POJ 2891 Strange Way to Express Integers

Description

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?

Input

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

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.

Sample Input

2
8 7
11 9

Sample Output

31

Hint

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

Source

POJ Monthly--2006.07.30, Static

线性同余方程组

像普通同余方程那样求出解，然后将解的通式带入下一个方程，最后求出的

有方程：a1*x-a2*y=b2-b1

设a=a1/t,b=a2/t,c=(b2-b1)/t t=gcd(a,b)

用exgcd得到a*x+b*y=c的解x0，通解x=x0+k*b,k为整数←目前这个有点理解不了

带入a1*x+b1=n

a1*b*k+a1*x0+b1=n

所以b1=b1+a1*x0，a1=a1*b

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define LL long long
using namespace std;
int n;
LL gcd(LL a,LL b){
    return (!b)?a:gcd(b,a%b);
}
void exgcd(LL a,LL b,LL &x,LL &y){
    if(!b){
        x=1;y=0;return;
    }
    exgcd(b,a%b,x,y);
    LL t=x;
    x=y;
    y=t-a/b*y;
    return;
}
int main(){
    bool flag;
    LL a1,a2,b1,b2;
    while(scanf("%d",&n)!=EOF){
        int i,j;
        scanf("%I64d%I64d",&a1,&b1);
        LL a,b,c,x,y;
        flag=0;
        for(i=1;i<n;i++){
            scanf("%I64d%I64d",&a2,&b2);
            if(flag)continue;
            a=a1;
            b=a2;
            c=b2-b1;
            LL tmp=gcd(a,b);
            if(c%tmp!=0){//无解
                printf("-1
");
                flag=1;
            }
            else{
                a/=tmp;b/=tmp;c/=tmp;
                exgcd(a,b,x,y);
                x=((c*x)%b+b)%b;//求出最小整数解 
                b1=b1+a1*x;//带入 
                a1=a1*b;// 
            }
        }
        if(!flag)printf("%I64d
",b1);
    }
    return 0;
}