Lightoj 1234 Harmonic Number (play table + tricks)

http://lightoj.com/volume_showproblem.php?problem=1234

Harmonic number Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%LLD &%llusubmit Status Practice Lightoj 1234

Description

In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers:

In this problem, you is given n, you have to find Hn.

Input

Input starts with an integer T (≤10000), denoting the number of test cases.

Each case is starts with a line containing an integer n (1≤n≤108).

Output

For each case, print the case number and the nth harmonic number. Errors less than 10-8 'll be ignored.

Sample Input

12

1

2

3

4

5

6

7

8

9

90000000

99999999

100000000

Sample Output

Case 1:1

Case 2:1.5

Case 3:1.8333333333

Case 4:2.0833333333

Case 5:2.2833333333

Case 6:2.45

Case 7:2.5928571429

Case 8:2.7178571429

Case 9:2.828968254

Case 10:18.8925358988

Case 11:18.9978964039

Case 12:18.9978964139

Main topic:

Request 1 + 1/3 + + + + 1/5 +...+ 1/n(1≤n≤108)

The harmonic progression part and, can use the formula, (alas, however I do not remember the formula high number did not learn-_-| | )

If the direct loop is bound to time out, then we open an array of 10^8/40 = 2.5 million to separate the

1 to 1/40 and, 1 to 1/80 and, 1 to 1/120 and, 1 to 1/160, and 、... 、 1 to 1/2500000 and

This saves time by looping up to 39 times per N

#include <stdio.h>#include<math.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespacestd;Const intN =2500010;Const intM = 1e8 +Ten; typedefLong Longll;DoubleA[n];intMain () {intT, N, p =0; Doubles =0;  for(inti =1; i < M; i++) {s+= (1.0/i); ifI +==0) A[i/ +] =s; } scanf ("%d", &t);  while(t--) {p++; scanf ("%d", &N); intx = N/ +; S=A[x];  for(inti = x * ++1; I <= N; i++) s+= (1.0/i); printf ("Case %d:%.10f\n", P, s); }    return 0;}

Lightoj 1234 Harmonic Number (play table + tricks)