HDU 5879 Cure

Cure

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1333    Accepted Submission(s): 440

Problem Description

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

 

Source

2016 ACM/ICPC Asia Regional Qingdao Online

 

 

 

解析：一定要注意这句话“The input file is at most 1M.”坑点所在。输入的长度可能达到1e6。这个极限为PI*PI/6，保留5位小数就是1.64493。n达到1000000左右以后，结果就不会再变了。

 

 

 

#include <cstdio>
#include <cstring>

const int MAXN = 1e6+5;
double sum[MAXN];
char s[MAXN];

void init()
{
    sum[0] = 0;
    for(int i = 1; i <= 1000000; ++i){
        sum[i] = sum[i-1]+1.0/(i*1.0*i);
    }
}

int main()
{
    init();
    while(~scanf("%s", s)){
        int len = strlen(s);
        if(len >= 7){
            printf("1.64493\n");
            continue;
        }
        int n = 0;
        for(int i = 0; i < len; ++i)
            n = n*10+s[i]-'0';
        printf("%.5f\n", sum[n]);
    }
    return 0;
}