Educational Codeforces Round 54 (Rated for Div. 2) B. Divisor Subtraction

观察易得

1.质数无1和自身外的因子 且只有本身既质又因 按题意直接一步减自身至零

2.若N是偶数则一直减2直到0

所有质数都是奇数 奇数减奇数易得偶数 再回到条件2 一步到位

所以操作次数不会太多

线筛打表 结合1 2 暴力模拟即可

/*
    Zeolim - An AC a day keeps the bug away
*/

//pragma GCC optimize(2)
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <sstream>
#include <map>
#include <ctime>
#include <vector>
#include <fstream>
#include <list>
#include <iomanip>
#include <numeric>
using namespace std;
typedef long long ll;

const int MAXN = 1e7 + 10;

bool check[MAXN] = {0};
 
int prime[MAXN] = {0};
 
int pos = 0;
 
int flag;

void initprime(int len)
{
	for(int i = 2; i < len; i++)
	{
		if(!check[i])
			prime[pos++] = i;
			
		for(int j = 0; j < len && i * prime[j] < len; j++)
		{
			check[i * prime[j] ] = 1;
			
			if(i % prime[j] == 0)
				break;
		}
	}
}

bool isprime(ll x)
{
    if(x < 2)
        return false;
    for(ll i = 2; i * i <= x; i++)
        if(x % i == 0)
            return false;
    return true;
}

int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);     cout.tie(0);
    //freopen("D://test.in", "r", stdin);
    //freopen("D://test.out", "w", stdout);
	//double start, stop;
	//start = clock();
	//stop = clock();
	//printf("%lf", (stop - start) / CLK_TCK);
	
	
    initprime(MAXN - 10);
	
	ll n, count = 0;

    scanf("%lld", &n);
    
    
	
	if(n > prime[pos - 1] && isprime(n) || prime[lower_bound(prime, prime + pos, n) - prime] == n)
        printf("1
");
        
    else
    {
        while(n)
        {
           	if(n > prime[pos - 1] && isprime(n) || prime[lower_bound(prime, prime + pos, n) - prime] == n)
            {
                ++count;
                break;
            }
            else if(n % 2 == 0)
            {
                count += n / 2;
                break;
            }
            else
            {
                for(ll i = 0; i < pos; i++)
                {
                    if(n % prime[i] == 0)
                    {
                    	n -= prime[i], ++count;
                    	break;
					}
                        
                }
            }
                
        }

        printf("%lld
", count);
    }

    return 0;
}