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 <= 1000000). 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
2 3 4 5 0
Sample Output
1 3 5 9
思路很简单，就是个求和。
#include <iostream> #include <string.h> using namespace std; long long int phi[1000001]; int main() { long long int n,sum,i,j; for(i=1;i<=1000000;i++) phi[i]=i; for(i=2;i<=1000000;i++) if(phi[i]==i) for(j=i;j<=1000000;j+=i) phi[j]=phi[j]-phi[j]/i; while(cin>>n) { if(n==0) break; sum=0; for(int i=2;i<=n;i++) sum=sum+phi[i]; cout<<sum<<endl; } return 0; }开始的时候while放在前面（紧跟在声明后面）的时候，一直超时。改成这个样子就好了。
有个使用memset的写法，在while循环内能过，算法原理都一样。
#include <iostream> #include <string.h> using namespace std; long long int p[1000001]; int main() { long long int n,sum; while(cin>>n) { if(n==0) break; memset(p,0,sizeof(int)); for(int i=2;i<=1000000;i++) { if(p[i]==0) { for(int j=i;j<=1000000;j=j+i) { if(p[j]==0) p[j]=j; p[j]=p[j]-p[j]/i; } } } sum=0; for(int i=2;i<=n;i++) sum=sum+p[i]; cout<<sum<<endl; } return 0; }