POJ 2478-Farey Sequence(筛选法求欧拉函数) - c++编程基础

Farey Sequence Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2478 Appoint description:

Description

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are
F2 = {1/2}
F3 = {1/3, 1/2, 2/3}
F4 = {1/4, 1/3, 1/2, 2/3, 3/4}
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5}

You task is to calculate the number of terms in the Farey sequence Fn.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10 ⁶). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn.

Sample Input

Sample Output

题意：给定一个数n，求小于或等于n的数中两两互质组成的真分数的个数。

思路：这个博客有关于这道题的推理---->点击打开链接

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             using namespace std; typedef long long LL; const int inf=0x3f3f3f3f; const double pi= acos(-1.0); LL phi[1000010]; LL res[1000010]; void Euler() { int i,j; memset(phi,0,sizeof(phi)); phi[1]=1; for(i=2;i<=1000010;i++) { if(!phi[i]) { for(j=i;j<=1000010;j+=i) { if(!phi[j]) phi[j]=j; phi[j]=phi[j]/i*(i-1); } } } } int main() { int n; Euler(); memset(res,0,sizeof(res)); res[1]=res[2]=1; for(int i=3;i<1000010;i++) res[i]=res[i-1]+phi[i]; while(~scanf("%d",&n)){ if(!n) break; printf("%lld\n",res[n]); } return 0; }