2016 ACM/ICPC Asia Regional Qingdao Online（Cure）

题目地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=5879

Cure

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1156 Accepted Submission(s): 391

Problem Description

Given an integer n, we only want to know the sum of1/

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

124815

Sample Output

1.000001.250001.423611.527421.58044

Source

2016 ACM/ICPC Asia Regional Qingdao Online

题意分析：计算1~1/

K²的和。(k从2到n)。

注意这题没给n的范围，可能是高精度问题，后来验证果然如此。此题要求四舍五入精确到第5位。分析问题发现，随着k的逐渐增大，1/

K²越来越小趋近于0，至少小数位第5位基本保持不变。注意测试数据可能达到1M。

^{源代码如下：}

#include<cstdio>#include<cstring>#define N 150000    //150000之后的结果均为1.64493using namespace std;double a[N];int main(void){    int i;    char num[1000000];  //输入的数字用字符串形式保存    a[1]  = 1.00000;    for(i = 2; i < N; i++)  //打表保存前150000个的结果        a[i] = a[i-1]+1.0/i/i;    while(gets(num) != NULL)    {        int len = strlen(num),now=0;        if(len <= 6 )        {            //转化数字            i = 0;            while(num[i])            {                now = now*10+num[i]-'0';i++;//把字符串转化成数字            }            if(now < N) printf("%.5f\n",a[now]);            else puts("1.64493");        }        else            puts("1.64493");    }    return 0;}