hdu 5494 Card Game 排序-优快云博客

本文介绍了一个卡片游戏场景下的胜负判断算法。游戏双方各自拥有数量相等的卡片，并从中选择一部分进行比较，胜者为所选卡片数值之和较大者。文章提供了一种通过排序和对比来确定一方是否能确保胜利的方法。

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Card Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 94 Accepted Submission(s): 79

Problem Description

Soda and Beta are good friends. They are going to play a card game today. Soda has

n cards with number

a1,a2,...,an while Beta has

n cards with number

b1,b2,...,bn.

First, they choose a number

m no larger than

n. Then they both randomly select

m cards from their own

n cards. The one with larger sum of the selected cards will win. Soda wants to know if he can always win no mater what cards will be randomly selected from him and Beta.

Input

There are multiple test cases. The first line of input contains an integer

T(1≤T≤100), indicating the number of test cases. For each test case:

The first line contains two integer

n and

(1≤m≤n≤500). The second line contains

n integers

a1,a2,...,an

(1≤ai≤1000) denoting Soda's cards. The third line contains

n integers

b1,b2,...,bn

(1≤bi≤1000) denoting Beta's cards.

Output

For each test case, output "YES" (without the quotes) if Soda can always win, otherwise output "NO" (without the quotes) in a single line.

Sample Input

Sample Output

YES
NO

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;

#define maxn 100007
int a[maxn],b[maxn];

int main(){
    int t,n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(int i = 0;i < n; i++){
            scanf("%d",&a[i]);
        }
        for(int i = 0; i < n; i++)
            scanf("%d",&b[i]);
        int ans = 0;
        sort(a,a+n);
        sort(b,b+n);
        for(int i = 0;i < m; i++){
            ans += a[i] - b[n-1-i];
        }
        if(ans > 0) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}