杭电oj —— 2058 (超时优化)

超时代码：思路易想到，但是数据量太大

package com.demo3;
import java.util.Scanner;
/*
 * Given a sequence 1,2,3,......N, 
 * your job is to calculate all the possible sub-sequences(子序列) that the sum of the sub-sequence is M.
 */
public class HDU_oj2058 {
	public static void main(String[] args) {
		Scanner sn = new Scanner(System.in);
		long n,m,sum,start,end=0;
		while(sn.hasNext()) {
			n = sn.nextLong();
			m = sn.nextLong();
			if(n == 0 && m == 0) {
				break;
			}
			for(long i = 1;i <= n;i++) {
				sum = 0; 
				start = i; //首先从i开始
				if(start <= m) { //如果刚好遍历到这个值等于m
					end = start;
					while(sum < m) { //遍历序列
						sum += end;
						end++;
					}
					if(sum == m) {
						end = end-1;
						System.out.println("[" + start + "," + end +"]");
					} else {
						continue;
					}
				}
			}
			System.out.println();
		}
		sn.close();
	}
}

正确代码如下：数据规模太大，需要整数分解，思路见代码注释

package com.demo3;
import java.util.Scanner;
/*
 * 整数分解：（首项+末项）*项数 /2 
 */
public class HDU_oj2058_1 {
	public static void main(String[] args) {
		Scanner sn = new Scanner(System.in);
		int len, temp;
		while (sn.hasNext()) {
			int N = sn.nextInt();
			int M = sn.nextInt();
			if(N == 0 && M == 0)
				break;
			/*
			 * len = (int) Math.sqrt(M * 2.0);
			 * 思路分析：
			 * 等差数列，公差为1，假设首项是a,项数是b；  等差数列求和公式：(a1+an)*n / 2;
			 * 则 M = (a + (a+b-1))*b /2; 
			 * 所以：2M = (a + (a+b-1))*b = (2a + b - 1)*b 
			 * 且:a>=1
			 * 所以：2M >= b^2
			 * 即：项数b <= 根号2M
			 */
			len = (int) Math.sqrt(M * 2.0);
			
			/*
			 * temp = M - (i * i + i) / 2;
			 * 思路分析：
			 * M = (a + (a+b-1))*b /2; 
			 * 2M = 2ab + b^2 -b
			 * 2ab = 2M - b^2 + b
			 * ab = (2M - b^2 + b) / 2 = M - (b^2 - b)/2;
			 * 令temp = ab;
			 * temp / a = b 项数
			 * temp / b = a 首项
			 */
			for (int i = len; i >= 1; --i) { //这里的i就是思路中表示的b(项数)
				temp = M - (i * i - i) / 2; 
				if (temp % i == 0) { //首项为整数
					// 首项：(temp / i) 
					// 末项：(temp / i + i- 1) 首项+项数-1
					System.out.println("[" + (temp / i) + "," + (temp / i + i - 1) + "]");
				}
			}
			System.out.println();
		}
		sn.close();
	}
}