zoj 3175 Number of Containers （nbut1375） 计算n /1+n/2+n/3+n/4....+n/n

http://acm.nbut.cn/Problem/view.xhtml?id=1375

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3216

hnust_xiehonghao

[1375] Guess The Number

时间限制: 1000 ms　内存限制: 65535 K

问题描述

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).

输入

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.

输出

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

样例输入

2
1
4

样例输出

0
4

提示

无

来源

ZOJ 3175 Number of Containers

 

 

题意：

求n*(1/1+1/2+1/3+……1/(n-2)+1/(n-1))-n
的值   

由于n非常大  所以不能直接暴力

 

画图  可以用   横坐标表示i      从改点画一条垂直的线   这条线上的所有整数点的个数就是 n/i 

那么n/1+n/2+n/3+……n/(n-2)+n/(n-1)+n/n  可以表示为i*(n/i)=n这条线

答案就是这条线与坐标轴围成的面积内的整数点的个数

画一条x=y的线与x*y=n相交  可以知道 面积关于x=y对称 

我们求n/1+n/2+n/3+……  只求到k=sqrt(n)处（1个梯形）     之后乘以2 （得到2个梯形的面积 其中有一个正方形的区域是重复的）  减去重复的区域k*k个 即可

 

#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((double)n);
         for(i=1;i<=t;i++)
             sum+=(n/i);
         sum*=2;
         sum=sum-t*t-n;
         printf("%lld\n",sum);
     }
     return 0;
 }

此外 我们可以用这种方法快速求n/1+n/2+n/3+......n/n

参考 http://www.cnblogs.com/XBWer/archive/2012/08/26/2657568.html