HDU 5879 Cure （数学）

大体题意：

给你一个数字k  求1 + 1/4 + 1/9 .... + 1/(k^2)?

思路：

忘记队友怎么推得了，这个式子有个极限是 pi*pi / 6，如果只取前5位的话，那么k = 1000000的时候就可以达到了！

因此我们判断那个数字如果大于1000000 直接输出pi*pi/6的前五位！

否则 就打表计算！

坑：

注意 用字符串读取整数，会很长的= =！

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
const int maxn = 1000000 + 10;
double sum[maxn];
void init(){
	sum[1] = 1.0;
	for (int i = 2; i < maxn; ++i){
		sum[i] = sum[i-1] + 1.0/((double)i*(double)i);
	}
}
char s[maxn];
int main(){
	init(); 
	while(scanf("%s",s) == 1){
		if (strlen(s) > 6){
			printf("%.5lf\n",sum[maxn-10]);
			continue;
		}
		int v = 0;
		for (int i = 0; s[i]; ++i)v = v*10 + s[i] - 48;
		printf("%.5lf\n",sum[v]);
	}
	return 0;
}

Cure

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 423    Accepted Submission(s): 144

Problem Description

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

 

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