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GCD迅速水过#include<iostream> using namespace std; int gcd(int i,int j) { int k; while(i%j) { k=j; j=i%j; i=k; } return j; } int main() { int n,k=1,i,a[1001]; a[1]=3,a[2]=5; for(n=3;n<=1000;n++) { for(i=1;i<n;i++) if(gcd(n,i)==1)k++; a[n]=k*2+3; } cin>>n; for(i=0;i<n;i++) { cin>>k; cout<<i+1<<" "<<k<<" "<<a[k]<<endl; } } Followed by: Post your reply here: |
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