Ciel and Flowers-----（一个非常简单的贪心水题呀呀呀呀呀~~~）

time limit per test1 second memory limit per test256 megabytes

贪心

来源：
codeforces 322 B Ciel and Flowers
vjudge C Ciel and Flowers

Description

Fox Ciel has some flowers: r red flowers, g green flowers and b blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:

• To make a "red bouquet", it needs 3 red flowers.
• To make a "green bouquet", it needs 3 green flowers.
• To make a "blue bouquet", it needs 3 blue flowers.

• To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.

  Help Fox Ciel to find the maximal number of bouquets she can make.
  

Input

The first line contains three integers r, g and b (0 ≤ r, g, b ≤ 109) — the number of red, green and blue flowers.

Output

Print the maximal number of bouquets Fox Ciel can make.

Examples

input

3 6 9

output

6

input

4 4 4

output

4

input

0 0 0

output

0

Note

In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.

In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.

题意：

有红、绿、蓝三种颜色的花，
要组成一束红花需要三朵红花
要组成一束绿花需要三朵绿花
要组成一束蓝花需要三朵蓝花
要组成一束混合花需要1朵红花,1朵绿花，1朵蓝花；
给你三种花的数量，让你求出最多可以组成多少束花

解题思路：

这个题非常简单，对吧~~呵呵哒，然而你忽略了：若当剩下的花为红花0朵，绿花和蓝花都分别为2朵时， 时，你可以减少一束红花，而增加两束混合花~~是不是啊！！！
唉~~你好zz哦~

完整代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    int a,b,c;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF)
    {
        int a1=a%3;
        int b1=b%3;
        int c1=c%3;
        int m1=min(a1,min(b1,c1));
        int num1=a1-m1+b1-m1+c1-m1;
        ///-----------------------///
        int m2=min(a,min(b,c));
        int a2=a-m2;
        int b2=b-m2;
        int c2=c-m2;
        int a3=a2%3;
        int b3=b2%3;
        int c3=c2%3;
        int num2=a3+b3+c3;
        ///-----------------------///
        int ans;
        if(num1>num2)
         {
             ans=a2/3+b2/3+c2/3+m2;
             if(m2!=0&&((a3==0&&b3==2&&c3==2)||(a3==2&&b3==0&c3==2)||(a3==2&&b3==2&c3==0)))
                ans+=1;
         }
        else
        {
            ans=a/3+b/3+c/3+m1;
            if((m2!=0)&&((a1==0&&b1==2&&c1==2)||(a1==2&&b1==0&&c1==2)||(a1==2&&b1==2&&c1==0)))
                ans+=1;
        }
        printf("%d\n",ans);
    }
    return 0;
}