HDU 2841 Visible Trees（容斥定理）

Visible Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1951    Accepted Submission(s): 792

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

 

         题意：有一个n*m的矩阵，左下角坐标为(1,1)，右上角坐标为(n,m);你现在的位置的坐标是(0,0),输出现在你能看到几个点.

         思路：你将矩阵仔细研究一下会发现如果两个点A(x1,y1),B(x2,y2)如果x2,y2有一个公约数k，使得x2/k == x1 && y2/k == y1 这样的话就两个点就会形成一个线段y=kx，正好(0,0)在这条线段中，故在这条线段上的矩阵上的点就只能看到一个。所以我们就得出结论如果坐标(x,y)中x,y互质，那么就能过被看到，如果我们将矩阵的一边固定住，假设是n，那么题意就转化成求1--m与区间[1,n]求互质的个数的题了，这样的话就是一个很简单的容斥定理的问题了，具体看代码吧。

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#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<vector>#include<queue>#include<stack>#include<map>#define N 1100using namespace std;int n,m,k;int prime[N],num[N];int t;__int64 IEP(int pn){   /// [n,m]区间求与k互质的个数    int pt = 0;    __int64 s = 0;    num[pt++] = -1;    for(int i=0;i<t;i++){        int l = pt;        for(int j=0;j<l;j++){            num[pt++] = num[j]*prime[i]*(-1);        }    }    for(int i=1;i<pt;i++){        s += pn/num[i];    }    return s;}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        __int64 sum = 0;        for(int i=1;i<=n;i++){            memset(prime,0,sizeof(prime));            memset(num,0,sizeof(num));            int pk = sqrt(i);            int nn = i;            t = 0;            for(int j=2;j<=pk;j++){                if(nn%j == 0){                    prime[t++] = j;                    while(nn%j == 0){                        nn = nn / j;                    }                }            }            if(nn!=1){                prime[t++] = nn;            }            sum += m - IEP(m);        }        printf("%I64d\n",sum);    }    return 0;}