B. Divisor Subtraction_what is the smallest prime divisor of 523 + 717?-优快云博客

本文详细解析了一种特殊算法，该算法通过不断减去输入整数n的最小素数因子直至n变为0，探讨了算法的执行过程，并提供了一种高效的方法来确定算法完全执行所需的步数。

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B. Divisor Subtraction

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer number n

. The following algorithm is applied to it:

if n=0

, then end algorithm;
find the smallest prime divisor d
of n
;
subtract d
from n and go to step 1
Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer n
(2≤n≤1010
).

Output

Print a single integer — the number of subtractions the algorithm will make.

Examples
Input
Copy
```
5
```
Output
Copy
```
1
```
Input
Copy
```
4
```
Output
Copy
```
2
```
Note

In the first example 5
is the smallest prime divisor, thus it gets subtracted right away to make a 0
.

In the second example 2

is the smallest prime divisor at both steps.

题意：给一个n，和一个算法，求这个算法能执行多少次。算法内容：

1. 如果n等于0，算法结束。

2. 找到n最小的素数因子d

3. n-=d，之后回到步骤1
分析：对于输入的n有3种情况：
1. 素数-------------输出1
2. 偶数-------------输出n/2
3. 奇数或合数-------------减去最小的素因子，变为偶数，再执行偶数的计算方法并加1

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

bool IsPrime(long long num)	//num为我们要判断的数
{
	for (long long i = 2; i <= sqrt(num); i++)		//最优化的判断方式
	{
		if (num%i == 0)
		{
			return false;
		}
	}
	return true;
}
long long fi(long long n){
    for(long long i = 2;i < n;i++){
        if(n % i == 0)
            return i;
    }
}

int main()
{
    long long n;
    while(cin>>n){
        long long t = 0;
        if(n % 2 == 0)
            cout<<n/2<<endl;
        else{
            if(IsPrime(n))
                cout<<1<<endl;
            else{
                t = n-fi(n);
                cout<<t/2+1<<endl;
            }
        }
    }
    return 0;
}