标签：style blog color os io for ar 问题 div
SPOJ Problem Set (classical)7001. Visible Lattice PointsProblem code: VLATTICE |
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
题意：GCD(a,b,c)=1, 0<=a,b,c<=N ;
莫比乌斯反演，十分的巧妙。
GCD(a,b)的题十分经典。这题扩展到GCD(a,b,c)加了一维，但是思想却是相同的。
设f(d) = GCD(a,b,c) = d的种类数 ；
F(n) 为GCD(a,b,c) = d 的倍数的种类数, n%a == 0 n%b==0 n%c==0。
即 ：F(d) = (N/d)*(N/d)*(N/d);
则f(d) = sigma( mu[n/d]*F(n), d|n )
由于d = 1 所以f(1) = sigma( mu[n]*F(n) ) = sigma( mu[n]*(N/n)*(N/n)*(N/n) );
由于0能够取到，所以对于a,b,c 要讨论一个为0 ，两个为0的情况 (3种).
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<cstdlib> 5 using namespace std; 6 7 typedef long long LL; 8 const int maxn = 1000000+3; 9 bool s[maxn]; 10 int prime[maxn],len = 0; 11 int mu[maxn]; 12 void init() 13 { 14 memset(s,true,sizeof(s)); 15 mu[1] = 1; 16 for(int i=2;i<maxn;i++) 17 { 18 if(s[i] == true) 19 { 20 prime[++len] = i; 21 mu[i] = -1; 22 } 23 for(int j=1;j<=len && (long long)prime[j]*i<maxn;j++) 24 { 25 s[i*prime[j]] = false; 26 if(i%prime[j]!=0) 27 mu[i*prime[j]] = -mu[i]; 28 else 29 { 30 mu[i*prime[j]] = 0; 31 break; 32 } 33 } 34 } 35 } 36 37 int main() 38 { 39 int n,T; 40 init(); 41 scanf("%d",&T); 42 while(T--) 43 { 44 scanf("%d",&n); 45 LL sum = 3; 46 for(int i=1;i<=n;i++) 47 sum = sum + (long long)mu[i]*(n/i)*(n/i)*3; 48 for(int i=1;i<=n;i++) 49 sum = sum + (long long)mu[i]*(n/i)*(n/i)*(n/i); 50 printf("%lld\n",sum); 51 } 52 return 0; 53 }
spoj 7001. Visible Lattice Points GCD问题 莫比乌斯反演
标签：style blog color os io for ar 问题 div
原文地址：http://www.cnblogs.com/tom987690183/p/3933335.html
