hdu1787——GCD Again(欧拉函数入门)
来源:互联网 发布:js修改css样式 生效 编辑:程序博客网 时间:2024/05/26 05:52
GCD Again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3583 Accepted Submission(s): 1575
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
题目大意:
给你一个整数N,求范围小于N中的整数中,与N的最大公约数大于1的整数的个数。
思路:
典型的欧拉函数变形。欧拉函数φ(N)是用来求小于N的整数中,与N的最大公约数为1的数的个数。
那么此题的答案ans = N - φ(N) - 1。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int oula(int n)
{
int res=1,i;
for(i=2; i*i<=n; i++)
if(n%i==0)//i是素数
{
n/=i;//因为要算ni-1次方,所以在这先除一下
res*=i-1;
while(n%i==0)
{
n/=i;//n越来越小
res*=i;
}
}
if(n>1)
res*=n-1;
return res;
}
int main()
{
int n;
while(cin>>n&&n)
{
cout<<n-oula(n)-1<<endl;
}
return 0;
}
1 0
- hdu1787——GCD Again(欧拉函数入门)
- hdu1787 GCD Again(数论:欧拉函数)
- HDU1787 GCD Again【欧拉函数】
- hdu1787 GCD Again 欧拉函数在线算法 待补完
- (hdu1787)GCD Again(欧拉函数)
- HDU1787——GCD Again
- 欧拉函数:HDU1787-GCD Again(欧拉函数的模板)
- HDU1787 GCD again 欧拉公式 的运用
- hdu1787-GCD Again
- hdu1787-GCD Again
- hdu1787 GCD Again
- HDU1787 GCD Again
- hdu1787(欧拉函数)
- 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)
- HDOJ GCD Again 1787【欧拉函数】
- Android Bitmap颜色、亮度、饱和度等变化
- 【JMeter 菜鸟实操之三】性能自动化集成方案实施(监控资源、分布式压测等)
- 进程的开始与终止(exit和_exit区别)
- 山东省第三届ACM省赛 Pick apples
- XAMPP安装常见问题以及解决?
- hdu1787——GCD Again(欧拉函数入门)
- Unity各类数据库的基本操作(五)-- PlayerPrefs
- 合并两个有序链表
- oracle 12C 创建用户失败 解决方案
- JS面向对象
- JS中面向对象简单入门
- mac android编译打包时OOM:java heap space的解决方法,亲测有效
- netstat命令(持续更新中)
- Git撤销&回滚操作