hdoj 5494 Card Game 【水题】

最新推荐文章于 2019-03-25 18:47:00 发布

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本文介绍了一个简单的算法问题，即如何判断玩家Soda是否能在一种特定的卡片游戏中始终获胜。问题中，两个玩家各自拥有一定数量的卡片，通过选择部分卡片进行比较来决定胜负。文章通过排序和比较的方法给出了明确的解决方案。

Card Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6 Accepted Submission(s): 6

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

很简单的题目，不多说了。。。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 500+10
using namespace std;
bool cmp(int a, int b)
{
    return a>b;
}
int a[MAXN], b[MAXN];
int main()
{
    int t;
    int 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]);
        sort(a, a+n);
        sort(b, b+n, cmp);
        int suma = 0, sumb = 0;
        for(int i = 0; i < m; i++)
            suma += a[i], sumb += b[i];
        if(suma > sumb)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}