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.

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.

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

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.

题解：这一题是一道贪心。策略是优先取最小的那堆，然后最大堆拿2个。看到题目数据非常大。不可能模拟。所以就从2堆的基础上推出一个公式：当最小的两堆相加乘以2比最大的那堆大时，结果就是3堆相加在整除3。否则就是两堆最小堆的和。（由于每次拿的都是3的倍数，并且最小的两堆并不能在最大的堆被拿完钱拿完，所以结果一定是相加后整除3）

#include <iostream>

#include <cstring>

#include <cstdio>

#include <cstdlib>

using namespace std;

long long r,g,b,ans;

int main()

{

    scanf("%I64d%I64d%I64d",&r,&g,&b);

    if (r>g) swap(r,g);

    if (r>b) swap(r,b);

    if (g>b) swap(g,b);

    if ((r+g)*2<=b)

    {

        printf("%I64d\n",r+g);

        return 0;

    }

    else ans=(r+g+b)/3;

    printf("%I64d\n",ans);

    return 0;

}