周赛--Heros and Swords

最新推荐文章于 2024-02-17 19:05:48 发布

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本文深入分析了在给定英雄力量与不同重量的剑的情况下，英雄们如何进行配剑，确保每把剑的重量不超过英雄的力量限制。通过具体的案例研究，展示了在特定条件下计算配剑方案的方法，以及如何利用排序和动态规划技巧解决此类问题。

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

#include <iostream>
#include <string.h>
#include <math.h>
#include <queue>
#include <algorithm>
#include <stdlib.h>
#include <map>
#include <set>
#include <stdio.h>
#include <time.h>
using namespace std;
#define LL long long
#define pi acos(-1.0)
#pragma comment(linker, "/STACK:1024000000")
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
const double eqs=1e-9;
const int MAXN=1e5+10;
int a[MAXN], b[MAXN];
int main()
{
        int t, n, i, j, num=0;
        LL ans;
        scanf("%d",&t);
        while(t--){
                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);
                ans=1;
                j=0;
                for(i=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;
}