Codeforces Round #273 (Div. 2)C. Table Decorations 数学

C. Table Decorations

 

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

 

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.

 题意：给你三种颜色的气球， 现在让你从中选出三个，满足：最多有两个气球的颜色是一样的。问你这样满足条件的方案数

题解：   我们来对三种气球的数量排序，从大到小a,b,c，  

          如果a/2>=b+c显然最多就是b+c；画个图，就知道了

         那么就只有  (a+b+c)/3这种解了

///1085422276
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define inf 1000000007
#define mod 1000000007
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){
        if(ch=='-')f=-1;ch=getchar();
    }
    while(ch>='0'&&ch<='9'){
        x=x*10+ch-'0';ch=getchar();
    }return x*f;
}
//************************************************
const int maxn=20000+5;
ll c[maxn];
ll ans;
int main(){


    for(int i=1;i<=3;i++){
        scanf("%I64d",&c[i]);
    }
    sort(c+1,c+3+1);
    if(c[1]+c[2]<=c[3]/2){
       ans=c[1]+c[2];
    }
    else if(c[1]+c[2]<=c[3]) {
           ans=(c[1]+c[2]+c[3])/3;
    }
    else if(c[1]+c[2]>c[3]){
        ans=(c[1]+c[2]+c[3])/3;
    }
    cout<<ans<<endl;
  return 0;
}