Hdu2841 Visible Trees

Visible Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3131    Accepted Submission(s): 1387

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

21 12 3

 

Sample Output

15

 

Source

2009 Multi-University Training Contest 3 - Host by WHU

 

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题目的 意思是从(1,1)开始有个m*n的矩阵每个整点有一颗树 问从(0,0)看过去可以看到多少克树

思路：说白了就是求[1,m],[1,n]有多少对数互质(不互质的肯定会被挡住) 可以枚举m,在n中找有多少互质的数，可以先把枚举的m质因数的分解，然后利用容斥定理计数

防止超时分解的质因数可以先打个表

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<string>#include<queue>#include<stack>#include<map>#include<set>using namespace std;#define LL long longconst int inf=0x3f3f3f3f;vector<LL>p[100010];int pr[100010];int a,b,x,y,k;LL ans;void init(){    memset(pr,0,sizeof pr);    for(int i=1;i<=100005;i++) p[i].clear();    for(int i=2;i<=100005;i+=2)        p[i].push_back(2);    for(int i=3;i<=100005;i+=2)    {        if(!pr[i])        {            for(int j=i;j<=100005;j+=i)            {                pr[j]=1;                p[j].push_back(i);            }        }    }}void dfs(int no,int pos,int xx,int cnt){    if(pos>=p[no].size())    {        if(cnt==0)            return;        if(cnt%2)        {            ans+=y/xx;        }        else        {            ans-=y/xx;        }        return;    }    dfs(no,pos+1,xx*p[no][pos],cnt+1);    dfs(no,pos+1,xx,cnt);}int main(){    init();    int T;    int q=1;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&x,&y);        if(x>y) swap(x,y);        LL ans2=0;        for(int i=1; i<=x; i++)        {            ans=0;            dfs(i,0,1,0);            ans2+=y-ans;        }        printf("%lld\n",ans2);    }    return 0;}