cf Educational Codeforces Round 54 B. Divisor Subtraction_find the smallest prime divisor d-优快云博客

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本文详细解析了一种求解整数n通过不断减去其最小质因子直至变为0的算法，探讨了算法的实现方式及优化策略，通过实例展示了算法的具体运行过程。

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原题：
B. Divisor Subtraction
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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≤10^10).

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

Examples
input
5
output
1
input
4
output
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.

中文：
给你一个数n,让你将这个数走一遍如下的算法，要求如果n等于0就结束，否则找到n的最小质因子，然后用n减去该因子。不断的重复该过程，问你这个算法会重复多少次？

代码：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

ll n;

int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n)
    {
        ll ans=0,flag;
        while(n)
        {
            flag=0;
            for(ll i=2;i<=sqrt(n);i++)
            {
                if(n%i==0)
                {
                    if(i==2)
                    {
                        flag=-1;
                        ans+=n/i;
                        break;
                    }
                    n-=i;
                    ans++;
                    flag=1;
                    break;
                }
            }
            if(flag==-1)
                break;
            if(!flag)
            {
                ans++;
                break;
            }
        }
        cout<<ans<<endl;
    }

    return 0;
}