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HDU 5879 Cure

2024-08-14 05:30:05   217人阅读

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;}

　　

HDU 5879 Cure


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