【数论】UVa 11526 - H(n)

What is the value this simple C++ function will return?

 long long H(int n)

 {
 　　long long res = ;
 　　for( int i = ; i <= n; i=i+ )

 　　{
 　　　　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 = ;
     LL pre = n, next_pos = n, tmp;
     for(LL i = ; 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 ;
 }