1245 - Harmonic Number (II)（规律题）

1245 - Harmonic Number (II)

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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：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #define mem(x,y) memset(x,y,sizeof(x))
 using namespace std;
 typedef long long LL;
 const int INF=0x3f3f3f3f;

 int main(){
     int T;
     int n,cnt=;
     LL res;
     scanf("%d",&T);
     while(T--){
         res=;
         scanf("%d",&n);
         int m=sqrt(n);
         for(int i=;i<=m;i++)res+=n/i;
         for(int i=;i<=m;i++)
             res+=(n/i-n/(i+))*i;
             if(m==n/m)res-=m;
         printf("Case %d: %lld\n",++cnt,res);
     }
     return ;
 }