NO.2000：Gold Coins

Gold Coins
Time Limit:1000MS  Memory Limit:30000K
Total Submit:3769 Accepted:2442

Description
The king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On each of the next two days (the second and third days of service), the knight receives two gold coins. On each of the next three days (the fourth, fifth, and sixth days of service), the knight receives three gold coins. On each of the next four days (the seventh, eighth, ninth, and tenth days of service), the knight receives four gold coins. This pattern of payments will continue indefinitely: after receiving N gold coins on each of N consecutive days, the knight will receive N+1 gold coins on each of the next N+1 consecutive days, where N is any positive integer.

Your program will determine the total number of gold coins paid to the knight in any given number of days (starting from Day 1).

Input
The input contains at least one, but no more than 21 lines. Each line of the input file (except the last one) contains data for one test case of the problem, consisting of exactly one integer (in the range 1..10000), representing the number of days. The end of the input is signaled by a line containing the number 0.

Output
There is exactly one line of output for each test case. This line contains the number of days from the corresponding line of input, followed by one blank space and the total number of gold coins paid to the knight in the given number of days, starting with Day 1.

Sample Input

10
6
7
11
15
16
100
10000
1000
21
22
0

 

Sample Output

10 30
6 14
7 18
11 35
15 55
16 61
100 945
10000 942820
1000 29820
21 91
22 98

 

Source
Rocky Mountain 2004



Java实现代码：

/**//*
 * Main.java
 *
 * Created on 2006年9月13日, 下午8:10
 *
 * To change this template, choose Tools | Template Manager
 * and open the template in the editor.
 */

/** *//**
 *
 * @author jailu
 */
import java.util.*;
import java.lang.*;

public class Main {
    
    public Main() {
    }
    
    public static void main(String[] args) {
        int [] input = new int[21];
        int [] output = new int[21];
        
        Scanner cin = new Scanner(System.in);
        int Day = 0;
        for (int i = 0;i < 21;i ++)
        {
            if (cin.hasNext())
            {
                Day = cin.nextInt();
                
                if (Day != 0)
                {
                    input[i] = Day;
                }
                else
                {
                    break;
                }
            }
        }
        
        System.out.println();
        long start = System.currentTimeMillis();
        for (int i = 0;i < input.length;i ++)
        {
            if (input[i] == 0)
            {
                break;
            }
            else
            {
                double Temp = (Math.pow(1 + 8 * input[i],0.5) / 2) - 0.5; 
                for (int j = 1;j <= Temp;j ++)
                {
                    output[i] += j * j;
                }

                int intA = (int)Temp;
                output[i] += (input[i] - intA * (intA + 1) / 2) * (intA + 1);
                System.out.println(input[i] + " " + output[i]);
            }
        }
        System.out.println("运行结束，共耗时：" + (System.currentTimeMillis() - start) + "ms");
    } 
}

批定结果：

35

1582592

jailu

1380K

468MS

Java

1286B

2006-09-13 22:05:20