CF 322B Ciel and Flowers 贪心水题

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.

题目意思是

做蓝的花束需要3朵蓝花，做红的花束需要3朵红花，绿的花束需要3朵绿花，混合花束需要3种颜色各一朵

现在给出红，绿，蓝花的数目，问最多有几个花束可以做出来。

贪心。对r,g,b分别除3取余数,为r1,g1,b1时，
然后排序一下，如果 r1==0 , g1 == 2, b1 == 2时，且原始的r答案再加1。减少一个r花的数目，换得2个混合花

/*
* @author ipqhjjybj
* @date  20130711
*
*/
#include <cstdio>
#include <cstdlib>
#include <algorithm>
void swap(int &a,int &b){
int t;
t=a,a=b,b=t;
}
int main(){
int r,g,b,ans=0;
scanf("%d %d %d",&r,&g,&b);
if(r==0||g==0||b==0){
printf("%d\n",r/3+b/3+g/3);
}else{
ans = r/3+g/3+b/3;
r%=3,g%=3,b%=3;
if(r>=g) swap(r,g);
if(r>=b) swap(r,b);
if(g>=b) swap(g,b);
// r<=g<=b
if(r==0&&g==2){
ans+=1;
}else{
ans+=r;
}
printf("%d\n",ans);
}
return 0;
}