poj2478Farey Sequence

Farey Sequence

链接：http://poj.org/problem?id=2478

Time Limit: 1000MS		Memory Limit: 65536K

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 <= 10⁶). 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：给定一个数n，求在[1,n]这个范围内两两互质的数的个数。（转化为给定一个数n，比n小且与n互质的数的个数）,
要用筛法求欧拉，不然会T

#include<cstdio>
#include<cstdlib>
using namespace std;
#define ll long long 
const int M = 1000005;
ll c[M], eular[M];
void init(){
    eular[1] = 0;
    
    for(int i = 2; i <= M; i++)
        if(!eular[i])
            for(int j = i; j <= M; j+=i){
                if(!eular[j])eular[j] = j;
                eular[j] = eular[j] / i * (i - 1);
            }
    
}
void add(int x, ll del){
    while(x <= M){
        c[x] += del;
        x += x&-x;
    }
}
ll sum(int x){
    ll ret = 0;
    while(x>0){
        ret += c[x];
        x -= x&-x;
    }
    return ret;
}
int main(){
    int n;
    init();
    for(int i = 1; i <= M; i++)
        add(i, eular[i]);
    while(scanf("%d",&n) == 1){
        if(!n)break;
        printf("%lld
",sum(n));
    }
    
}