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.
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\n",sum(n)); } }