codeforces 680A Bear and Five Cards

最新推荐文章于 2018-10-03 18:36:47 发布

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小熊Limak正在玩一个游戏，手中持有五张带有正整数的卡片。本任务的目标是最小化剩余卡片上的数值之和，允许一次丢弃两到三张相同数字的卡片。

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A little bear Limak plays a game. He has five cards. There is one number written on each card. Each number is a positive integer.

Limak can discard (throw out) some cards. His goal is to minimize the sum of numbers written on remaining (not discarded) cards.

He is allowed to at most once discard two or three cards with the same number. Of course, he won't discard cards if it's impossible to choose two or three cards with the same number.

Given five numbers written on cards, cay you find the minimum sum of numbers on remaining cards?

Input

The only line of the input contains five integers t1, t2, t3, t4 and t5 (1 ≤ ti ≤ 100) — numbers written on cards.

Output

Print the minimum possible sum of numbers written on remaining cards.

   Examples
 
     input 
    
      Copy
    
7 3 7 3 20

     output 
    
      Copy
    
26

     input 
    
      Copy
    
7 9 3 1 8

     output 
    
      Copy
    
28

     input 
    
      Copy
    
10 10 10 10 10

     output 
    
      Copy
    
20

Note

In the first sample, Limak has cards with numbers 7, 3, 7, 3 and 20. Limak can do one of the following.

Do nothing and the sum would be 7 + 3 + 7 + 3 + 20 = 40.
Remove two cards with a number 7. The remaining sum would be 3 + 3 + 20 = 26.
Remove two cards with a number 3. The remaining sum would be 7 + 7 + 20 = 34.

You are asked to minimize the sum so the answer is 26.

In the second sample, it's impossible to find two or three cards with the same number. Hence, Limak does nothing and the sum is 7 + 9 + 1 + 3 + 8 = 28.

In the third sample, all cards have the same number. It's optimal to discard any three cards. The sum of two remaining numbers is 10 + 10 = 20.

水题

#include<bits/stdc++.h>  
using namespace std;  
typedef long long ll;  
const int inf = 0x3f3f3f3f;  
int main()   
{  
	// freopen("shuju.txt","r",stdin);
    int a[10];
    memset(a,0,sizeof(a));
    int sum=0;
    for(int i=1;i<=5;i++)
    {
    	cin>>a[i];
    	sum+=a[i];
    }
    sort(a+1,a+1+5);
    int mi=sum;
    for(int i=5;i>1;i--)
    {
    	if(a[i]==a[i-1]&&a[i]==a[i-2])
    	{
    		mi=min(mi,sum-a[i]*3);
    	}
    	if(a[i]==a[i-1])
    	{
    		mi=min(mi,sum-a[i]*2);
    	}
    }  
    cout<<mi<<endl;
    return 0;  
}