大牛来讨论一下哦

[komai908]

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]

http://acm.hdu.edu.cn/showproblem.php?pid=2058

[捧剑者]

up

[hua_zhixing_]

应该比较简单吧，会不会超时就不知道。

[bigbug9002]

假设从i开始到x之间的整数之和为M,由等比数据求和公式：
(i+x)(x-i+1)/2=M
化简为：x^2+x-i*i-2*M+i=0
由一元二次方程的求根公式，x=(-1+sqrt(1+4*(i*i+2*M-i)))/2,另一个负根舍去。
所以在已知i的情况下可以直接求出x.

做一个i从1到min(N,M)的循环，对每个i求x就可以了。
用数学函数开方的时间效率极低，所以可以自己写一个方法来判断1+4*(i*i+2*M-i)是不是完全平方数。

[abcdef0966]

#include <stdio.h>
#define MAX 20
int main()
{
	int buf1[MAX],buf2[MAX];
	int c;
	printf("enter your n and m:\n");

	for (c = 0; c < MAX; c++)
	{
		scanf("%d %d",&buf1[c],&buf2[c]);
		if ( !buf1[c] && !buf2[c])
			break;
		c++;
	}



	int r;
	for ( r = 0; r < c; r++)
	{
		int end,n,m;
		n = buf1[r];
		m = buf2[r];

		end = m < n ? m:n;
		int i,j;
		for (i = 1; i <= end; i++)
		{
			int sum = 0;
			int k;
			k = j = i;
			while (sum < m)
				sum += j++;
			if (sum == m)
				printf("[%d %d]\n",k,j-1);
		}

		printf("\n");
	}

	return 0;		
	    
}