[POJ 2891] Strange Way to Express Integers (扩展中国剩余定理)

题面

传送门：POJ

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Solution

就是裸的扩展中国剩余定理嘛qwq

注意几点：一定要时时刻刻去膜取模，否则一定会爆long long，尤其是算出来的\(k_0\)

这里给出几组易锅数据：(第三组容易爆long long)

input:

4
18373 16147
8614 14948
8440 17480
22751 21618
6
19576 8109
18992 24177
9667 17726
16743 19533
16358 12524
8280 22731
4
9397 38490
22001 25094
33771 38852
19405 35943

output

13052907343337200
-1
78383942913636233

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Code

//POJ2891 Strange Way to Express Integers
//Jan,14th,2019
//扩展中国剩余定理
#include<iostream>
#include<cstdio>
using namespace std;
long long read()
{
	long long x=0,f=1; char c=getchar();
	while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}
	while(isdigit(c)){x=x*10+c-'0';c=getchar();}
	return x*f;
}
long long exgcd(long long A,long long B,long long &x,long long &y)
{
	if(B==0) 
	{
		x=1,y=0;
		return A;
	}
	long long t=exgcd(B,A%B,x,y),tx=x;
	x=y;
	y=tx-(A/B)*y;
	return t;
}
long long lcm(long long A,long long B)
{
	long long tx,ty;
	return (A*B)/exgcd(A,B,tx,ty);
}
const int N=100000+1000;
long long r[N],p[N];
int n;
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=1;i<=n;i++)
			p[i]=read(),r[i]=read();
		
		long long R=r[1],P=p[1];
		bool OK=true;
		for(int i=2;i<=n;i++)
		{
			long long x,y,gcd=exgcd(P,p[i],x,y);
			if((r[i]-R)%gcd!=0)
			{
				OK=false;
				break;
			}
			long long t_P=P,t=p[i]/gcd;
			x*=(r[i]-R)/gcd,P=lcm(P,p[i]),x=(x%t+t)%t;
			R=((t_P*x)%P+R)%P;
		}
		
		if(OK==true)
			printf("%lld\n",(R%P+P)%P);
		else
			printf("-1\n");
	}
	return 0;
}