UVa 11526 H(n) (数论)

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 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                      Output for Sample Input

2 5 10

10 27

怎么计算sum{ [n/i] }？(1<=i<=n)(n<=2147483647)

n太大，硬算肯定不行，我们先观察一个例子，看能否得出一些结论。

当n=20时，和式展开为

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

注意到后面相同的数太多，不妨化简下：

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

也许，我们可以

这样，复杂度就从O(n)降为O(√n)了。

完整代码：

/*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;
}

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