UESTC 1273 God Qing's circuital law

Description

As we all know,God Qing is a very powerful ACMer. She very like junior sister apprentice,but in UESTC-ACM team,there is no junior sister apprentice,we can use a mathematical formula to express it.

But God Qing is a life winner , she has many beautiful girls,every girl has its BeautifulValue V ,and he want to take the most beautiful girl among all girls. For every girl,if God Qing chooses her,she can get a value called Life-Winner Point. We can use a mathematical formula to express it.

LifeWinnerPoint = HandsomeValue * BeautifulValue

When other people who come from UESTC-ACM team(such as Liao772002,Final-Pan,Final-Zhu and so on) know God Qing want to choose the most beautiful girl,they can't stand it! ,They also want a girl,for every person,he/she has its own HandsomeValue P. But God Qing wants to make her Life-Winner Point most,so she wants you to tell her there are how many ways can GodQing get the most Life-Winner Point.

In general. Thre are N competitors come from UESTC-ACM team(God Qing's ID is always 1) and N Girls, for every competitor,he/she has a HandsomeValue P[i] , for every girl , she has a BeatuifulValue V[i],every competitor can only choose a girl,and every girl can be only chose once. you should calculate there are how many ways that make GodQing get the most Life-Winner Point. (God Qing’s Life-Winner Point strictly greater than any other competitors).

Because the answer can be very large,you only need to print the answer modulo 10^9 + 7.

Input

Line 1: an integet N(the number of competitors and girls).

Line 2: N numbers p[i](the HandsomeValue of every competitor)

Line 3: N numbers v[i](the BeautifulValue of every girl).

1 <= N <= 100, 1 <= p[i] <= 10^6 , 1 <= v[i] <= 106.

Output

the number of ways that GodQing can get the most Life-Winner Point.

Sample Input

4 
5 8 4 8 
19 40 25 20

Sample Output

2

Hint

Valid assignment #1: (5,40), (8,19), (4,25), (8,20).

Valid assignment #2: (5,40), (8,20), (4,25), (8,19).

In assignment #1, the Life-Winner Point of God Qing is 5 x 40 = 200. The other three units have Life-Winner Point 8 x 19 = 152, 4 x 25 = 100, and 8 x 20 = 160. This is a valid assignment because the number 200 is strictly greater than each of the numbers 152, 100, and 160.

这题第一眼还以为是二分图匹配什么的，结果想了半天也没想法。

后来突然间就想到了，这其实就是个暴力枚举。

简单来说，先把除了godqing以为的人从大到小排，所有女生从小到大排，

对于godqing来说，枚举所有的女生和他组合，对于每个组合来说

统计最大的人能选的数量，然后在统计第二大的，在算进答案里的时候要减去被第一大的人选过的那1个

之后的同理，这样就能算出所有的组合了。

#include<iostream>  
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<string>
#include<functional>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 2;
const int base = 1e9 + 7;
int T, n;
LL a[maxn], b[maxn], ans;

int main(){
	//scanf("%d", &T);
	while (~scanf("%d", &n))
	{
		for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
		for (int i = 0; i < n; i++) scanf("%lld", &b[i]);
		sort(a + 1, a + n, greater<int>());	sort(b, b + n);
		ans = 0;
		for (int i = 0; i < n; i++)
		{
			LL now = a[0] * b[i], res = 1, cnt = 0;
			for (int j = 1, k = 0; j < n; j++)
			{
				while (k < n)
				{
					if (k == i) k++;
					if (k < n&&a[j] * b[k] < now) { cnt++; k++; }
					else break;
				}
				(res *= cnt--) %= base;
			}
			(ans += res) %= base;
		}
		printf("%lld\n", ans);
	}
	return 0;
}