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最新推荐文章于 2022-03-02 13:12:04 发布

原创最新推荐文章于 2022-03-02 13:12:04 发布 · 263 阅读

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本文探讨了1234-HarmonicNumber问题，并提供了一种高效的求解方法。通过优化循环范围和避免重复计算，该方法显著提高了代码效率。

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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

题解：先在大脑里面模拟一下，当i取某个值时n/i可能等于n/(i+1),然后需要把出现这种情况的i的范围确定；设小数：k,若存在：int(x/kx)==int(n/(k+1));那么x/k<x/(k+1)+1;解方程可得k约大于等于根号下n，那么也就是说整数i取1到n的0.5次时，n/i！=n/(i+1);最后ret+=n/i;ret+=(n/i-n/(i+1))*i;跑1到根号n的氛围循环就行了，如果i*i==n就要多减一个i.

#include<stdio.h>
#include<cstdio>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long ll;
int main()
{
    int t;
    while(scanf("%d",&t)!=EOF)
    {
        int kase=0;
        while(t--)
        {
            ll i,n,ret=0;
            scanf("%lld",&n);
            for(i=1;i*i<=n;i++)
            {
                ret+=n/i;
                ret+=(n/i-n/(i+1))*i;
            }
            if((i-1)==n/(i-1)) ret-=(i-1);
            printf("Case %d: %lld\n",++kase,ret);
        }
    }
}