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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
2
3
4
5
0
Sample Output
1
3
5
9

题意:

欧拉函数+前缀和.暴力打表.
数组不小心越界了.
RE了一次.

代码:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e6+10000;typedef long long LL;LL phi[maxn];void phi_table(int n,LL phi[]){    for(int i=2;i<=n;i++) phi[i]=0;    phi[1]=1;    for(int i=2;i<=n;i++) if(!phi[i])        for(int j=i;j<=n;j+=i)    {        if(!phi[j]) phi[j]=j;        phi[j] = phi[j]/i*(i-1);    }    for(int i=3;i<n;i++)        phi[i]+=phi[i-1];}int main(){    phi_table(maxn-100,phi);    int n;    while(cin>>n&&n)    {       printf("%lld\n",phi[n]);    }    return 0;}