Strange Way to Express Integers（扩展欧几里得解同余方程）

 

时间限制(普通/Java):1000MS/10000MS     内存限制:65536KByte
总提交: 55            测试通过:26

描述

 

Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:

Choose k different positive integers a₁, a₂, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_kare properly chosen, m can be determined, then the pairs (a_i, r_i) can be used to express m.

“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”

Since Elina is new to programming, this problem is too difficult for her. Can you help her?

输入

 

The input contains multiple test cases. Each test cases consists of some lines.

• Line 1: Contains the integer k.
• Lines 2 ~ k + 1: Each contains a pair of integers a_i, r_i (1 ≤ i ≤ k).

输出

Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.

样例输入

 

样例输出

 

提示

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

题目来源

POJ Monthly 2006.7

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#define ll long long
using namespace std;

ll exgcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0){
        x=1,y=0;
        return a;
    }
    ll r=exgcd(b,a%b,x,y);
    ll t=x;
    x=y,y=t-a/b*y;
    return r;
}

int main()
{
    ll n,a1,r1,a2,r2,ans,a,b,c,d,x0,y0;
    while(scanf("%lld",&n)!=EOF){
        int flag=1;
        scanf("%lld%lld",&a1,&r1);
        for(ll i=1;i<n;i++){
            scanf("%lld%lld",&a2,&r2);
            a=a1,b=a2,c=r2-r1;
            ll d=exgcd(a,b,x0,y0);
            if(c%d!=0){//不存在可能值
                flag=0;
            }
            int t=b/d;
            x0=(x0*(c/d)%t+t)%t;//x0为正
            r1=a1*x0+r1;
            a1=a1*(a2/d);
        }
        if(!flag){
            printf("-1
");
        }
        else{
            printf("%lld
",r1);
        }
    }
    return 0;
}