HDU 5879---cure

Posted 2020-09-26 kimsimple

 

 

Cure

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 293    Accepted Submission(s): 96

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

 

 

大数据量，想到打表预处理+数学|规律

 

规律：k大到一定程度，保留五位小数就不变了

 

坑：n没有给出范围，意思就是默认无限大。

RE原因，输入过大

 

 

#include "string"
#include "cstdio"
#include "iostream"
using namespace std;
#define MAX 120005
#define LL long long
double a[MAX]={0.0,1.0};
void init()
{
    for(int i=2;i<MAX;i++)
    {
        a[i]=a[i-1]+1.0/i/i;
    }
}
int main()
{
    init();
    int n,len;
    string s;
    while(cin>>s)
    {
        len=s.length();
        if(len>=7){
            printf("1.64493\\n");
        }else
        {
            n=0;
            for(int i=0;i<len;i++){
                n=n*10+s[i]-\'0\';
                if(n>120000)
                {
                    n=120000;
                    break;
                }
            }
            printf("%.5f\\n",a[n]);
        }
    }
    return 0;
}

 

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