Monkey Tradition（中国剩余定理）

In ‘MonkeyLand‘, there is a traditional game called "Bamboo Climbing". The rules of the game are as follows:

1)       There are N monkeys who play this game and there are N bamboos of equal heights. Let the height be L meters.

2)       Each monkey stands in front of a bamboo and every monkey is assigned a different bamboo.

3)       When the whistle is blown, the monkeys start climbing the bamboos and they are not allowed to jump to a different bamboo throughout the game.

4)       Since they are monkeys, they usually climb by jumping. And in each jump, the i^th monkey can jump exactly p_i meters (p_i is a prime). After a while when a monkey finds that he cannot jump because one more jump may get him out of the bamboo, he reports the remaining length r_i that he is not able to cover.

5)       And before the game, each monkey is assigned a distinct p_i.

6)       The monkey, who has the lowest r_i, wins.

Now, the organizers have found all the information of the game last year, but unluckily they haven‘t found the height of the bamboo. To be more exact, they know N, all p_i and corresponding r_i, but notL. So, you came forward and found the task challenging and so, you want to find L, from the given information.

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 12). Each of the next n lines contains two integers p_i (1 < p_i < 40, p_i is a prime) and r_i (0 < r_i < p_i). All p_i will be distinct.

For each case, print the case number and the minimum possible value of L that satisfies the above conditions. If there is no solution, print ‘Impossible‘.

2

3

5 4

7 6

11 3

4

2 1

3 2

5 3

7 1

Case 1: 69

Case 2: 113

题解：

用扩展GCD求；剩下的就是模版；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
LL p[20], r[20];
void ex_gcd(LL a, LL b, LL &x, LL &y){
    if(!b){
        x = 1;
        y = 0;
        return;
    }
    ex_gcd(b, a%b, x, y);
    LL temp = x;
    x = y;
    y = temp - a/b * y;
}
int main(){
    int T, n, kase = 0;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        LL MOD = 1;
        for(int i = 0; i < n; i++){
            scanf("%lld%lld", &p[i], &r[i]);
            MOD *= p[i];
        }
        LL x, y;
        LL ans = 0;
        for(int i = 0; i < n; i++){
            ex_gcd(MOD/p[i], p[i], x, y);
            ans = (ans + MOD/p[i]*x*r[i] + MOD) % MOD;
        }
        printf("Case %d: %lld\n",++kase, (ans + MOD) % MOD);
    }
    return 0;
}