Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46844 Accepted Submission(s): 21489
1 #include <bits/stdc++.h> 2 using namespace std; 3 inline double gcd(int n) 4 { 5 int sum=1; 6 for(int i=1;i<=n;i++) 7 sum*=i; 8 return sum; 9 } 10 int main() 11 { 12 cout<<‘n‘<<" "<<‘e‘<<endl; 13 cout<<"- -----------"<<endl; 14 cout<<0<<" "<<1<<endl; 15 cout<<1<<" "<<2<<endl; 16 cout<<2<<" "<<2.5<<endl; 17 double sum=2.5; 18 for(int i=3;i<=9;i++) 19 { 20 sum+=(1.0/(double)gcd(i)); 21 printf("%d %.9lf\n",i,sum); 22 } 23 return 0; 24 }
HDU 1012 u Calculate e【暴力打表,水】
原文:http://www.cnblogs.com/ECJTUACM-873284962/p/7214781.html