hdu2841 Visible Trees 容斥原理-优快云博客

本文介绍了一个基于容斥原理的问题：在一个m*n的网格中计算农民Sherlock能看见多少树木。当三棵树在一条直线上时，只能看见最近的一棵。通过预处理质因数分解和初始化相关数据结构，利用容斥原理高效地解决了该问题。

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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.

容斥原理裸题

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include<math.h>
 5 using namespace std;
 6 typedef long long ll;
 7 
 8 const int maxn=1e5+5;
 9 
10 int pnum[maxn],good[maxn];
11 
12 void init(){
13     memset(pnum,0,sizeof(pnum));
14     memset(good,0,sizeof(good));
15     for(int i=2;i<=100000;++i){
16         if(!pnum[i]){
17             pnum[i]=1;
18             for(int j=2;i*j<=100000;++j){
19                 if(!(j%i))good[i*j]=1;
20                 else pnum[i*j]++;
21             }
22         }
23     }
24 }
25 
26 int main(){
27     init();
28     int T;
29     scanf("%d",&T);
30     while(T--){
31         int n,m;
32         scanf("%d%d",&n,&m);
33         if(n>m)swap(n,m);
34         ll ans=0;
35         for(int i=2;i<=n;++i){
36             if(!good[i]){
37                 if(pnum[i]%2)ans+=(n/i)*((ll)m/i);
38                 else ans-=(n/i)*((ll)m/i);
39             }
40         }
41         printf("%lld\n",n*(ll)m-ans);
42     }
43     return 0;
44 }