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
11
1
2
3
4
5
6
7
8
9
10
2147483647
Sample Output
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
思路： 找规律这件事，emmm.....
注意sqrt(n)这个数，数之间的差与后面数的个数。。。。。
写几个完整的例子，努力寻找规律！
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <queue>
#include <map>
#include <stack>
#include <deque>
#include <iostream>
using namespace std;
typedef long long LL;
const LL N = ; map<int, bool> check;
int prime[]; long long H( int n ) {
long long res = ;
for( int i = ; i <= n; i++ )
res = res + n / i;
return res;
} int main()
{
LL i, p, j, n, t, cnt = ;
LL sum; scanf("%lld", &t);
while(t--) {
sum = ;
scanf("%lld", &n);
for(i = ; i <= (LL)sqrt(n); i++) {
// cout << "i: " << i << endl;
sum += (n / i - n / (i + )) * i;
sum += n / i;
}
if(n / (LL)sqrt(n) == (LL)sqrt(n)) {
// sum -= (n / (LL)(sqrt(n)) - n / (LL)(sqrt(n) + 1)) * (i - 1);
sum -= (LL)sqrt(n);
}
printf("Case %lld: %lld\n", cnt ++, sum);
}
return ; //2 147 483 648
}