HDU2841：Visible Trees（莫比乌斯函数）

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本文探讨了一个可视树的问题，即从特定位置观察一个由树木组成的网格，可以看见多少棵树。通过数学方法，特别是莫比乌斯反演的应用，提供了高效的算法解决方案。

Visible Trees

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3078 Accepted Submission(s): 1360

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

Sample Output

1
5

Source

2009 Multi-University Training Contest 3 - Host by WHU

题意：人站在(0,0)点，有个n*m的树林，左下角在(1,1)处，问人可以看见多少棵树。

思路：gcd(i,j)不为1的点都不能看见，直接容斥即可，可以使用莫比乌斯反演。

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# define ll long long
using namespace std;
const int maxn = 1e5;
short mu[maxn+3];
int prime[maxn+3];
bool bo[maxn+3];

void get_table()
{
    mu[1] = 1;
    memset(bo, 0, sizeof(bo));
    for(int i=2; i<=maxn; ++i)
    {
        if(!bo[i])
        {
            prime[++prime[0]] = i;
            mu[i] = -1;
        }
        for(int j=1; j<=prime[0]&&prime[j]*i<=maxn; ++j)
        {
            bo[i*prime[j]] = 1;
            if(i % prime[j] == 0)
            {
                mu[i*prime[j]] = 0;
                break;
            }
            else
                mu[i*prime[j]] = -mu[i];
        }
    }
    for(int i=1; i<=maxn; ++i)
        mu[i] += mu[i-1];//前缀和，用于分块。
}

int main()
{
    get_table();
    int t, n, m, j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        if(n > m) swap(n, m);
        ll ans = 0;
        for(int i=1; i<=n; i=j+1)
        {
            j = n/(n/i);//分块优化。
            ans += (ll)(mu[j]-mu[i-1])*(n/i)*(m/i);
        }
        printf("%lld\n",ans);
    }
    return 0;
}