LightOJ 1234 Harmonic Number(打表 + 技巧)
2015-11-05 14:36
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http://lightoj.com/volume_showproblem.php?problem=1234
Harmonic Number
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit 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 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
题目大意:
求1 + 1/2 + 1/3 + 1/4 + 1/ 5 +...+ 1/ n(1 ≤ n ≤ 108)
调和级数部分和,可以利用公式,(唉,然而我并不记得公式高数没学好-_-||)
如果直接循环的肯定会超时,那么我们开一个10^8/40 = 250万的数组用来分别存
1到1/40的和、1到1/80的和、1到1/120的和、1到1/160的和、... 、1到1/2500000的和
这样对于每一个n最多循环39次,节省了时间
Harmonic Number
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit 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 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
题目大意:
求1 + 1/2 + 1/3 + 1/4 + 1/ 5 +...+ 1/ n(1 ≤ n ≤ 108)
调和级数部分和,可以利用公式,(唉,然而我并不记得公式高数没学好-_-||)
如果直接循环的肯定会超时,那么我们开一个10^8/40 = 250万的数组用来分别存
1到1/40的和、1到1/80的和、1到1/120的和、1到1/160的和、... 、1到1/2500000的和
这样对于每一个n最多循环39次,节省了时间
#include<stdio.h> #include<math.h> #include<string.h> #include<stdlib.h> #include<algorithm> using namespace std; const int N = 2500010; const int M = 1e8 + 10; typedef long long ll; double a ; int main() { int t, n, p = 0; double s = 0; for(int i = 1 ; i < M ; i++) { s += (1.0 / i); if(i % 40 == 0) a[i / 40] = s; } scanf("%d", &t); while(t--) { p++; scanf("%d", &n); int x = n / 40; s = a[x]; for(int i = x * 40 + 1 ; i <= n ; i++) s += (1.0 / i); printf("Case %d: %.10f\n", p, s); } return 0; }
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