B - 皇马

Description

There are n swords of different weights W_iand n heros of power P_i.

Your task is to find out how many ways the heros can carry the swords so that each hero carries exactly one sword.

Here are some rules:

(1) Every sword is carried by one hero and a hero cannot carry a sword whose weight is larger than his power.

(2) Two ways will be considered different if at least one hero carries a different sword.

Input

The first line of the input gives the number of test cases T(1 ≤ T ≤ 50).

Each case starts with a line containing an integer n (1 ≤ n ≤ 10⁵) denoting the number of heros and swords.

The next line contains n space separated distinct integers denoting the weight of swords.

The next line contains n space separated distinct integers denoting the power for the heros.

The weights and the powers lie in the range [1, 10⁹].

Output

For each case, output one line containing "Case #x: " followed by the number of ways those heros can carry the swords.

This number can be very big. So print the result modulo 1000 000 007.

题意：t组数据，有n把剑和n个人，每把剑都有自己的重量，每个人都有自己的能力，只有此人的能力大于等于此剑的重量才能拿起这把剑，把n把剑分给n个人，每人都能拿起各自的剑，求总共有几种分法。

Sample Input

3

5

1 2 3 4 5

1 2 3 4 5

2

1 3

2 2

3

2 3 4

6 3 5

Sample Output

Case #1: 1

Case #2: 0

Case #3: 4

#include <cstdio>#include <algorithm>using namespace std;int w[100002],p[100002];int main(){    int T,n;    scanf("%d",&T);    for(int k=1; k<=T; k++)    {        scanf("%d",&n);        for(int i=0; i<n; i++)            scanf("%d",&w[i]);        for(int i=0; i<n; i++)            scanf("%d",&p[i]);        sort(w,w+n);        sort(p,p+n);        long long int cnt=1;        int j=0;        for(int i=0; i<n; i++)        {                    for(;j<n;j++)                    {                       if(w[j]>p[i])                       break;;                    }                    cnt=cnt*(j-i)%1000000007;        }        printf("Case #%d: %lld\n",k,cnt);    }    return 0;}