Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.
Input
There are multiple cases.
For each test case, there is a single line, containing a single positive integer n.
The input file is at most 1M.
Output
The required sum, rounded to the fifth digits after the decimal point.
Sample Input
1
2
4
8
15
Sample Output
1.00000
1.25000
1.42361
1.52742
1.58044

求级数的前n项和，只要精确到五位小数，打表发现（其实百度发现）n到1e6就会收敛接近一个值，所以就可以了 注意读入用字符串然后一位一位计算

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=1e6+5;
char a[N];
double sum[N];
void init(){
    for(int i=1;i<N;i++){
        sum[i]=sum[i-1]+(double)1/i/i;
    }
    // for(int i=1;i<1000;i++){
   
    // printf("%.5lf\n",sum[i]);
    // }
}
int main(void){
    init();
    while(~scanf("%s",a)){
        int l=strlen(a);
        if(l>=7){
            printf("%.5lf\n",sum[1000000]);
        }
        else{
            int n=0;
            for(int i=0;i<l;i++){
                n=n*10+a[i]-'0';
            }
            printf("%.5lf\n",sum[n]);
        }
    }
    return 0;
}