light oj 1245 - Harmonic Number (II)【规律】

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

题意 可参考题目中所给的函数~ 这里就不多说了；

下面 模拟一下，找规律：

输入 n = 10 输出sum = 27；

i = 1 2 3 4 5 6 7 8 9 10

n/i = 10 5 3 2 2 1 1 1 1 1. 这几个数求和得 27；

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；

可以推出m=i的个数 为 (n/i -n/(i+1)) (1 <= i <=sqrt(n) )

利用两次for循环，统计所有个数，最后去掉重复的即可

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
int t;
long long sum;
scanf("%d", &t);
int ca = 1;
while(t--)
{
int n,m;
scanf("%d", &n);
sum = 0;
m = sqrt(n);
int i,j;
for(i = 1; i <= m; i++)
sum += n/i;
for(i = 1; i <= m; i++)
sum += ((n/i) - (n/(i+1)))*i;
if(m == n/m)
sum -= m;
printf("Case %d: %lld\n", ca++, sum);
}
return 0;
}