UVa 11526 H (n) (number theory)

11526-h (N)

Time limit:5.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=2521

What is the value of this simple C + + function would return?

Long long H (int n) {

Long long res = 0;

for (int i = 1; I <= n; i=i+1) {

Res = (res + n/i);

}

return res;

}

Input
The the "a" of input is a integer t (T <= 1000) that indicates the number of test cases. Each of the next T-line would contain a single signed bit integer n.

Output
For the each test case, output would be a single line containing H (n).

Sample input Output for sample input

2 5 10

10 27

How do I calculate sum{[n/i]}? (1<=i<=n) (n<=2147483647)

n too big, hard to calculate certainly not, we first observe an example, see whether can draw some conclusions.

When n=20, the and type expands to

20+10+6+5+4+3+2+2+2+2+1+1+1+1+1+1+1+1+1+1

Note that the same number of the following is too many, may wish to simplify the following:

20+10+6+5+1* (20-10) +2* (10-6) +3* (6-5) +4* (5-4)

= (20+10+6+5) + (20+10+6+5) -4*4

=2 (20+10+6+5) -4*4

Maybe, we can

This article URL address: http://www.bianceng.cn/Programming/sjjg/201410/45351.htm

Thus, the complexity is reduced from O (n) to O (√n).

Complete code:

/*0.206s*/
    
#include <cstdio>  
#include <cmath>  
typedef long Long ll;  
    
inline ll ans (ll N)  
{  
    ll r = 0, m = sqrt (n), I;  
    for (i = 1; I <= m; ++i) R + = n/i;  
    Return (r << 1)-M * m;  
}  
    
int main ()  
{  
    int t;  
    ll N;  
    scanf ("%d", &t);  
    while (t--)  
    {  
        scanf ("%lld", &n);  
        printf ("%lld\n", ans (n));  
    }  
    return 0;  
}