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HDOJ 2058 The sum problem

登高且赋 2022-05-14 阅读 62


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]

题目的意思:输入两个整数N,M。 N, M( 1 <= N, M <= 1000000000),如果在范围[1,M]内连续整数的和为N,按从小到大次序输出所有这样的连续段,当输入的M,N都为0时结束。

计算的思路:

不考虑子列的终点,而是考虑子列的起点和子列元素的个数,分别记为i,j。由等差数列求和公式,得(i+(i+j-1))*j/2==M ,即(2*i+j-1)*j/2==M(2式),故得i=(2*M/j-j+1)/2,将i,j代回2式,成立则[i,i+j-1]满足条件。注意j最小为1,而由2式,得(j+2*i)*j=2*M,而i>=1,故j*j<=(int)sqrt(2*M).

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();
if(m==0&&n==0){
return ;
}

int j =(int)Math.pow(2.0*m, 0.5);
for(j=j;j>0;j--){
int i;
i = (2*m/j-j+1)/2;
if(j*(j+2*i-1)/2==m){
System.out.println("["+i+","+(i+j-1)+"]");
}
}
System.out.println();
}
}
}



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