POJ 2478Farey Sequence(筛选法求欧拉函数)

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下面是小编收集整理的POJ 2478Farey Sequence(筛选法求欧拉函数)，本文共2篇，供大家参考借鉴，希望可以帮助到有需要的朋友。

篇1：POJ 2478Farey Sequence(筛选法求欧拉函数)

Farey SequenceTime Limit:1000MSMemory Limit:65536KB64bit 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 <= 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

23450

Sample Output

1359

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，

POJ 2478Farey Sequence(筛选法求欧拉函数)

，

思路：这个博客有关于这道题的推理---->点击打开链接

#include#include#include#include#include#include#include #include#include#include#includeusing 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;}

篇2：POJ 2407Relatives(求一个整数的欧拉函数值)

RelativesTime Limit:1000MSMemory Limit:65536KB64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2407 Appoint description:

Description

Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x >1, y >0, z >0 such that a = xy and b = xz.

Input

There are several test cases. For each test case, standard input contains a line with n <= 1,000,000,000. A line containing 0 follows the last case.

Output

For each test case there should be single line of output answering the question posed above.

Sample Input

7120

Sample Output

题意：给一个数n，求出不大于n且与n互素的数的个数，

POJ 2407Relatives(求一个整数的欧拉函数值)

，

#include#include#include#include#include#include#include #include#include#include#includeusing namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const double pi= acos(-1.0);int Euler(int n){ int m=floor(sqrt(n+0.5)); int ans=n; for(int i=2;i<=m;i++){ if(n%i==0){ans=ans/i*(i-1);while(n%i==0){ n/=i;} } } if(n>1) ans=ans/n*(n-1); return ans;}int main(){ int n; while(~scanf(“%d”,&n)){ if(!n) break; printf(“%d\n”,Euler(n)); } return 0;}

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