题目大意:求1到n-1与n不互质数的个数
题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1787
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!
这题是欧拉函数的模板题,答案是n-1-φ(n),φ(n)是欧拉函数即1到n中与n互质的数个数。
欧拉函数:点击打开链接
要注意下result/i*(i-1)那里,刚开始我写的result*(i-1)/i,WA了,后来看了别人的代码才知道这样写溢出了。
#include <cstdio> #include <iostream> #include <cstring> using namespace std; int main(){ int n; while(scanf("%d",&n) != EOF && n){ int t = n; int result = n; //计算欧拉函数 //找n的所有质因数 for(int i = 2;i*i <= n;i++) { if(n % i == 0) { result = result / i * (i-1); //找到质因数 } while(n % i == 0) n /= i; } if(n != 1){ result = result / n * (n-1); } printf("%d\n",t - 1 -result); } return 0; }