HDU 1787 GCD Again (欧拉函数)
来源:互联网 发布:南京未来网络待遇 编辑:程序博客网 时间:2024/05/18 01:21
Problem Description
Do you have spent some time to think and try to solve those unsolved problem after one ACM contest?
No? Oh, you must do this when you want to become a "Big Cattle".
Now you will find that this problem is so familiar:
The greatest common divisor GCD (a, b) of two positive integers a and b, sometimes written (a, b), is the largest divisor common to a and b. For example, (1, 2) =1, (12, 18) =6. (a, b) can be easily found by the Euclidean algorithm. Now I am considering a little more difficult problem:
Given an integer N, please count the number of the integers M (0<M<N) which satisfies (N,M)>1.
This is a simple version of problem “GCD” which you have done in a contest recently,so I name this problem “GCD Again”.If you cannot solve it still,please take a good think about your method of study.
Good Luck!
No? Oh, you must do this when you want to become a "Big Cattle".
Now you will find that this problem is so familiar:
The greatest common divisor GCD (a, b) of two positive integers a and b, sometimes written (a, b), is the largest divisor common to a and b. For example, (1, 2) =1, (12, 18) =6. (a, b) can be easily found by the Euclidean algorithm. Now I am considering a little more difficult problem:
Given an integer N, please count the number of the integers M (0<M<N) which satisfies (N,M)>1.
This is a simple version of problem “GCD” which you have done in a contest recently,so I name this problem “GCD Again”.If you cannot solve it still,please take a good think about your method of study.
Good Luck!
Input
Input contains multiple test cases. Each test case contains an integers N (1<N<100000000). A test case containing 0 terminates the input and this test case is not to be processed.
Output
For each integers N you should output the number of integers M in one line, and with one line of output for each line in input.
Sample Input
240
Sample Output
01
Author
lcy
Source
2007省赛集训队练习赛(10)_以此感谢DOOMIII
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int euler(int n){int res=n,i;for(i=2;i*i<=n;i++) {if(n%i==0) {res=res/i*(i-1);while(n%i==0) n=n/i;}}if(n>1) res=res/n*(n-1);return res;}int main(){int n;while(scanf("%d",&n)==1 && n) {printf("%d\n",n-1-euler(n));}return 0;}
0 0
- hdu 1787 GCD Again 欧拉函数
- HDU 1787 GCD Again 欧拉函数
- HDU 1787 GCD Again (欧拉函数)
- HDU 1787 GCD Again (欧拉函数)
- HDU 1787 GCD Again (欧拉函数)
- hdu GCD Again(欧拉函数)
- hdu 1787 GCD Again 欧拉函数小水水 数论
- HDU 1787 GCD Again 【欧拉函数模板】
- GCD Again HDU杭电1787【欧拉函数】
- hdu 1787 GCD Again (欧拉函数在线模板)
- HDOJ GCD Again 1787【欧拉函数】
- HDOJ-1787 GCD Again(欧拉函数)
- GCD Again(欧拉函数)
- hdoj 1787 GCD Again(欧拉函数)
- HDOJ 题目1787 GCD Again(欧拉函数)
- HDOJ 1787 GCD Again(欧拉函数)
- HDOJ 1787 GCD Again (欧拉函数)
- hdoj GCD Again 1787 (欧拉函数)
- lookupfile.vim插件详解
- iOS开发实践之GET和POST请求
- Android studio 项目上传至GitHub
- 最简单的Lambda入门教程
- struts2访问web元素
- HDU 1787 GCD Again (欧拉函数)
- Altium Designer阵列黏贴
- On-Demand Resources Essentials
- Linux常用命令总结A-Z
- A-Deeper-Understanding-of-Spark-Internals(Spark内核深入理解)
- 智能称体脂称实现(datasheet篇)
- iOS中的事件产生与传递
- 指针基础知识
- Androidx学习笔记(31)--- android-smart-image-view查看网络图片