The sum problem

Posted 2020-10-26 ixummer的博客

Problem Description

Given a sequence 1,2,3,......N, your job is to calculate all the possible sub-sequences that the sum of the sub-sequence is M.

 

 

Input

Input contains multiple test cases. each case contains two integers N, M( 1 <= N, M <= 1000000000).input ends with N = M = 0.

 

 

Output

For each test case, print all the possible sub-sequence that its sum is M.The format is show in the sample below.print a blank line after each test case.

 

 

Sample Input

20 10 50 30 0 0

 

 

Sample Output

[1,4] [10,10] [4,8] [6,9] [9,11] [30,30]

 思路：

1.由高斯求和公式文字表述：和=(首项 + 末项）x项数 /2可知：

区间[a,b]的各项和且b=a+i-1

 

2.转化该式子求出a，由a即可推导出b

     (a+b)*i/2 = m

     (a+b)*i = 2*m

     (a+ a+i-1)*i = 2*m

      2*a = 2*m/i + 1 – i

               a = m/i –(i-1)/2

 

3.限定循环次数，条件需满足a>=1，即m/i –(i-1)/2>=1，化简可得i <= (int)Math.sqrt(2 * m)

 

AC代码：

import java.util.Scanner;

public class Main {
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            //暴力法，但会LTE
//            if (m == 0 && n == 0) {
//                break;
//            }
//            for (int i = 1; i <= m; i++) {
//                int sum = 0;
//                for (int j = i; j <= m; j++) {
//                    sum += j;
//                    if (sum == m) {
//                        System.out.println("[" + i + "," + j + "]");
//                    }
//                }
//            }
//            System.out.println();
            //高斯公式法
            if(n == 0 && m == 0){
                break;
            }
            int a,b;
            for(int i = (int)Math.sqrt(2 * m); i > 0; i--){
                a = m / i - (i - 1) / 2;
                b = a + i - 1;
                if((a + b) * i == 2 * m){
                    System.out.println("[" + a + "," + b +"]");
                }
            }
            System.out.println();
        }
    }
}