    m[] = ;
    int cnt = ;
    while(scanf("%d%d%d%d",&a[],&a[],&a[],&d))
    {
        if(a[] == - && a[] == - && a[] == - && d == -)   break;
        int ans = CRT();
        if(ans <= d)    ans += ;
        ans -= d;
        printf("Case %d: the next triple peak occurs in %d days.\n",cnt++,ans);
    }
    return ;
}