Cattleya number and guard of honor

Guard of honor this topic

Theoretically

Euler need to use the function

But if this title was changed to a rectangular square

Euler function it is not easy to use

When the GCD abscissa and ordinate of a point will be blocked is greater than 1

So we can determine the number of horizontal and vertical (not greatest common divisor) is the number i of Convention coordinates

f[i] = (n / i) * (m / i)f[i]=(n/i)∗(m/i)

Since some point be counted twice, we're going to look at weight

minus

 

 

The complexity is then O (nlogn) of

#include<iostream>
#include<cstdio>
using namespace std;
int py[10000002];
int main()
{
    freopen("matrix.in","r",stdin);
    freopen("matrix.out","w",stdout);
    int n,m;
    ios_base::sync_with_stdio(false);
    cin>>n>>m;
    if(n==1&&m == 
        cout) {1<<0;
        return 0;
    }
    if(n==1||m==1) {
        cout<<1;
        return 0;
    }
    int ans=(n-1)*(m-1);
    n--;m--;
    int l=min(n,m);
    for(int i=l;i>=2;i--)
    {
        py[i]=(n/i)*(m/i);
        for(int j=2* i; j <= l; j + = i) 
        py [i] - = py [j]; 
        years - = py [i]; 
    } 
    Cout << years + 2 ; 
}

Cattleya number:

So that h (0) = 1, h (1) = 1, catalan a number satisfying the recurrence formula:

h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2)

Two recursive formula:
H (n-) = H (. 1-n-) * (. 4-* n-2) / (n-+. 1);

Through 项公 formula:

h(n)=(2n!)/n!*(n+1)!