HDU2058 The sum problem【数学计算+枚举】

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 26343    Accepted Submission(s): 7884

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]

Author

8600

Source

校庆杯Warm Up

问题链接：HDU2058 The sum problem。

题意简述：参见上文。

问题分析：

这个问题用数学方法+暴力法来解决，先做数学推导再编写枚举程序。

因为Sn=(a1+an)*n/2=a1*n+d(n-1)*n/2=a1*n+(n-1)*n/2（其中d=1），所以a1*n = m - (n-1)*n/2。

考虑最小的a1=1，则有Sn=n+(n-1)*n/2=(n+1)n/2。

那么n<=sqrt(2Sn)=sqrt(2m)，即n最大值为sqrt(2m)。

基于以上推导，对n进行逐个试探即可。

程序说明：（略）

题记：（略）

AC的C语言程序如下：

/* HDU2058 The sum problem */

#include <stdio.h>
#include <math.h>

int main(void)
{
    int n, m, len, a1n, i;

    while(scanf("%d%d", &n, &m) != EOF && (n || m)) {
        len = (int)sqrt(m * 2.0);
        for(i=len; i>0; i--) {
            a1n = m - (i - 1) * i / 2;     //a1*n = m-(n+1)*n/2;
            if(a1n % i == 0)
                printf("[%d,%d]
", a1n / i, a1n / i + i - 1);
        }
        putchar('
');
    }

    return 0;
}