# [POJ 2891] Strange Way to Express Integers【扩展中国剩余定理】

• 2018-02-20
• 0
• 0

## Problem:

 Time Limit: 1000MS Memory Limit: 131072K

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, …, ak. For some non-negative m, divide it by every ai (1 ≤ ik) to find the remainder ri. If a1, a2, …, ak are properly chosen, m can be determined, then the pairs (ai, ri) 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 ai, ri (1 ≤ ik).

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

## Solution:

• x ≡ a1 (mod m1)
• x ≡ a2 (mod m2)

• x = a1 + m1 * p
• x = a2 + m2 * q

• m1 * (-p) + m2 * q = a1 - a2

d = exgcd(m1, m2, x, y)，则当且仅当 d | (a1 - a2) 时可合并，

• x ≡ a1 + m1 * p (mod LCM(m1, m2))

## Code: O(TklogV), V为值域 [696K, 16MS]

```#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

int k;
ll a, r;

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

int main(){
while(scanf("%d", &k) != EOF){
for(register int i = 1; i <= k; i++) scanf("%lld%lld", a + i, r + i);
ll M = a, R = r, x, y, d;
bool nosol = 0;
for(register int i = 2; i <= k; i++){
d = exgcd(M, a[i], x, y);
if((R - r[i]) % d) {puts("-1"), nosol = 1; break;}
x = (R - r[i]) / d * (-x % a[i]) % a[i];  // Divide first to avoid overflowing
R += M * x;
M = M / d * a[i];  // Divide first to avoid overflowing
R %= M;
}
if(!nosol) printf("%lld\n", R < 0 ? R + M : R);
}
return 0;
}
```

#### 评论

darkleafin.cf
(该域名已过期且被抢注。。)
darkleafin.github.io

49750

https://github.com/Darkleafin

OPEN AT 2017.12.10

Please refresh the page if the code cannot be displayed normally.

https://visualgo.net/en

- Theme by Qzhai