Codeforces Round #243 (Div. 2) A~C

题目链接

A. Sereja and Mugs

time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output

Sereja showed an interesting game to his friends. The game goes like that. Initially, there is a table with an empty cup and n water mugs on it. Then all players take turns to move. During a move, a player takes a non-empty mug of water and pours all water from it into the cup. If the cup overfills, then we assume that this player lost.

As soon as Sereja's friends heard of the game, they wanted to play it. Sereja, on the other hand, wanted to find out whether his friends can play the game in such a way that there are no losers. You are given the volumes of all mugs and the cup. Also, you know that Sereja has (n - 1) friends. Determine if Sereja's friends can play the game so that nobody loses.

Input
The first line contains integers n and s (2 ≤ n ≤ 100; 1 ≤ s ≤ 1000) — the number of mugs and the volume of the cup. The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 10). Number ai means the volume of the i-th mug.

Output
In a single line, print "YES" (without the quotes) if his friends can play in the described manner, and "NO" (without the quotes) otherwise.

Sample test(s)

input

3 4
1 1 1

output

YES

input

3 4
3 1 3

output

YES

input

3 4
4 4 4

output

NO

题意 ： 大意大概是有很多n个小杯，1大杯，每个小杯的容积以及大杯的容积都知道了，游戏规则是，每个人拿一个非空的小杯，把水全部倒大杯里，如果谁倒的时候大杯满了，谁就输，然后她的（n-1）个朋友要玩儿，问你有没有一种倒水的方法使得没有人会输，如果没有输出NO。

思路 ： 一开始以为她自己也玩儿，就不知道第二组样例为什么对了，后来又看了一遍题，觉得应该是光朋友玩儿，就试了一下儿，结果对了。。。。。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std ;

int a[1010] ;

int main()
{
int n ,k;
while(~scanf("%d %d",&n,&k))
{
for(int i = 1 ; i <= n ; i++)
scanf("%d",&a[i]) ;
int maxx = -9999999,sum ;
for(int i = 1 ; i <= n ; i++)
{
for(int j = i ; j <= n ; j++)
{
sum = 0 ;
vector<int > v ,u;
for(int h = i ; h <= j ; h++)
{
v.push_back(a[h]) ;
sum += a[h] ;
}
maxx = max(maxx,sum) ;
for(int h = 1 ; h <= n ; h++)
{
if(h < i || h > j)
u.push_back(a[h]) ;
}
sort(v.begin(),v.end()) ;
sort(u.begin(),u.end()) ;
reverse(u.begin(),u.end()) ;
for(int h = 1 ; h <= k && h <= v.size() && h <= u.size() ; h++)//枚举交换次数
{
if(v[h-1] < u[h-1])
{
sum -= v[h-1] ;
sum += u[h-1] ;
maxx = max(sum,maxx) ;
}
}
}
}
printf("%d\n",maxx) ;
}
return 0 ;
}

View Code