Codeforces 872C Maximum splitting【思维】

C. Maximum splitting

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

You are given several queries. In the i-th query you are given a single positive integer ni.
You are to represent ni as
a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.

An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and
the integer itself.

Input

The first line contains single integer q (1 ≤ q ≤ 105) —
the number of queries.

q lines follow. The (i + 1)-th
line contains single integer ni (1 ≤ ni ≤ 109) —
the i-th query.

Output

For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.

Examples

input

1
12

output

3

input

2
6
8

output

1
2

input

31
23

output

-1
-1
-1

Note

12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.

8 = 4 + 4, 6 can't be split
into several composite summands.

1, 2, 3 are less than any composite number, so they do not have valid splittings.

题目大意：

问我们将n分成尽可能多的合数的加和，问最多可以分成多少个合数。

思路：

尽可能分4，那么对于n来讲，其有四种情况：

%4==0

%4==1

%4==2

%4==3
那么对其分类讨论即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
int n;
int t;scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
if(n%4==0)printf("%d\n",n/4);
else if(n%4==1)
{
if(n<9)printf("-1\n");
else printf("%d\n",(n-9)/4+1);
}
else if(n%4==2)
{
if(n<6)printf("-1\n");
else printf("%d\n",(n-6)/4+1);
}
else
{
if(n<=11)printf("-1\n");
else printf("%d\n",(n-9)/4+1);
}
}
}