[CSP-S Simulation Test]: metaphysical questions / c (math)

Topic Portal (internal title 40)

---

Input Format

The first line: two positive integers $ n $, $ m $.

---

Output Format

The first line: an integer value representing the expression.

---

Sample

Sample input 1:

4 5

Sample output 1:

0
Sample Input 2:

799 8278

Sample Output 2:

-11

---

Data range and tips

---

answer

For each $ i $, its contribution is its ability to answer a few numbers with $ 1 \ sim m $ is obtained by multiplying a perfect square.

Obviously we need to enumerate the $ i $, then $ \ Theta (1) $ count contributions.

So this time there are two cases:

　　$ \ Alpha.i $ is a perfect square, this time we only required a $ 1 \ sim m $ has several perfect square, and this number is $ \ sqrt m $.

　　$ \ Beta.i $ is not a square, is well known, perfect square $ \ times $ $ $ = perfect square perfect squares. So consider the decomposition $ i $, will be broken down into $ p_1 \ times p_2 \ times ... \ times q $, where $ p $ is a perfect square, $ q $ non-perfect square, we request that $ q \ times j $ a few are a perfect square, and this number is the $ \ sqrt {\ frac {m} {q}} $.

Pretreatment out of every $ I $ $ q $, to $ \ Theta (1) $ calculated contribution.

Time complexity: $ \ Theta (n) $.

Expectations score: $ 100 $ points.

Actual score: $ 100 $ points.

---

Code time

#include<bits/stdc++.h>
using namespace std;
int n;
long long m;
int cnt[10000001];
bool pfs[10000001];
int ans;
int main()
{
	scanf("%d%lld",&n,&m);
	for(int i=1;i*i<=n;i++)
	{
		for(int j=1;i*i*j<=n;j++)
			cnt[i*i*j]=j;
		pfs[i*i]=1;
	}
	int sqr=sqrt(m);
	for(int i=1;i<=n;i++)
		if(pfs[i])
		{
			if(sqr&1)ans--;
			else ans++;
		}
		else
		{
			if((int)sqrt(m/cnt[i])&1)ans--;
			else ans++;
		}
	printf("%d",ans);
	return 0;
}

---

rp ++