[周赛] Heros and Swords-优快云博客

本文介绍了一个剑英雄匹配问题，即如何让每个英雄携带一把不超过其力量的剑，并计算出所有可能的组合方式。通过排序和计数的方法，解决了这个问题，并提供了一个C++实现示例。

Problem Description

There are n swords of different weights W_i and 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

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;
int a[100010],b[100010];
int main()
{
    int t,n,i,j;
    int num=0;
    scanf("%d",&t);
    while (t--)
    {
        long long ans=1;
        scanf("%d",&n);
        for (i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for (i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
        }
        sort(a,a+n);
        sort(b,b+n);
        for (i=0,j=0;i<n;i++)
        {
            while (j<n&&a[j]<=b[i])
            {
                j++;
            }
            ans = ans*(j-i)%mod;
        }
        printf("Case #%d: %lld\n",++num,ans);
    }
    return 0;
}