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.
Example
1
12
3
2
6
8
1
2
3
1
2
3
-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.
题意给你一个数让你算出,这个数最多能由多少个合数组成,这里首先要明确什么是合数
合数的定义:
数学用语,指自然数中除了能被1和本身整除外,还能被其他的数整除的数。"0"“1”既不是质数也不是合数。
合数素数
折叠概念
除了2之外,所有的偶数都是合数。反之,除了2之外,所有的素数都是奇数。但是奇数包括了合数和素数。合数根和素数根的概念就是用来区分任何一个大于9的奇数属于合数还是素数。任何一个奇数都可以表示为2n+1(n是非0的自然数)。我们将n命名为数根。当2n+1属于合数时,我们称之为合数根;反之,当2n+1是素数时,我们称之为素数根。
折叠规律
任何一个奇数,如果它是合数,都可以分解成两个奇数的乘积。设2n+1是一个合数,将它分解成两个奇数2a+1和2b+1的积(其中a、b都属于非0的自然数),则有
2n+1=(2a+1)(2b+1)=4ab+2(a+b)+1=2(2ab+a+b)+1
可见,任何一个合数根都可以表示为"2ab+a+b",反之,不能表示为"2ab+a+b"的数根,就称为素数根。由此可以得到合数根表。判断一个大奇数属于合数还是素数,只需在合数根表中查找是否存在它的数根就知道了。
折叠合数根表
表中第一行表示a的取值,第一列表示b的取值,其余表示2ab+a+b
2ab+a+b | a=1 | a=2 | a=3 | a=4 | a=5 | a=6 | a=7 | a=8 | a=9 | a=10 | … | a=n |
b=1 | 4 | 7 | 10 | 13 | 16 | 19 | 22 | 25 | 28 | 31 | … | 1+3n |
b=2 | 7 | 12 | 17 | 22 | 27 | 32 | 37 | 42 | 47 | 52 | … | 2+5n |
b=3 | 10 | 17 | 24 | 31 | 38 | 45 | 52 | 59 | 66 | 73 | … | 3+7n |
b=4 | 13 | 22 | 31 | 40 | 49 | 58 | 67 | 76 | 85 | 94 | … | 4+9n |
b=5 | 16 | 27 | 38 | 49 | 60 | 71 | 82 | 93 | 104 | 115 | … | 5+11n |
b=6 | 19 | 32 | 45 | 58 | 71 | 84 | 97 | 110 | 123 | 136 | … | 6+13n |
b=7 | 22 | 37 | 52 | 67 | 82 | 97 | 112 | 127 | 142 | 157 | … | 7+15n |
b=8 | 25 | 42 | 59 | 76 | 93 | 110 | 127 | 144 | 161 | 178 | … | 8+17n |
b=9 | 28 | 47 | 66 | 85 | 104 | 123 | 142 | 161 | 180 | 199 | … | 9+19n |
b=10 | 31 | 52 | 73 | 94 | 115 | 136 | 157 | 178 | 199 | 220 | … | 10+21n |
…… | … | … | … | … | … | … | … | … | … | … | … | …… |
b=n | 1+3n | 2+5n | 3+7n | 4+9n | 5+11n | 6+13n | 7+15n | 8+17n | 9+19n | 10+21n | … | n^2+2n |
折叠意义
通过研究合数根表,对研究素数的规律会有深远的意义。
这道题主要是理解题意,还有几个特判的数1,2,3,5,7,11,
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { int q; int p; int i,j; int ans; int num; scanf("%d",&q); while(q--) { scanf("%d",&p); num = p / 4; ans = p % 4; if(ans == 0) printf("%d ",num); else if(ans == 1) { if(num > 1) printf("%d ",num-1); else printf("-1 "); } else if(ans == 2) { if(num > 0) printf("%d ",num); else printf("-1 "); } else { if(num > 2) printf("%d ",num-1); else printf("-1 "); } } return 0; }