1245 - Harmonic Number (II)    PDF (English)StatisticsForum 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

 

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PROBLEM SETTER: JANE ALAM JAN 嗯，给一个数n求它除以从1到n的整数解的和， n太大不能for循环，找规律… 1的个数是8-4；2的个数是4-2 如下 /*找规律： i 1 2 3 4 5 6 7 8 ans 8 4 2 2 1 1 1 1 i 1 2 3 4 5 6 7 8 9 10 ans 10 5 3 2 2 1 1 1 1 1*/ #include<cstdio> #include<cmath> #define LL long long LL h(LL n) { LL sum,m; sum=0; m=(LL)sqrt(n*1.0);//sqrt(n)对半分 for(int i=1;i<=m;i++) { sum+=n/i; sum+=(n/i-(n/(i+1)))*i; } if(m==n/m)//对应10那种情况 sum-=m; return sum; } int main() { int t,k=1; LL n; scanf("%d",&t); while(t--) { scanf("%lld",&n); printf("Case %d: %lld\n",k++,h(n)); } return 0; }