1319 - Monkey Tradition

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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 knowN, all p_i and corresponding r_i, but not L. So, you came forward and found the task challenging and so, you want to find L, from the given information.

Input

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.

Output

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‘.

SPECIAL THANKS: JANE ALAM JAN

贴码先， 定理稍后学；

//分析：首先我们先学习下中国剩余定理：中国剩余定理
//在模数mi两两互素的情况下，可以用中国剩余定理求n使其满足：
//n=b1(mod m1)
//n=b2(mod m2)
//n=b3(mod m3)
//....
//我感觉这分析好像没什么卵...
#include <cstdio>
#define LL long long
LL p[45], r[45];
void Exgcd(LL a, LL b, LL &xx, LL &yy)
{
    if(b==0)
    {
        xx=1; yy=0;
        return;
    }
    Exgcd(b, a%b, xx, yy);
    LL t=xx; xx=yy; yy=t-a/b*yy;
    return;
}
LL China(LL s[], LL b[], int k)// s: 模数； b: 余数
{
    LL n=1, xx, yy;
    LL ans=0;
    for(int i=0; i<k; i++)
        n*=s[i];
    for(int i=0; i<k; i++)
    {
        LL t=n/s[i];
        exgcd(t, s[i], xx, yy);
        ans=(ans+xx*t*b[i])%n;
    }
    return (ans%n+n)%n;
}
int main()
{
    int t, n, Q=1;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        for(int i=0; i<n; i++)
            scanf("%lld%lld", &p[i], &r[i]);
        LL ans=China(p, r, n);
        printf("Case %d: %lld\n", Q++, ans);
    }
    return 0;
}