POJ1006

题目大意

略。。。有中文。。。

题解

就是解同余方程组

x≡(p-d)(mod 23)

x≡(e-d)(mod 28)

x≡(i-d)(mod 33)

最简单的中国剩余定理应用。。。。

代码：

#include<iostream>
#include<cstdio>
using namespace std;
void gcd(int a,int b,int &d,int &x,int &y)
{
    if(!b)
    {
        d=a,x=1,y=0;
    }
    else
    {
        gcd(b,a%b,d,y,x),y-=x*(a/b);
    }
}
int main()
{
    int p,e,d,i,a1,a2,a3,M1,M2,M3,t=0;
    while(cin>>p>>e>>i>>d&&p+e+i+d>=0)
    {
        int x=0,y,q;
        a1=p-d,a2=e-d,a3=i-d;
        M1=924,M2=759,M3=644;
        gcd(23,M1,q,q,y),x=(x+y*M1*a1)%21252;
        gcd(28,M2,q,q,y),x=(x+y*M2*a2)%21252;
        gcd(33,M3,q,q,y),x=(x+y*M3*a3)%21252;
        x=(x%21252+21252)%21252,x=x?x:21252;
        printf("Case %d: the next triple peak occurs in %d days.
",++t,x);
    }
    return 0;
}