問題描述

方程組：
\(x \equiv a_1 ( mod\:m_1)\)
\(x \equiv a_2(mod\:m_2)\)
····
\(x \equiv a_k(mod\: m_k)\)
求解x

求解

首先考慮前兩個式子
\(x=m_1\cdot y_1+a_1=m_2\cdot y_2+a_2\)
則\(m_1y_1-m_2y_2=a_2-a_1\)（有解當且僅當\((m_1,m_2)|a_2-a_1)\)
\(y_1=y_0+\frac{km_2}{(m_1,m_2)}(k\in Z)\)
則\(x=m_1y_0+\frac{km_1m_2}{(m_1,m_2)}+a_1\),設\(x'=m_1y_0+a_1\)

poj2891
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define ll long long
using namespace std;
const int maxn=10000000+101;
inline ll read(){
	ll x=0,f=1;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	return x*f;
}
int k;
void exgcd(ll a,ll b,ll &d,ll &x,ll &y){
	if(!b){d=a;x=1;y=0;return ;}
	exgcd(b,a%b,d,x,y);
	ll t=x;x=y;y=t-(a/b)*y;
	return ;
}
int main(){
	while(scanf("%d",&k)!=EOF){
		ll a1=read(),r1=read(),a2,r2;
		bool fa=false;
		for(int i=2;i<=k;i++){
			a2=read();r2=read();
			if(fa)continue;
			ll d,x,y,c=r1-r2;
			exgcd(a2,a1,d,x,y);
			if(c%d){fa=true;continue;}
			x=x*(c/d);
			a1/=d;
			x=(x%a1+a1)%a1;//求最小整數解
			x=(x*a2+r2); 
			a1=a1*a2;
			r1=((x%a1)+a1)%a1;
		}
		if(fa)printf("-1\n");
		else printf("%lld\n",r1);
	}
	return 0;
}