A - Visible Lattice Points SPOJ - VLATTICE 容斥原理/莫比乌斯反演

Posted 2020-09-17 dy blog

Consider a N*N*N lattice. One corner is at (0,0,0) and the opposite one is at (N,N,N). How many lattice points are visible from corner at (0,0,0) ? A point X is visible from point Y iff no other lattice point lies on the segment joining X and Y. 
 
Input : 
The first line contains the number of test cases T. The next T lines contain an interger N 
 
Output : 
Output T lines, one corresponding to each test case. 
 
Sample Input : 
3 
1 
2 
5 
 
Sample Output : 
7 
19 
175 
 
Constraints : 
T <= 50 
1 <= N <= 1000000

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int N=1e6+10;
 5 const int INF=0x3f3f3f3f;
 6 int cas=1,T;
 7 int c[N];
 8 void init()
 9 {
10     memset(c,0,sizeof(c));
11     c[1]=1;
12     for(int i=1;i<N;i++)
13     {
14         if(c[i]) for(int j=i<<1;j<N;j+=i) c[j]-=c[i];
15         c[i]+=c[i-1];
16     }
17 }
18 int p[N],vis[N],pn;
19 void getMu(int *mu)  //O(n)
20 {  
21     memset(vis,0,sizeof(vis));  
22     mu[1]=1;  
23     pn=0;  
24     for(int i=2; i<N; i++)  
25     {  
26         if(!vis[i]) { p[pn++]=i;mu[i]=-1; }  
27         for(int j=0; j<pn&&i*p[j]<N; j++)  
28         {  
29             vis[i*p[j]]=1;  
30             if(i%p[j]) mu[i*p[j]]=-mu[i];  
31             else { mu[i*p[j]]=0;break;}  
32         }  
33         mu[i]+=mu[i-1];
34     }  
35 }  
36 int main()
37 {
38     //freopen("1.in","w",stdout);
39     //freopen("1.in","r",stdin);
40     //freopen("1.out","w",stdout);
41     //init();
42     getMu(c);
43     scanf("%d",&T);
44     while(T--)
45     {
46         int n;
47         scanf("%d",&n);
48         LL ans=0;
49         //for(int i=1;i<=n;i++) if(c[i]) ans+=c[i] * ((LL) (n/i+1)*(n/i+1)*(n/i+1)-1);
50         for(int i=1;i<=n;)
51         {
52             int x=n/i;
53             int y=n/x;
54             ans+=(c[y]-c[i-1]) * ((LL) (n/i+1)*(n/i+1)*(n/i+1)-1);
55             i=y+1;
56         }
57         printf("%lld\n",ans);
58     }
59     //printf("time=%.3lf\n",(double)clock()/CLOCKS_PER_SEC);
60     return 0;
61 }

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