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spoj 7001 Visible Lattice Points莫比乌斯反演

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 : 




 
Sample Output : 

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 }