題意:

x === p ( mod 23 )

x === e ( mod 28 )

x === i ( mod 33 )

給p e i ，求x 。

解析：

23 28 33互素，所以用中國剩餘定理做。

用之前的解同餘方程組的方法也試了試，很噁心，樣例裡面有0 0 0 0 這樣的輸入，會爆除0錯。

程式碼：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <cassert>
#define LL long long

using namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = 4 * atan(1.0);
const double ee = exp(1.0);

const int maxn = 30 + 10;

void exgcd(int a, int b, int& d, int& x, int& y)
{
    if (b == 0)
    {
        d = a;
        x = 1;
        y = 0;
    }
    else
    {
        exgcd(b, a % b, d, y, x);
        y -= a / b * x;
    }
}

int a[5], m[5], M;
int CRT(int r)
{
    M = 1;
    int res = 0;
    for (int i = 1; i <= r; i++)
    {
        M *= m[i];
    }
    for (int i = 1; i <= r; i++)
    {
        int Mi = M / m[i];
        int d, x, y;
        exgcd(Mi, m[i], d, x, y);
        res = (res + Mi * x * a[i]) % M;
    }
    if (res < 0)
        res += M;
    return res;
}

int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif // LOCAL
    int ca = 1;
    int p, e, i, d;
    while (scanf("%d%d%d%d", &p, &e, &i, &d) == 4)
    {
        if (p + d + i + d == -4)
            break;
        a[1] = p, a[2] = e, a[3] = i;
        m[1] = 23, m[2] = 28, m[3] = 33;
        int ans = CRT(3);
        while (ans <= d)
            ans += M;
        printf("Case %d: the next triple peak occurs in %d days.\n", ca++, ans - d);
    }

    return 0;
}