lightoj 1319 - Monkey Tradition （中国剩余定理）

1319 - Monkey Tradition

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Time Limit: 2 second(s)Memory Limit: 32 MB

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'.

Sample Input

Output for Sample Input

2

3

5 4

7 6

11 3

4

2 1

3 2

5 3

7 1

Case 1: 69

Case 2: 113

 这道题可以采用中国剩余定理，但是只能处理两两互质的情况，看了别人代码有些地方还是不特别的懂。

先说一下我的理解举个例子：

5     4

7     6

11    3

设要求的数是X：

X=5（mod 4）

x=7 (mod 6)

x=11 (mod 3)

在中国剩余定理中有这样情况

我们先从n1这个角度出发，已知n1满足除以5余4，能不能使得 n1+n2 的和仍然满足除以5余4？进而使得n1+n2+n3的和仍然满足除以5余4？

这是成立的即：

1. 为使n1+n2+n3的和满足除以5余4，n2和n3必须是5的倍数。
2. 为使n1+n2+n3的和满足除以7余6，n1和n3必须是7的倍数。
3. 为使n1+n2+n3的和满足除以11余3，n1和n2必须是11的倍数。

    因此，为使n1+n2+n3的和作为“孙子问题”的一个最终解，需满足：

1. n1除以5余4，且是11和7的公倍数。
2. n2除以7余6，且是5和11的公倍数。
3. n3除以11余3，且是7和5的公倍数。 

这里有个技巧比如说对于5来说从7和11公倍数中找到一个最小的数除以5余数为1，再乘以4，每个都是这样算然后将算出得数相加，这样求出的并不是最小的X，需要对X处理求余，即X%（5*7*11）；

<span style="font-family:Courier New;font-size:24px;">#include<algorithm>#include<iostream>#include<cstring>#include<cmath>using namespace std;int pi[20],ri[20];int n;long long sum,ans;void gcd(long long a,long long b,long long &x,long long &y){if(b==0){x=1;y=0;return ;}gcd(b,a%b,x,y);long long t=x;x=y;y=t-y*(a/b);return ;}long long CRT(){long long x,y;for(int i=1;i<=n;i++){long long s=sum/pi[i];gcd(s,pi[i],x,y);ans=(ans+s*ri[i]*x)%sum;}return (ans%sum+sum)%sum;}int main(){int t,cas=1;scanf("%d",&t);while(t--){scanf("%d",&n);sum=1;for(int i=1;i<=n;i++){scanf("%d%d",&pi[i],&ri[i]);sum*=pi[i];}ans=0;ans=CRT();printf("Case %d: %lld\n",cas++,ans);}return 0;}</span>