【AtCoder-4247】解题报告(水题)

原始题目

B - 105

  • Time limit : 2sec / Memory limit : 1000MB

  • Score: 200 points

Problem Statement

The number 105 is quite special - it is odd but still it has eight divisors. Now, your task is this: how many odd numbers with exactly eight positive divisors are there between 1 and N (inclusive)?

Constraints

  • N is an integer between 1 and 200 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print the count.

Sample Input 1

105

Sample Output 1

1

  • Among the numbers between 1 and 105, the only number that is odd and has exactly eight divisors is 105.

Sample Input 2

7

Sample Output 2

0

  • 1 has one divisor. 3, 5 and 7 are all prime and have two divisors. Thus, there is no number that satisfies the condition.

题目大意

给定n,求1到n恰好又8个因数的数字有多少个

集体思路

  • 水题,预处理一下因子个数,再求一下前缀和就行。

解题代码

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#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define mp make_pair
#define np next_permutation
#define pb push_back
#define all(x) (x.begin(),x.end())
#define rep(i,a,n) for(int i=a;i<n;++i)
#define per(i,a,n) for(int i=n-1;i>=a;--i)

#define fi first
#define se second
using namespace std;
const int maxn=1e4+5;
const int maxl=26;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int,int> pii;


int cnt[maxn],ans[maxn];
void solve(){
rep(i,1,300) cnt[i]++;
rep(i,2,300){
for(int j=1;j*i<=300;j++){
cnt[i*j]++;
}
}
int ccnt=0;ans[0]=0;
rep(i,1,300) {
if(cnt[i]==8 and (i&1)){
ans[i]=++ccnt;
}
else ans[i]=ccnt;
}
rep(i,1,300) {
// cout<<"ans["<<i<<"]="<<ans[i]<<endl;
}
}

int a,b;
int main(){
ios::sync_with_stdio(false);
solve();
int n;
while(cin>>n)
cout<<ans[n]<<endl;
}

收获与反思