【CODEFORCES】 C. Table Decorations

C. Table Decorations

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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.

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。看到题目数据很大，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。（因为每次拿的都是3的倍数，而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;
}