GCD Again
Time limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 2673 Accepted Submission (s): 1123
Problem Descriptiondo you had spent some time to think and try to solve those unsolved problem after one ACM contest?
No? Oh, you must does this if you want to become a "Big cattle".
Now you'll find the This problem are so familiar:
The greatest common divisor GCD (A, b) of both 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 is 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 was a simple version of the problem "GCD" which you had done with 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!
Inputinput contains multiple test cases. Each test case is contains an integers N (1<n<100000000). A test case containing 0 terminates, the input and this test are not processed.
Outputfor each integers N-should output the number of integers M in one line, and with one line of output for each Lin E in input.
Sample Input240
Sample Output01 Test Instructions: The number of 2 to n-1 and N does not coprime, because the Euler function is calculated with n coprime number of numbers so with N-el (n) can be, because more than one so also minus 1
#include <stdio.h> #include <string.h>int el (int n) {int i;int ans=n;for (i=2;i*i<=n;i++)//i*i for increased computational efficiency { if (n%i==0) ans=ans/i* (i-1), while (n%i==0) n/=i;} if (n>1)//In order to avoid a situation where there is no operation to 1 ans=ans/n* (n-1); return ans;} int main () {int n,m,j,i;while (scanf ("%d", &m), m) {printf ("%d\n", M-el (m)-1);} return 0;}
Hdoj 1787 GCD Again "Euler function"
