Since the three cycles have different periods, the peaks of the three cycles generally occur at different times. We would like to determine when a triple peak occurs (the peaks of all three cycles occur in the same day) for any person. For each cycle, you will be given the number of days from the beginning of the current year at which one of its peaks (not necessarily the first) occurs. You will also be given a date expressed as the number of days from the beginning of the current year. You task is to determine the number of days from the given date to the next triple peak. The given date is not counted. For example, if the given date is 10 and the next triple peak occurs on day 12, the answer is 2, not 3. If a triple peak occurs on the given date, you should give the number of days to the next occurrence of a triple peak.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Case 1: the next triple peak occurs in 1234 days.

Use the plural form ``days'' even if the answer is 1.

#include<iostream>

#include<stdio.h>

using namespace std;

int ext_gcd(int a,int b,int *x,int *y)

{

    if(b==)

    {

        *x=,*y=;

        return a;

    }

    int r = ext_gcd(b,a%b,x,y);

    int t =*x;

    *x=*y;

    *y=t-a/b**y;

    return r;

}

int chinese_remainder(int a[],int w[],int len)//a存放余数，w存放两两互质的数

{

    int i,d,x,y,m,n,ret;

    ret=;

    n=;

    for(i=; i<len; i++)

    {

        n*=w[i];

    }

    for(i=; i<len; i++)

    {

        m=n/w[i];

        d=ext_gcd(w[i],m,&x,&y);

        ret=(ret+y*m*a[i])%n;

    }

    return(ret%n+n)%n;

}

int main()

{

    int a[];

    int w[]= {,,};

    int t;

    scanf("%d",&t);

    while(t--)

    {

        int cas=;

        int d;

        while(scanf("%d%d%d%d",&a[],&a[],&a[],&d))

        {

            if(a[]==-) break;

            for(int i=; i<; i++)

                a[i]%=w[i];

            int ans=chinese_remainder(a,w,);

            ans=ans-d;

            if(ans<=) ans+=;

            printf("Case %d: the next triple peak occurs in %d days.\n",cas++,ans);

        }

    }

    return ;

}

点这里看我整理中国剩余定理的分析

点这里看注释代码