codeforces-478C-Table Decorations【贪心】

codeforces-478C-Table Decorations

[code]                time limit per test1 second     memory limit per test256 megabytes

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.

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.

题目链接：cf-478C

题目大意：有三种颜色气球，问最多有几种方案满足条件。条件：三个气球一组，一组气球不同都是同一种颜色

题目思路：我模拟了很久，最后发现。。。最多有(r + g + b) / 3种方案。如果存在某两种颜色相加比最优方案还要小，那答案是此两种颜色相加。

以下是代码：

[code]#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
using namespace std;

int main(){
    long long r,g,b;
    cin >> r >> g >> b;

    long long ans = (r + g + b) / 3;
    ans = min(ans,r + g);
    ans = min(ans,r + b);
    ans = min(ans,g + b);

    cout << ans << endl;
    return 0;
}