A. Compote
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Nikolay has a lemons, b apples and c pears. He decided to cook a compote. According to the recipe the fruits should be in the ratio 1: 2: 4. It means that for each lemon in the compote should be exactly 2 apples and exactly 4 pears. You can’t crumble up, break up or cut these fruits into pieces. These fruits — lemons, apples and pears — should be put in the compote as whole fruits.
Your task is to determine the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. It is possible that Nikolay can’t use any fruits, in this case print 0.
Input
The first line contains the positive integer a (1 ≤ a ≤ 1000) — the number of lemons Nikolay has.
The second line contains the positive integer b (1 ≤ b ≤ 1000) — the number of apples Nikolay has.
The third line contains the positive integer c (1 ≤ c ≤ 1000) — the number of pears Nikolay has.
Output
Print the maximum total number of lemons, apples and pears from which Nikolay can cook the compote.
Examples
input
2
5
7
output
7
input
4
7
13
output
21
input
2
3
2
output
0
Note
In the first example Nikolay can use 1 lemon, 2 apples and 4 pears, so the answer is 1 + 2 + 4 = 7.
In the second example Nikolay can use 3 lemons, 6 apples and 12 pears, so the answer is 3 + 6 + 12 = 21.
In the third example Nikolay don’t have enough pears to cook any compote, so the answer is 0.
题意：一个柠檬，两个苹果，四个梨可以构成一个水果拼盘，现在输入有a个柠檬，b个苹果，c个梨，问做完整的水果拼盘要用到几个水果。如果不能做直接输出0
题解：直接除每个水果需要个数后得到份数，取三种水果中的最小值作为最终能做的份数，输出最终份数乘每个水果需要的个数求和。
#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
int a,b,c;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
if(a<1||b<2||c<4)
{
printf("0
");
}
else
{
b=b/2;
c=c/4;
int num=min(a,min(b,c));
printf("%d
",num+num*2+num*4);
}
}
}