Codeforces Round #440 (Div. 2, based on Technocup 2018 Elimination Round 2) C. Maximum splitting

地址:

题目:

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
3
1
2
3
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.

 

思路:

  最小合数是4,所以用4去凑就行了

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define MP make_pair
 6 #define PB push_back
 7 typedef long long LL;
 8 typedef pair<int,int> PII;
 9 const double eps=1e-8;
10 const double pi=acos(-1.0);
11 const int K=1e6+7;
12 const int mod=1e9+7;
13 
14 
15 int main(void)
16 {
17     int n,x,ans;cin>>x;
18     while(x--)
19     {
20         scanf("%d",&n);
21         if(n%4==0)
22             ans=n/4;
23         else if(n%4==1)
24         {
25             if(n<9)
26                 ans=-1;
27             else if(n==9)
28                 ans=1;
29             else
30                 ans=(n-13)/4+2;
31         }
32         else if(n%4==2)
33         {
34             if(n<6)
35                 ans=-1;
36             else if(n==6)
37                 ans=1;
38             else
39                 ans=n/4;
40         }
41         else
42         {
43             if(n<15)
44                 ans=-1;
45             else
46                 ans=(n-15)/4+2;
47         }
48         printf("%d\n",ans);
49     }
50     return 0;
51 }

 

 

posted @ 2017-10-15 19:43  weeping  阅读(180)  评论(0编辑  收藏  举报