ZOJ3175 Number of Containers[数学题]

Number of Containers

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Time Limit: 1 Second      Memory Limit: 32768 KB

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For two integers m and k, k is said to be a container of m if k is divisible by m. Given 2 positive integers n and m (m < n), the function f(n, m) is defined to be the number of containers of m which are also no greater than n. For example, f(5, 1)=4, f(8, 2)=3, f(7, 3)=1, f(5, 4)=0...

Let us define another function F(n) by the following equation:

Now given a positive integer n, you are supposed to calculate the value of F(n).

Input

There are multiple test cases. The first line of input contains an integer T(T<=200) indicating the number of test cases. Then T test cases follow.

Each test case contains a positive integer n (0 < n <= 2000000000) in a single line.

Output

For each test case, output the result F(n) in a single line.

Sample Input

2
1
4

Sample Output

0
4

题目就是要我们求:n*(1/1+1/2+1/3+……1/(n-2)+1/(n-1))-n
由于题目的数据太大，显然暴力是会tle，于是建立如上的图，发现：
x*y=n这条直线是关于y=x对称，于是
1+2+1就是结果了。
能想到根号n就几乎是知道答案了。


code:

#include<stdio.h>
#include<math.h>

int main()
{
    int t;
    scanf("%d",&t);
    int n;
    while(t--)
    {
        int i;
        int t;
        long long sum=0;
        scanf("%d",&n);
        t=(int)(sqrt(n)+0.000001);
        for(i=1;i<=t;i++)
            sum+=(n/i);
        sum*=2;
        sum=sum-t*t-n;
        printf("%lld\n",sum);
    }
    return 0;
}