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spoj 7001. Visible Lattice Points GCD问题 莫比乌斯反演
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 else29 {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