Visible Lattice Points
Time Limit:7000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Submit Status
Description
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
题目大意：从点(0,0,0)出发的直线可看到多少个点（只能看到第一个，后面的视为挡住了看不见）。
解题思路：求gcd(x,y,z)=1的点有多少个，F(n) 表示满足条件的 gcd(x,y,z)==n的 (x,y,z) 对数;G(n) 表示满足 n | gcd(x,y,z) 的(x,y,z)对数，即 gcd(x,y,z)%n==0 的(x,y,z) 对数;
由定义：G(n)=sigma(F(d)),F(n)=sigma(U(d/n)*G(d))
这题就是求F(1)。G(d)=(n/d)*(n/d)(n/d)。
当3个坐标为0时有0个点;
2坐标为0的时候可见点在三条坐标轴上一共3个；
1坐标为0的时候3*ans(ans=sigma(u(d)*(n/i)*(n/i)))；
坐标都不为0的时候ans=ans=sigma(u(d)*(n/i)*(n/i)*(n/i))
提示：提交代码时不能用__int64，只能用long long
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5
6 typedef __int64 LL;
7 const int maxn=1000005;
8 int prime[maxn],mu[maxn],num;
9 bool flag[maxn];
10
11 void init()
12 {
13 int i,j;num=0;mu[1]=1;
14 memset(flag,true,sizeof(flag));
15 for(i=2;i<maxn;i++)
16 {
17 if(flag[i])
18 {
19 prime[num++]=i;mu[i]=-1;
20 }
21 for(j=0;j<num&&prime[j]*i<maxn;j++)
22 {
23 flag[i*prime[j]]=false;
24 if(i%prime[j]==0)
25 {
26 mu[i*prime[j]]=0;break;
27 }
28 else mu[i*prime[j]]=-mu[i];
29 }
30 }
31 }
32
33 int main()
34 {
35 init();
36 int t,i,n;
37 scanf("%d",&t);
38 while(t--)
39 {
40 scanf("%d",&n);
41 LL ans=3;
42 for(i=1;i<=n;i++)
43 ans+=(LL)mu[i]*(n/i)*(n/i)*(n/i+3);
44 printf("%I64d\n",ans);
45 }
46 return 0;
47 }
spoj 7001 Visible Lattice Points莫比乌斯反演