2016 ACM/ICPC Asia Regional Qingdao Online 1002Cure

CureTime Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0Problem DescriptionGiven an integer n, we only want to know the sum of 1/k2 where k from 1 to n.InputThere 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.OutputThe required sum, rounded to the fifth digits after the decimal point.Sample Input124815Sample Output1.000001.250001.423611.527421.58044

这道题其实就是精度问题，1/k^2，k越大，值就越小，我们打个表会发现，到12万左右值就卡在1.64493，
这里我打到了100000000试了一下，依旧是这个值，所以我们就可以打个12w左右的表。
注意这里范围没有给，应该是字符串输入。

#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cctype>#include<cmath>#include<ctime>#include<string>#include<stack>#include<deque>#include<queue>#include<list>#include<set>#include<map>#include<cstdio>#include<limits.h>#define fir first#define sec second#define fin freopen("/home/ostreambaba/文档/input.txt", "r", stdin)#define fout freopen("/home/ostreambaba/文档/output.txt", "w", stdout)#define mes(x, m) memset(x, m, sizeof(x))#define pii pair<int, int>#define Pll pair<ll, ll>#define INF 1e9+7#define Pi 4.0*atan(1.0)#define MOD 1000000007#define lowbit(x) (x&(-x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ls rt<<1#define rs rt<<1|1typedef long long ll;typedef unsigned long long ull;const double eps = 1e-12;const int maxn = 50010;using namespace std;//#define TIMEconst int N=100000+5;int a[N];inline int read(){    int x(0),f(1);    char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}double Sum[120010];int main(){  //fin;  //fout;    int j;    double long sum = 0;    Sum[0] = 0;    for(double long i = 1; i <= 120000; ++i){        sum += (double long)1/(i*i);        j = (int)i;        Sum[j] = sum;    }    string str;    int len;    int num;    while(cin >> str){        len = str.size();        if(len >= 7){            printf("1.64493\n");        }        else{            sscanf(str.c_str(), "%d", &num);            if(num > 120000){                printf("1.64493\n");            }            else{                printf("%.5lf\n", Sum[num]);            }        }    }    return 0;}