[CSP-S Simulation Test]: Divisors (Mathematics)

Title Description

Given $ m $ distinct positive integers $ a_1, a_2, ..., a_m $, $ 0 to the request for $ m $ $ $ k $ each calculation, a positive number in the interval $ [1, n] $ li a is an integer divisor of $ $ $ k $ exactly in number.

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Input Format

The first line contains two positive integers $ n, m $, respectively, and the size of the interval range $ A $ array.
The second line contains $ m $ distinct positive integers $ a_1, a_2, ..., a_m $, $ A $ array represents.

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Output Format

$ M + 1 $ output lines, each an integer, wherein the first output line $ I $ $ k = i $ answer.

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Sample

Sample input 1:

10 3
4 6 7

Sample output 1:

4
4
1
1

Sample input 2:

5 1
8

Sample Output 2:

2
3

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Data range and tips

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answer

To explain the meaning of problems (when the exam questions for half an hour before reading read ......), to this problem is to require $ [1, n] $ how much the number is $ a_1 \ sim a_m $ in $ 0 $ number about about the number, the number of $ 1 $ number of ......

Then consider how we should do.

We can enumerate only $ 1 \ sim \ sqrt {a_i} $ can be drawn about the number of all $ a_i $, and consider how storage.

My approach is to use the $ map $, storing each number about a divisor of several numbers $ a_1 \ sim a_i $ in.

Then we stored directly answer, set up an array $ sum [j] $ represent $ a_1 \ sim a_i $ is the number of divisors of which $ j $ number.

Whenever sweep times about a number just to $ sum [mp [j]] -; mp [j] ++; sum [mp [j]] ++ $ to.

Time complexity: $ \ Theta (m \ times \ sqrt {a_i}) ​​$.

Expectations score: $ 100 $ points.

Actual score: $ 100 $ points.

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Code time

#include<bits/stdc++.h>
using namespace std;
int n,m;
int sum[201];
map<int,int> mp;
int main()
{
	scanf("%d%d",&n,&m);
	sum[0]=n;
	for(int i=1;i<=m;i++)
	{
		int x;scanf("%d",&x);
		for(int j=1;j*j<=x&&j<=n;j++)
			if(!(x%j))
			{
				sum[mp[j]]--;
				mp[j]++;
				sum[mp[j]]++;
				if(j*j!=x&&x/j<=n)
				{
					sum[mp[x/j]]--;
					mp[x/j]++;
					sum[mp[x/j]]++;
				}
			}
	}
	for(int i=0;i<=m;i++)printf("%d\n",sum[i]);
	return 0;
}

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rp ++