Maximum splitting
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.
思路如下
- 题意:这一题让判断一个数 最多能够被分成几个
合数
相加,能被分成几个答案就是 输出 几 ,但是如果某个数没法被分成几个合数相加 那么直接输出 - 1 - 既然是让求最多可以被分成几个合数之和,那么我们考虑,每一份分的越小越好,而最小的合数是
4
,那么我们先分尽量多的4,最后对 余数 进行一一讨论即可
题解如下
#include <bits/stdc++.h>
using namespace std;
int solve(int x)
{
int t = x/4; //最多分成t个4
int rest = x%4; //求出剩余部分
if (rest==0) //刚好整除4 t就是最大值返回
return t;
if (rest==1) //余1 咋办 取出一个4 发现4+1=5 不是合数 那就取出两个 4*2+1=9 是合数 那么就9了
{
if (t>=2)
{
t-=2;
}else
return -1;
t++;
return t;
}
if (rest==2) //同理 余2 取出一个4 发现4+2=6是合数 那就取出一个就行
{
if (t>=1)
{
t--;
}else return -1;
t++;
return t;
}
if (rest==3) //同理 取出3个4 3*4+3=15是合数,这里切记 15 还可以 分成两个合数 6与9.因此 在下面的代码中 t += 2;
{
if (t>=3)
{
t-=3;
}else return -1;
t+=2;
return t; //返回t
}
}
int main()
{
int q;
scanf("%d",&q);
for (int i = 1;i <= q;i++)
{
int x;
scanf("%d",&x);
printf("%d\n",solve(x));
}
return 0;
}