Strange Way to Express Integers

 

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_k are 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?

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).

 

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.

 

2
8 7
11 9

31

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

#include<stdio.h>  
#include<string.h>  
#include<algorithm>  
using namespace std;  
  
  
#define LL long long  
  
  
LL exgcd(LL a,LL b,LL &x,LL &y)  
{  
    if(b==0){  
        x=1,y=0;  
        return a;  
    }  
  
  
    LL d=exgcd(b,a%b,y,x);  
    y-=(a/b)*x;  
    return d;  
}  
  
  
int main()  
{  
    LL k,a1,a2,r1,r2;  
    LL x,y,d;  
   // freopen("in.txt","r",stdin);  
    while(scanf("%I64d",&k)!=EOF)  
    {  
        scanf("%I64d%I64d",&a1,&r1);  
        k--;  
        int flag=0;  
        while(k--)  
        {  
            scanf("%I64d%I64d",&a2,&r2);  
            if(flag)  
                continue;  
  
  
            //扩展欧几里得  
            LL a=a1,b=a2;  
            LL c=r2-r1;  
            d=exgcd(a,b,x,y);//最小公约数  
            if(c%d){  
                flag=1;  
                continue;  
            }  
              
            LL t=b/d;//除以d表示方程两边都除以d  
            x=(x*(c/d)%t+t)%t;  
            r1=a1*x+r1;//同上面所说的式子5  
            a1=a1*(a2/d);//a1 a2最大公倍数  
        }  
  
  
        if(flag)  
            printf("-1\n");  
        else  
            printf("%I64d\n",r1);  
    }  
    return 0;  
}