C. Table Decorations（Codeforces Round 273)，codeforces273

C. Table Decorations（Codeforces Round 273)，codeforces273
C. Table Decorationstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

You have r red, g green and b blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number t of tables can be decorated if we know number of balloons of each color?

Your task is to write a program that for given values r, g and b will find the maximum number t of tables, that can be decorated in the required manner.

Input

The single line contains three integers r, g and b (0 ≤ r, g, b ≤ 2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.

Output

Print a single integer t — the maximum number of tables that can be decorated in the required manner.

Sample test(s)input

5 4 3

output

4

input

1 1 1

output

1

input

2 3 3

output

2

Note

In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively.

首先要明白，當最大的氣球數量的一半小於另外兩種顏色的數量之和，肯定可以組成全部顏色數量之和/3,當最大的

氣球數量的一半大於等於另外兩種顏色的數量之和，全部組成為2+1，2為最多數量的顏色。

代碼：

#include <iostream>#include <cstdio>#include <cstdio>#include <algorithm>using namespace std;int main(){    long long a[3];    scanf("%I64d%I64d%I64d",&a[0],&a[1],&a[2]);    sort(a,a+3);    long long ans=0;    if((a[0]+a[1])<=a[2]/2)    {       ans=a[0]+a[1];    }    else    {        ans=(a[0]+a[1]+a[2])/3;    }    printf("%I64d\n",ans);    return 0;}

codeforces怎才可以改變id顏色（程式設計的）

不是北京時間,晚4個小時這樣吧.顏色的話不是很清楚...我只知道兩三百名就是藍名