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1245 - Harmonic Number (II)(规律题)

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1245 - Harmonic Number (II)
Time Limit: 3 second(s) Memory Limit: 32 MB

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

Output for Sample Input

11

1

2

3

4

5

6

7

8

9

10

2147483647

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,自己有思路,却写不出来,看了答案,倒是挺简单的;

先看两个例子
1.
n = 10    sqrt(10) = 3     10/sqrt(10) = 3
i        1   2   3         4   5   6   7   8   9   10
n/i    10  5   3         2   2   1   1   1   1    1
 
m =  n/i
sum += m;
m = 1的个数10/1-10/2 = 5;
m = 2的个数10/2-10/3 = 2;
m = 3的个数10/3-10/4 = 1;
 
2.
n = 20     sqrt(20) = 4     20/sqrt(20) = 5
i        1   2   3   4       5   6   7   8   9   10   11   12   13   14   15   16   17   18   19   20
n/i    20  10 6   5       4   3   2   2   2    2     1     1     1     1     1     1     1     1    1    1
 
m =  n/i
sum += m;
m = 1的个数20/1-20/2 = 10;
m = 2的个数20/2-20/3 = 4;
m = 3的个数20/3-20/4 = 1;
m = 4的个数20/4-20/5 = 1;
...
m = i的个数20/i - 20/(i + 1)(1<= i <= sqrt(n))
 
这样我们可以得出:sqrt(n)之前的数我们可以直接用for循环来求
sqrt(n)之后的sum += (n/i - n/(i + 1)) * i;
当sqrt(n) = n / sqrt(n)时(如第一个例子10,sum就多加了一个3),sum多加了一个sqrt(n),减去即可;
无语,这里输出%I64d竟然wa。。。
代码:
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #define mem(x,y) memset(x,y,sizeof(x))
 7 using namespace std;
 8 typedef long long LL;
 9 const int INF=0x3f3f3f3f;
10 
11 int main(){
12     int T;
13     int n,cnt=0;
14     LL res;
15     scanf("%d",&T);
16     while(T--){
17         res=0;
18         scanf("%d",&n);
19         int m=sqrt(n);
20         for(int i=1;i<=m;i++)res+=n/i;
21         for(int i=1;i<=m;i++)
22             res+=(n/i-n/(i+1))*i;
23             if(m==n/m)res-=m;
24         printf("Case %d: %lld\n",++cnt,res);
25     }
26     return 0;
27 }

 

1245 - Harmonic Number (II)(规律题)

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原文地址:http://www.cnblogs.com/handsomecui/p/4946231.html

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