Codeforces-Round-#356-(Div.-2)-Bear-and-Five-Cards

A. Bear and Five Cards

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题意：给你5个数，让你消除且只能消除有2个或者3个相同数字的数字，使最后数字之和最小。

题目链接：Bear and Five Cards

解题思路：循环比较就行了，然后取最小值

代码：

//A. Bear and Five Cards
//题目链接：http://www.codeforces.com/problemset/problem/680/A
#include
#include
#include
#include
using namespace std;
int cnt[105];
int main()
{
    int t1,t2,t3,t4,t5,ans,t,i,l;
    memset(cnt,0,sizeof(cnt));
    while(cin>>t1>>t2>>t3>>t4>>t5){
        cnt[t1]++,cnt[t2]++,cnt[t3]++,cnt[t4]++,cnt[t5]++;//记录5个数字出现的个数
        t=0,ans=t1+t2+t3+t4+t5;        //记录总和
        for(i=0;i<=100;i++){
            if(cnt[i] >= 2){           //每次只能取某个数的2个或者三个
                if(cnt[i] > 3) cnt[i]=3;
                t=max(i*cnt[i],t);     //选取值最大的取法
            }
        }
        printf("%d\n",ans-t);
        cnt[t1]=cnt[t2]=0,cnt[t3]=cnt[t4]=0,cnt[t5]=0;
    }
    return 0;
}