HDU 2841 Visible Trees

 

Visible Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 535    Accepted Submission(s): 197

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
   
   
   
   

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

 

感觉出的不错。

题目是以前poj某道题的变种。POJ上面是n=n的特殊情况，所以直接求欧拉函数就OK了

这里n,m不一样，所以是转化为求1到n的每一个数，在1-m这段区间内所有互质的数的个数和

举个例子

n=5,m=10

答案

n=1时有10个与1互质

n=2时有5个与2互质

n=3时有7个与3互质

n=4时有5个与4互质

n=5时有8个与5互质

所以一共有35个

 

注意到，其实互质的本意就是指最大公约数为1

那么就是指没有公共的素因子。

---------

n=5,m=10的时候

ans=10/1+(10-10/2)+(10-10/3)+(10-10/2)+(10-10/5) （m-m/每一个数的质因子+m/两个质因子的积-m/三个质因子的积。。。。）

由于n的质因数个数不定。所以可以用容斥原理来计算重合部分

欧拉函数也要自己用快速预处理出来

 

我的代码：

#include<stdio.h>
#include<algorithm>
#include<vector>
#define READ(x) (scanf("%I64d",&x))
#define COUT(x) (printf("%I64d\n",x))

using namespace std;

typedef __int64 ll;

ll prime[100005];
bool flag[100005];
ll phi[100005];
vector<ll>link[100005];

void init()//得到素数以及欧拉函数值
{
    ll i,j,num=0;
    phi[1]=1;
    for(i=2;i<=100000;i++)
    {
        if(!flag[i])
        {
            prime[num++]=i;
            phi[i]=i-1;
        }
        for(j=0;j<num&&prime[j]*i<=100000;j++)
        {
            flag[prime[j]*i]=true;
            if(i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }
            else
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
        }
    }
    for(j=1;j<=100000;j++)//得到所有数包含的素因子
    {
        ll tmp=j;
        for(i=0;prime[i]*prime[i]<=tmp;i++)
        {
            if(tmp%prime[i]==0)
            {
                link[j].push_back(prime[i]);
                tmp=tmp/prime[i];
                while(tmp%prime[i]==0)
                    tmp=tmp/prime[i];
            }
            if(tmp==1)
                break;
        }
        if(tmp>1)
            link[j].push_back(tmp);
    }
}

ll dfs(ll x,ll b,ll now)//容斥原理
{
    ll i,res=0;
    for(i=x;i<link[now].size();i++)
        res=res+b/link[now][i]-dfs(i+1,b/link[now][i],now);
    return res;
}

int main()
{
    init();
    ll i,t,T,ans,n,m;
	READ(T);
	for(t=1;t<=T;t++)
	{
		ans=0;
		READ(n);
		READ(m);
		if(n>m)
			swap(n,m);
		for(i=1;i<=n;i++)
			ans=ans+m-dfs(0,m,i);
		COUT(ans);
    }
    return 0;
}