周赛-Heros and Swords 分类： 比赛 2015-08-02 08:30 11人阅读 评论(0) 收藏

Heros and Swords

Time Limit: 6000/3000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Submit Statistic Next Problem

Problem Description

There are n swords of different weights Wi and n heros of power Pi.

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

一道贪心的题，在贪心中处理组合数用的2*n的时间复杂度，精妙

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <algorithm>
#define LL long long
#define RR freopen("output.txt","r",stdoin)
#define WW freopen("input.txt","w",stdout)

using namespace std;

const int MAX = 100100;

const int MOD = 1000000007;

int n;

int w[MAX],p[MAX];

int num[MAX];

int main()
{
    int T;
    int W=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&w[i]);
        }
        for(int i=0;i<n;i++)
        {
            scanf("%d",&p[i]);
        }
        sort(w,w+n);
        sort(p,p+n);
        for(int i=0,j=0;i<n;i++)//贪心
        {
            while(j<n&&w[j]<=p[i])
            {
                j++;
            }//此时这个人可以拿加j-1个兵器，而前面的i-1个人已经拿了i-1个，所以他还可以拿（j-1-(i-1)）=j-i个；
            num[i]=j-i;
        }
        LL ans=1;
        for(int i=0;i<n;i++)
        {
            ans=(ans*num[i])%MOD;
            if(!ans)
            {
                break;
            }
        }
        printf("Case #%d: %lld
",W++,ans);
    }
    return 0;
}

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