<题目链接>
题目大意:
Given n, a positive integer, how many positive integers less than n are relatively prime to n? Two integers a and b are relatively prime if there are no integers x > 1, y > 0, z > 0 such that a = xy and b = xz.
解题分析:
其实只要看懂题目就会发现这道题是欧拉函数的模板题,即求小于n且与n互质的数的个数。
欧拉函数的基本性质: >>>
#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> using namespace std; //直接求解欧拉函数 int euler(int n){ int res=n,a=n; for(int i=2;i*i<=a;i++){ if(a%i==0){ res=res/i*(i-1);//先进行除法是为了防止中间数据的溢出 while(a%i==0) a/=i; } } if(a>1) res=res/a*(a-1); return res; } /* //筛选法打欧拉函数表 #define Max 1000001 int euler[Max]; void Init(){ euler[1]=1; for(int i=2;i<Max;i++) euler[i]=i; for(int i=2;i<Max;i++) if(euler[i]==i) for(int j=i;j<Max;j+=i) euler[j]=euler[j]/i*(i-1);//先进行除法是为了防止中间数据的溢出 } */ int main() { int n; while(scanf("%d",&n)!=EOF,n) { printf("%d ",euler(n)); } return 0; }
2018-07-30