中国剩余定理poj1006

题意就是给出23,28,33三个周期，输入三个余数和起始时间，问下一个共同周期。

中国剩余定理模板题

中国剩余定理:

mi=n1*n2*...*n(i-1)*n(i+1)*...*ni

ci=mi ( mi^-1 mod ni )

a≡(a1c1+a2c2+...+akck)(mod n)

#include<iostream>
#include<stdio.h>
using namespace std;
#define ll long long

long long extend_gcd(long long a,long long b,long long &x,long long&y)
{
    if(a==0&&b==0) return -1;//无最大公约数
    if(b==0)
    {
        x=1;
        y=0;
        return a;
    }
    long long d=extend_gcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}
//ax = 1(mod n)
long long mod_reverse(long long a,long long n)
{
    long long x,y;
    long long d=extend_gcd(a,n,x,y);
    if(d==1) return (x%n+n)%n;
    else return -1;
}
int main()
{
    ll p,e,i,dd,n1,n2,n3,m1,m2,m3;
    n1=23,n2=28,n3=33;
    m1=n2*n3;
    m2=n1*n3;
    m3=n1*n2;
    ll n = n1*n2*n3;
    ll m11=mod_reverse(m1,23);
    ll m22=mod_reverse(m2,28);
    ll m33=mod_reverse(m3,33);
    ll c1=m1*m11;
    ll c2=m2*m22;
    ll c3=m3*m33;
    int acm=1;
    while(~scanf("%lld%lld%lld%lld",&p,&e,&i,&dd))
    {
        if(p==-1&&e==-1&&i==-1&&dd==-1)
        {
            break;
        }
        ll ans=(p*c1+e*c2+i*c3)%n;
        dd%=21252;
        ans=(ans-dd+21252)%21252;
        if(ans==0)ans=21252;
        printf("Case %d: the next triple peak occurs in %lld days.\n",acm++,ans);
    }
    return 0;
}