CodeForces 680ABear and Five Cards（取数字）

http://codeforces.com/problemset/problem/680/A

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：

给定五个数字，最多取出两个相同的数字，取出之后找出最小的和。

Code：

#include<stdio.h> #include<string.h> const int MYDD=1103; int main() { int num[MYDD];// num[j] 数字 j 的个数是num[j] int sum=0; memset(num,0,sizeof(num)); for(int j=0; j<5; j++) { int n; scanf("%d",&n); num[n]++; sum+=n; } int max=0; for(int j=1; j<=100; j++) { int temp=0; if(num[j]==2) temp=j*2; if(num[j]>2) temp=j*3; if(temp>max) max=temp; } int ans=sum-max; printf("%d\n",ans); return 0; } /*By: Shyazhut */