HDOJ 5494 Card Game（水）

Card Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 58    Accepted Submission(s): 52

Problem Description

Soda and Beta are good friends. They are going to play a card game today. Soda hasn 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 integerT(1≤T≤100), indicating the number of test cases. For each test case:

The first line contains two integer n and m(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

23 14 5 61 2 35 23 4 7 8 93 4 5 2 3

 

Sample Output

YESNOn个数取m个，判断是否Soda可以一直赢，直接判断Soda的m个最小的卡片和和Beta的m个最大的卡片的和的大小就行了。又水了一道题，汗= =ac代码：
#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define MAXN 550#define INF 0xfffffff#define MIN(a,b) a>b?b:ausing namespace std;int a[MAXN];int b[MAXN];int main(){int t,i,n,m;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);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);int sum1=0,sum2=0;for(i=0;i<m;i++)sum1+=a[i],sum2+=b[n-i-1];if(sum1>sum2)printf("YES\n");elseprintf("NO\n");}return 0;}