湘潭 A simple problem

A simple problem
Accepted : 30		Submit : 303
Time Limit : 15000 MS		Memory Limit : 655360 KB

Problem Description

There is a simple problem. Given a number N. you are going to calculate
N%1+N%2+N%3+...+N%N.

Input

First line contains an integer T, there are T(1≤T≤50) cases. For each
case T. The length N(1≤N≤10¹²).

Output

Output case number first, then the answer.

Sample Input

1
5

Sample Output

Case 1: 4

Source

daizhenyang

 1 import java.io.*;

 2 import java.awt.*;

 3 import java.math.BigInteger;

 4 import java.util.Scanner;

 5

 6 public class Main {

 7

 8     public static void main(String[] args) {

 9         Scanner cin = new Scanner(System.in);

10         int T = cin.nextInt();

11         for(int t=1;t<=T;t++)

12         {

13             long n =cin.nextLong();

14             BigInteger m =solve(n);

15             System.out.println("Case "+t+": "+m);

16         }

17     }

18     static BigInteger solve(long n)

19     {

20         BigInteger sum = BigInteger.ZERO;

21         long k = (long)Math.sqrt(n*1.0);

22         if(k*k!=n) k++;

23         long y = n;

24         long a1,d=1,x;

25         while(y>=k+1)

26         {

27             a1=n%y;

28             x=(y-a1+d)/(d+1);

29             sum = sum.add(get(a1,x,d));

30             y=y-x;

31             d++;

32         }

33         for(int i=2;i<=y;i++)

34         {

35             sum=sum.add(BigInteger.valueOf(n%i));

36         }

37         return sum;

38     }

39     static BigInteger get(long a1,long x,long d)

40     {

41         BigInteger sum = BigInteger.ZERO;

42         BigInteger a = BigInteger.valueOf(a1);

43         BigInteger cur = BigInteger.valueOf(d);

44         BigInteger an = a.add(BigInteger.valueOf(x-1).multiply(cur));

45

46         an = an.add(a);

47         sum = an.multiply(BigInteger.valueOf(x));

48         sum = sum.divide(BigInteger.valueOf(2));

49         return sum;

50     }

51 }

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时间： 2024-08-12 00:56:48

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