标签：pos ini mem logs main stdout ++ space time
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 }
A - Visible Lattice Points SPOJ - VLATTICE 容斥原理/莫比乌斯反演
标签：pos ini mem logs main stdout ++ space time
原文地址：http://www.cnblogs.com/cdyboke/p/6828137.html
