1319 - Monkey Tradition

1319 - Monkey Tradition

PDF (English)

Statistics

Forum

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 know N, 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

---

Problem Setter: Tanvir Hassan

Special Thanks: Jane Alam Jan

思路：中国剩余定理;扩展欧基里德。模板题。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<queue>
 6 #include<vector>
 7 using namespace std;
 8 typedef long long LL;
 9 typedef unsigned long long ll;
10 pair<LL ,LL>P(LL n,LL m);
11 typedef struct pp
12 {
13     long long x;
14     long long y;
15 } ss;
16 ss aa[100];
17 int main(void)
18 {
19     LL i,j,k,p,q;int ca;
20     int ba;int s;
21     ll sum;
22     scanf("%d",&ca);
23     for(ba=1; ba<=ca; ba++)
24     {
25         sum=1;
26         scanf("%d",&s);
27         for(i=0; i<s; i++)
28         {
29             scanf("%lld %lld",&aa[i].x,&aa[i].y);
30             sum*=aa[i].x;
31         }
32         LL ans=0;
33         for(i=0; i<s; i++)
34         {
35             LL mn=sum/aa[i].x;
36             LL x;
37             LL y;
38            pair<LL,LL>xx;
39            xx=P(mn,aa[i].x);
40            x=xx.first;y=xx.second;
41             LL z=(aa[i].x+x%aa[i].x)%aa[i].x;
42             ans=((ans+((aa[i].y*mn)%sum)*z)%sum)%sum;
43         }
44         ans=(ans+sum)%sum;
45         printf("Case %d:",ba);
46         printf(" %lld\n",ans);
47     }return 0;
48 }
49
50 pair<LL ,LL>P(LL n,LL m)
51 {
52       if(m==0)
53       {
54         pair<LL,LL> c=make_pair(1,0);
55          return c;
56       }
57       else
58       {
59          pair<LL,LL>x=P(m,n%m);
60           LL k=x.second;
61           LL kk=x.first;
62           x.first=k;
63           x.second=kk-(n/m)*k;
64          return x;
65       }
66 }