POJ 2891 Strange Way to Express Integers【模線性方程組】(中國剩餘定理非互質套用)

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.

/*
    題意:給出k個模方程組:x mod ai = ri.求x的最小正值.不存在輸出-1.
         ai之間可能不滿足兩兩互質的性質
    類型:比中國剩餘定理更強的方法QAQ一般模線性方程組
    分析:新模版,漲姿勢啊
         新模版是求解A[i]x = B[i] (mod M[i]),總共n個線性方程組 的x的,令A[i]=1,就是中國剩餘定理的升級版本了
*/
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef pair<LL, LL> PLL;
LL a[100005], b[100005], m[100005];
LL gcd(LL a, LL b){
    return b ? gcd(b, a%b) : a;
}
void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d){
    if (!b) {d = a, x = 1, y = 0;}
    else{
        ex_gcd(b, a % b, y, x, d);
        y -= x * (a / b);
    }
}
LL inv(LL t, LL p){//如果不存在，返回-1 
    LL d, x, y;
    ex_gcd(t, p, x, y, d);
    return d == 1 ? (x % p + p) % p : -1;
}
PLL linear(LL A[], LL B[], LL M[], int n) {//求解A[i]x = B[i] (mod M[i]),總共n個線性方程組 
    LL x = 0, m = 1;
    for(int i = 0; i < n; i ++) {
        LL a = A[i] * m, b = B[i] - A[i]*x, d = gcd(M[i], a);
        if(b % d != 0)  return PLL(0, -1);//答案，不存在，返回-1 
        LL t = b/d * inv(a/d, M[i]/d)%(M[i]/d);
        x = x + m*t;
        m *= M[i]/d;
    }
    x = (x % m + m ) % m;
    return PLL(x, m);//返回的x就是答案，m是最後的lcm值 
}
int main(){
    int n;
    while(scanf("%d", &n) != EOF){
        for(int i = 0; i < n; i ++){
            a[i] = 1;
            scanf("%d%d", &m[i], &b[i]);
        }
        PLL ans = linear(a, b, m, n);
        if(ans.second == -1) printf("-1\n");
        else printf("%I64d\n", ans.first);
    }
}