想法題——Codeforces Round #190 (Div. 2)——B. Ciel and Flowers
阿新 • • 發佈:2018-12-24
B. Ciel and Flowers
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Sample test(s)
input
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.
InputThe first line contains three integers r, g and b (0 ≤ r, g, b ≤ 109) — the number of red, green and blue flowers.
OutputPrint the maximal number of bouquets Fox Ciel can make.
3 6 9output
6input
4 4 4output
4input
0 0 0output
0Note
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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比賽時想搓了。。。
#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;
}
}