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

来源：互联网发布：软件开发报价编辑：程序博客网时间：2024/05/17 23:08

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

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.

————————————————————————————————————————————————————————————————

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;    }}