B - 皇马

 

ABCDEFGHIJ

B - 皇马
Time Limit:3000MS    Memory Limit:64000KB    64bit IO Format:%lld & %llu
SubmitStatusPracticeACdream
1199

Description

There are n swords of different weights Wiand n heros of power
Pi.

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.

Sample Input
对剑的重量和人物的pow都从小到大排序；后面人物可以买前面人物的剑；也就是（j-i）把剑；

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 <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<stack>
using namespace std;
int sword[100010];
int hero[100010];
int num[100010];
const int MOD=1000000007;
int main()
{
int n,t;
while(~scanf("%d",&t))
{
int k=0;
while(t--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&sword[i]);
}
for(int j=0;j<n;j++)
{
scanf("%d",&hero[j]);
}
sort(sword,sword+n);
sort(hero,hero+n);
long long ans=1;
for(int i=0,j=0;i<n;i++)
{
while(j<n&&hero[i]>=sword[j])
j++;
ans=(ans*(j-i))%MOD;
}
printf("Case #%d: %lld\n",++k,ans);
}
}
return 0;
}

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