周赛一 ACdream 1199 排列组合

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<algorithm>

using namespace std;

int p[100010];
int T,n;
int a[100010];
int main()
{
    while(~scanf("%d",&T))
    {
        int Case=0;
        while(T--)
        {
            Case++;
            scanf("%d",&n);
            for(int i=0; i<n; ++i)
                scanf("%d",&a[i]);
            for(int j=0; j<n; ++j)
                scanf("%d",&p[j]);
            sort(a,a+n);
            sort(p,p+n);
           long long sum=1;
         int j=0;
           for(int i=0;i<n;i++) //组合数 求排列组合的种类
		   {
		   	while(j<n&&a[j]<=p[i])j++;
		   	sum=sum*(j-i)%1000000007;
		   }
            printf("Case #%d: %lld\n",Case,sum%1000000007);
        }
    }
}

/*
struct node
{
    int power;
    int num;
} p[100010];
int T,n;
int a[100010];
int cmp(node p1,node p2)
{
    return p1.power<p2.power;
}
int main()
{
    while(~scanf("%d",&T))
    {
        int Case=0;
        while(T--)
        {
            Case++;
            scanf("%d",&n);
            for(int i=0; i<n; ++i)
                scanf("%d",&a[i]);
            for(int j=0; j<n; ++j)
            {
                scanf("%d",&p[j].power);
                p[j].num=0;
            }
            sort(a,a+n);
            sort(p,p+n,cmp);
            int j=0;

            for(int i=0; i<n; ++i)  //组合数
            {
                if(i==0)
                {
                    for(; j<n; j++)
                    {
                        if(p[i].power>=a[j])
                            p[i].num++;
                        else
                            break;
                    }
                }
                else
                {
                    if(p[i-1].power==p[i].power)
                    {
                        p[i].num=p[i-1].num-1;
                    }
                    else
                    {
                        p[i].num=p[i-1].num;
                        for(; j<n; j++)
                        {
                            if(p[i].power>=a[j])
                                p[i].num++;
                            else
                            {
                                p[i].num-=1;
                                break;
                            }
                        }
                        if(j==n)
                            p[i].num-=1;
                    }
                }
            }
            long long int sum=1;
            for(int i=0; i<n; i++)
            {
                sum*=p[i].num%1000000007;
                sum%=1000000007;   //勿漏<span id="transmark"></span>
                if(sum==0)
                    break;
            }
            printf("Case #%d: %lld\n",Case,sum%1000000007);
        }
    }
}
*/