Technocup 2018
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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.
题意
给出n组询问,问一个数最多由多少复合数组成(复合数是大于一的素数)
题解
最基础的复合数就是4,6,9,然后对m==1||m==2||m==3||m==5||m==7||m==11的时候进行特判。
当m%4等于0的时候说明自然m/4就是最大个数
当m%4等于2的时候那么可以将一个4和余下的2组合成6,那么也是m/4个
当m%4等于1的时候则需要让出两个4和余下的1组成9,所以是m/4+1个
当m%4等于3的时候则需要3个4和余下的3组成一个6一个9,所以也是m/4-1个
CODE
#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <cmath>using namespace std;int main(){ int n,m; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&m); if(m==1||m==2||m==3||m==5||m==7||m==11) cout << "-1" << endl; else { int x=m%4; int y=m/4; if(x==2||x==0) cout << y << endl; else cout << y-1 << endl; } } return 0;}
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