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


[align=left]Problem Description[/align]
Given an integer n, we only want to know the sum of 1/k2 where k from 1 to n.

[align=left]Input[/align]
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.

[align=left]Output[/align]
The required sum, rounded to the fifth digits after the decimal point.

[align=left]Sample Input[/align]

1

2

4

8

15

[align=left]Sample Output[/align]

1.00000

1.25000

1.42361

1.52742

1.58044

[align=left]Source[/align]
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
);
}
return 0;
}