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| Time Limit: 3000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Description
In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers:
In this problem, you are given n, you have to find Hn.
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each case 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 will 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: 2.450
Case 7: 2.5928571429
Case 8: 2.7178571429
Case 9: 2.8289682540
Case 10: 18.8925358988
Case 11: 18.9978964039
Case 12: 18.9978964139
Source
/** 题意：如题 做法：因为n为1e8比较大，然后可以进行打表，每100个存储，然后在进行剩余的 **/ #include <iostream> #include <stdio.h> #include <string.h> #include <cmath> #include <algorithm> #include <queue> const int maxn = 1e8 + 110; using namespace std; double mmap[maxn/100+10]; void init() { mmap[0] = 0; mmap[1] = 1; double sum = 1.0; for(int i=2;i<=maxn;i++) { sum += 1.0 / (i*1.0); if(i % 100 == 0) mmap[i/100] = sum; } } int main() { int Case = 1; int T; scanf("%d",&T); init(); while(T--) { long long n; scanf("%lld",&n); int res = n/100; double tot = mmap[res]; for(int i=res*100+1;i<=n;i++) { tot += (1.0)/(i*1.0); } printf("Case %d: %.10f\n",Case++,tot); } return 0; }
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原文地址：http://www.cnblogs.com/chenyang920/p/4474730.html
