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 < 2).
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 ;
}