poj 2891 Strange Way to Express Integers

题目：http://poj.org/problem?id=2891

思路：扩展欧几里得

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
#define maxn 100010
int n;
long long a[maxn],b[maxn];
long long exgcd(long long a,long long b,long long &x,long long &y)
{if(b==0){x=1;y=0;return a;}else{long long ans=exgcd(b,a%b,x,y);long long t=x;x=y;y=t-a/b*y;return ans;}
}
long long gcd(long long a,long long b)
{if(b==0)return a;return gcd(b,a%b);
}
bool solve()
{a[0]=a[1],b[0]=b[1];for(int i=2;i<=n;i++){long long a0=b[0];long long b0=b[i];long long c0=a[i]-a[0];long long x0,y0;long long r=gcd(a0,b0);if(c0%r!=0)return false;else{a0/=r;b0/=r;c0/=r;exgcd(a0,b0,x0,y0);x0*=c0;x0=(x0%b0+b0)%b0;a[0]=b[0]*x0+a[0];b[0]=b[0]*b0;}}return true; // a[0]为小于[b1,b2...bn]的非负整数解
}
int  main()
{while(scanf("%d",&n)!=EOF){for(int i=1;i<=n;i++)scanf("%lld%lld",&b[i],&a[i]);if(solve())printf("%lld\n",a[0]);elseprintf("-1\n");}return 0;
}