2016 ACM/ICPC Asia Regional Qingdao Online 1002 Cure

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstdlib>
 5 #include<cstring>
 6 using namespace std;
 7 double a[1000005];
 8 char ch[1000005];
 9 int n;
10 int main()
11 {
12 
13     for(int i=1000000;i>=1;i--)
14      a[i]=a[i+1]+1.0/pow((double)i,2.0);
15 
16     while(~scanf("%s",&ch))
17     {
18         int len=strlen(ch);
19         int n=0;
20         for(int i=0;i<len;i++) {n=n*10+ch[i]-‘0‘; if (n>1000000) break;}
21 
22         if(n>=1000000) printf("%.5lf\n",a[1]);
23           else printf("%.5lf\n",a[1]-a[n+1]);
24     }
25     return 0;
26 }