POJ 1006 Biorhythms 中国剩余定理 模版

#include<iostream>
using namespace std;
#define ll long long
const int maxn=1e5+5;
const int mod = 21252;
int n;
ll exgcd(ll a,ll b,ll &x,ll &y) {
	if(!b) {
		x=1,y=0;
		return a;
	}
	ll re=exgcd(b,a%b,x,y),tmp=x;
	x=y,y=tmp-(a/b)*y;
	return re;
}
ll m[maxn],a[maxn];
ll work() {
	ll M=m[1],A=a[1],t,d,x,y;
	int i;
	for(i=2; i<=n; i++) {
		d=exgcd(M,m[i],x,y);//解方程
		if((a[i]-A)%d)return -1;//无解
		x*=(a[i]-A)/d,t=m[i]/d,x=(x%t+t)%t;//求x
		A=M*x+A,M=M/d*m[i],A%=M;//日常膜一膜 防爆
	}
	A=(A%M+M)%M;
	return A;
}
int main() {
	int st,ans,cnt=1;// %m[i] 余 a[i]
	while(1) {
		m[1] = 23;
		m[2] = 28;
		m[3] = 33;
		n = 3;
		for(int i=1; i<=3; i++)scanf("%lld",&a[i]);
		scanf("%lld",&st);
		if(a[1]==-1&&a[2]==-1&&a[3]==-1&&st==-1)break;
		ans = (work() - st)%mod;
		if(ans<=0)
			ans+=mod;

		printf("Case %d: the next triple peak occurs in %lld days.\n",cnt++,ans);
	}
	return 0;
}