想法题——Codeforces Round #190 (Div. 2)——B. Ciel and Flowers

http://codeforces.com/contest/322/problem/B

B. Ciel and Flowers

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Sample test(s)

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.

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比赛时想搓了。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include<cstdlib>
#include <algorithm>
using namespace std;

int main()
{
    int a[100],b[100];
    int ans;
    while(cin>>a[0] >>a[1]>>a[2]){
        sort(a,a+3);
        b[0]=a[0];
        b[1]=a[1];
        b[2]=a[2];

        ans=0;
        ans += a[0];

        a[1]-=a[0];
        a[2]-=a[0];

        a[0]=0;

        ans += a[1] / 3;
        ans += a[2] / 3;

        if(a[0] == 0&&b[0]>0 && a[1]%3+a[2]%3==4)
            ans++;
        else if(a[1]==0&&b[1]>0 && a[0]%3+a[2]%3==4)
            ans++;
        else if(a[2]==0&&b[2]>0 && a[1]%3+a[0]%3==4)
            ans++;
            
        cout<<ans;
    }
}