SDUT2015暑假集训14级周赛1 B - 皇马(组合数)

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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.

Swords

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

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;
}