HDOJ 1787 GCD Again (欧拉函数)
来源:互联网 发布:c2c网络购物须知要点 编辑:程序博客网 时间:2024/06/05 14:38
GCD Again
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3169 Accepted Submission(s): 1400
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
题意:
你求和n不互质且小于n的数的个数。欧拉函数的对立,但是要注意1也不符合,所以要减去一个1。
代码:
#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<cstdlib>using namespace std;const int n=7000000;int main(){ int n,i,temp; while(scanf("%d",&n)!=EOF&&n) { int nn=n; temp=n; for(i=2;i*i<=n;i++) { if(n%i==0) { while(n%i==0) n=n/i; temp=temp/i*(i-1); } if(n<i+1) break; } if(n>1) temp=temp/n*(n-1); printf("%d\n",nn-temp-1); } return 0;}
0 0
- HDOJ GCD Again 1787【欧拉函数】
- HDOJ-1787 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)
- HDOJ 1787 GCD Again (欧拉函数)
- hdoj 1787 GCD Again (欧拉函数模板 )
- hdu 1787 GCD Again 欧拉函数
- HDU 1787 GCD Again 欧拉函数
- HDU 1787 GCD Again (欧拉函数)
- GCD Again(欧拉函数)
- HDU 1787 GCD Again (欧拉函数)
- 【杭电oj】1787 - GCD Again(欧拉函数)
- 杭电1787GCD Again(欧拉函数)
- HDU 1787 GCD Again (欧拉函数)
- hdoj GCD 2588 (欧拉函数)
- cpu调度算法
- HDOJ 2504 又见GCD (水题)
- 关于wamp修改www目录的问题
- QtOpenCV: error while loading shared libraries: libopencv_core.so.3.2: cannot open shared object fil
- 批量更改图片大小程序
- HDOJ 1787 GCD Again (欧拉函数)
- java.net.SocketTimeoutException: connect timed out
- HDOJ 2035 人见人爱A^B (快速幂)
- Shader自学笔记1.1 Hello Shader
- linux lcd设备驱动剖析一
- http://stackoverflow.com/questions/39998330/selenium-common-exceptions-webdriverexception-message-se
- Group Shifted Strings
- linux lcd设备驱动剖析二
- linux lcd设备驱动剖析三