LightOJ 1245 Harmonic Number (II) <思维---数学>

最新推荐文章于 2019-03-20 16:40:25 发布

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最新推荐文章于 2019-03-20 16:40:25 发布

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本文介绍了一个谐数计算问题，通过优化算法实现高效计算。对于给定的整数n，使用特定算法计算H(n)，并提供了代码实现及样例输入输出。优化方案包括预计算前sqrt(n)个值和按n/i分组求和。

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Harmonic Number (II)

Time Limit: 3000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status uDebug

Description

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 < 2³¹).

Output

For each case, print the case number and H(n) calculated by the code.

Sample Input

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

Source

Problem Setter: Jane Alam Jan

先找前sqrt( n ) 个---剩下的数i ，n / i 的值一定属于前sqrt（n）个数---相同的n / i 在一起加

代码：

#include<cstdio>
#define __int64 long long
void solve(int ca)
{
    __int64 n;scanf("%lld",&n);
    __int64 k=0,s=0,ji=0,shu=n;
    __int64 i;
    for (i=1;i*i<=n;i++)
    {
        s+=n/i;
        k++;
    }
    i--;
    __int64 chu=i;
    for (i;i>0;i--)
    {
        s+=i*(n/i-chu);
        chu=n/i;
    }
    printf("Case %d: %lld\n",ca,s);
}
int main()
{
    int t;scanf("%d",&t);
    for (int i=1;i<=t;i++)
        solve(i);
    return 0;
}