Poj 1006

就是中国剩余定理的应用……

幸好我学了数学基础

#include<iostream>
using namespace std;
int main(){
    int p,e,i,d;
    int k;
    int count=1;
    while(cin>>p>>e>>i>>d){
        if(p==-1&&e==-1&&i==-1&&d==-1){
            break;
        }
        k=0;
        p=p%23;
        e=e%28;
        i=i%33;
        k=(k+(p*5544))%21252;
        k=(k+(e*14421))%21252;
        k=(k+(i*1288))%21252;
        if(k==0){
            k+=21252;
        }
        if(d==k){
            cout<<"Case "<<count<<": the next triple peak occurs in "<<21252<<" days."<<endl;
        }
        else if(k>d){
            cout<<"Case "<<count<<": the next triple peak occurs in "<<k-d<<" days."<<endl;
        }
        else{
            cout<<"Case "<<count<<": the next triple peak occurs in "<<k-d+21252<<" days."<<endl;
        }
        count++;
    }
}