B. Ciel and Flowers
time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputFox 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 ",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 ",ans); } return 0; }