Light oj 1319 - Monkey Tradition【CRT】

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通过给定的猴子爬竹游戏中每只猴子的跳跃能力和未覆盖长度，利用中国剩余定理求解最小竹子高度。

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1319 - Monkey Tradition

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

5 4

7 6

11 3

2 1

3 2

5 3

7 1

Case 1: 69

Case 2: 113

题意：

给出 n 对 p 和 r ,使得n %p= r ,求满足题意的最小值

题解：

中国剩余定理，详细解释请点这里

个人是不会的，虽然知道是这个知识点，但不会写，比赛确实是卡住了.....

后来见到的大神的详细代码，想膜拜大神来点这里

说实话，这道题不会做！

数学确实伤不起，感觉需要学的东西太多了，数学没学好，真伤心啊

第一次做这样的题，算是整理出一个模板吧，希望以后能真正理解这个解法的本质！

/*
http://blog.youkuaiyun.com/liuke19950717
*/
#include<cstdio>
typedef long long ll;
void extgcd(ll a,ll b,ll &d,ll &x,ll &y)
{
	if(!b)
	{
		d=a;x=1;y=0;
		return;
	}
	extgcd(b,a%b,d,y,x);
	y-=x*(a/b);
}
ll crt(ll n,ll p[],ll r[]) 
{
	ll s=1,ans=0;
	for(int i=0;i<n;++i)
	{
		s=s*p[i];
	}
	for(int i=0;i<n;++i)
	{
		ll m=s/p[i],d,x,y;
		extgcd(p[i],m,d,x,y);
		ans=(ans+y*m*r[i])%s;
	}
	return (ans%s+s)%s;
}
int main()
{
	int t;
	scanf("%d",&t);
	for(int k=0;k<t;++k)
	{
		ll n,p[15],r[15];
		scanf("%lld",&n);
		for(int i=0;i<n;++i)
		{
			scanf("%lld%lld",&p[i],&r[i]);
		}
		printf("Case %d: %lld\n",k+1,crt(n,p,r));
	}
	return 0;
}