★【16.6.9】Codeforces Round #356 (Div. 2) A. Bear and Five Cards

本文链接：https://blog.youkuaiyun.com/SatenRuiko/article/details/51619498

本文介绍了一个关于小熊Limak的游戏策略问题。Limak拥有五张卡片，每张卡片上写有一个正整数。文章探讨了如何通过丢弃相同数值的两张或三张卡片来最小化剩余卡片上的数字之和的方法。

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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
7 3 7 3 20

output
26

input
7 9 3 1 8

output
28

input
10 10 10 10 10

output
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 is7 + 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 is10 + 10 = 20.

注：每次只能丢掉两张或三张

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstring>

using namespace std;

int i,j;
int a[9999];
int sum,max1,sum1,flag;
int main()
{
    while(~scanf("%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5]))
    {
        sum1=0;
        max1=0;
        for(i=1;i<=5;i++)
        {
            sum=0;
            sum1+=a[i];
            flag=0;
            for(j=i;j<=5;j++)
            {
                if(a[i]==a[j])
                {
                    sum+=a[j];
                    flag++;
                }
            }
            if(sum>=max1&&flag>=2&&flag<=3)max1=sum;
        }
        printf("%d\n",sum1-max1);
    }
    return 0;
}