hdoj Visible Trees 2841 （容斥原理）

Visible Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2209    Accepted Submission(s): 906

Problem Description

There are many trees forming a m * n grid, the grid starts from (1,1). Farmer Sherlock is standing at (0,0) point. He wonders how many trees he can see.

If two trees and Sherlock are in one line, Farmer Sherlock can only see the tree nearest to him.

Input

The first line contains one integer t, represents the number of test cases. Then there are multiple test cases. For each test case there is one line containing two integers m and n(1 ≤ m, n ≤ 100000)

Output

For each test case output one line represents the number of trees Farmer Sherlock can see.

Sample Input

2
1 1
2 3

Sample Output

1 5

#include<stdio.h>
#include<string.h>
#include<math.h>
#define N 100010
#define ll long long
int p[N];
int q[N];
int k;
void getp(int n)
{
	int i,j;
	k=0;
	for(i=2;i*i<=n;i++)
	{
		if(n%i==0)
		{
			p[k++]=i;
			while(n%i==0)
				n/=i;
		}
	}
	if(n>1)
		p[k++]=n;
}
int solve(int n)
{
	int i,j,kk,t=0;
	ll sum=0;
	q[t++]=-1;
	for(i=0;i<k;i++)
	{
		kk=t;
		for(j=0;j<kk;j++)
			q[t++]=p[i]*q[j]*-1;
	}
	for(i=1;i<t;i++)
		sum+=n/q[i];
	return sum;
}
int main()
{
	int t,n,m,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		ll ans=n;
		for(i=2;i<=m;i++)
		{
			getp(i);
			ans+=n-solve(n);
		}
		printf("%lld\n",ans);
	}
	return 0;
}