【数论】UVa 11526 - H(n)

What is the value this simple C++ function will 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 first line of input is an integer T (T ≤ 1000) that indicates the number of test cases. Each of the next T line will contain a single signed 32 bit integer n.

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

Sample Input
2
5
10

Sample Output
10
27

题意：求上述函数的返回值；

分析：简单讲就是求出一个tmp=[n/i]后计算一下这个数会出现多少次，方法就是n/tmp,求得的数是满足n/i==tmp的最大i值，然后继续更新i值即可。算法复杂度会由O(n)降为O(sqrt(n));

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
LL H(LL n)
{
    LL res = 0;
    LL pre = n, next_pos = n, tmp;
    for(LL i = 1; i <= n; i++)
    {
        res += n/i;
        tmp = n/i;
        if(tmp != pre)
        {
            next_pos = n/tmp;
            res += (next_pos-i)*tmp;
            pre = tmp;
            i = next_pos;
        }
    }
    return res;
}
int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        LL n; scanf("%lld", &n);
        printf("%lld\n", H(n));
    }
    return 0;
}