中国剩余定理模板poj1006

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 20;
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1; 
        y = 0;
        return a;
    }
    ll r = exgcd(b, a % b, x, y);
    ll t = x;
    x = y;
    y = t - a / b * y;
    return r;
}
ll china_remainder(int a[], int w[], int n)//w存放除数,a存放余数
{
    ll M = 1, ans = 0, x, y;
    for (int i = 0; i < n; i++)
        M *= w[i];
    for (int i = 0; i < n; i++)
    {
        ll m = M / w[i];
        exgcd(m, w[i], x, y);
        ans = (ans + x * m * a[i]) % M;
    }
    return (ans % M + M) % M;
}
int main()
{
    int T, m, a[maxn], w[maxn];
    w[0] = 23; w[1] = 28; w[2] = 33;
    ll n = 3;
    int p, e, i, d, kase = 0;
    while (~scanf("%d %d %d %d", &p, &e, &i, &d))
    {
        if (p == -1 && e == -1 && i == -1 && d == -1)
            break;
        a[0] = p;
        a[1] = e;
        a[2] = i;
        ll ans = china_remainder(a, w, n);
        ans -= d;
        int mod = w[0] * w[1] * w[2];
        ans = (ans % mod + mod) % mod;
        if (ans == 0)
            ans = mod;
        printf("Case %d: the next triple peak occurs in %lld days.
", ++kase, ans);

    }
    return 0;
}