SDUT2015暑假集训14级周赛1 B - 皇马(组合数)

B - 皇马

Time Limit:3000MS     Memory Limit:64000KB     64bit IO Format:%lld & %llu

Submit   Status   Practice   ACdream 1199

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.

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

当前人能拿的当前剑，之前的情况。 HUST  VJ可以用long long - -


#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define mod 1000000007
#define L long long

using namespace std;
L w[100010],p[100010];
int main()
{
    int n,m,i,j,cla;
    ios::sync_with_stdio(false);
    cin>>cla;
    for(int k=1;k<=cla;k++)
    {
        cin>>n;
        for(i=0;i<n;i++)
            cin>>w[i];
        for(i=0;i<n;i++)
            cin>>p[i];
        sort(w,w+n);
        sort(p,p+n);
        printf("Case #%d: ",k);
        L ans=1;
        j=0;
        for(i=0;i<n;i++)//枚举各个剑的重量是否在当前人所承受的范围之内若是j-i然后垒成在ans
        {
            while(j<n&&w[j]<=p[i])
            {
                j++;
            }
            ans=ans*(j-i)%mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}