I was trying to solve problem ‘1234 - Harmonic Number’, I wrote the following code
long long H( int n )
{
long long res = 0;
for( int i = 1; i <= n; i++ )
res = res + n / i;
return res;
}
Yes, my error was that I was using the integer divisions only. However, you are given n, you have to find H(n) as in my code.
Input
Input starts with an integer T (≤ 1000), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n < 2^31).
Output
For each case, print the case number and H(n) calculated by the code.
Sample Input
11
1
2
3
4
5
6
7
8
9
10
2147483647
Sample Output
Case 1: 1
Case 2: 3
Case 3: 5
Case 4: 8
Case 5: 10
Case 6: 14
Case 7: 16
Case 8: 20
Case 9: 23
Case 10: 27
Case 11: 46475828386
题意:
根据题中所给代码,可知该题是求 n/1+n/2+n/3+…+n/n 的和。
思路:
基于题目上给出的数据过大,使用题中所给代码超时是必然的,所以我们需要更改一下这个求解方法,降低运行时间,多算几项找规律吧!
总结公式就是2 * [ (n/1+n/2+…+n/sqrt(n) ] - [ sqrt(n)*sqrt(n) ] ,自己也动手推算一下吧!
代码:
#include<stdio.h>
#include<math.h>
int main()
{
int t;
long long n;
int k=0;
scanf("%d",&t);
while(t--)
{
k++;
int i;
long long s,s1=0;
scanf("%lld",&n);
s=(int)sqrt(n);
for(i=1;i<=s;i++)
s1=s1+n/i;
s1=s1*2;
s1=s1-s*s;
printf("Case %d: %lld\n",k,s1);
}
return 0;
}