The sum problem 2058

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The sum problem

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8287 Accepted Submission(s): 2530

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 1050 300 0

Sample Output

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

我本来的想法是考虑子列的起点和终点，分别以s和e表示，由等差数列求和公式有（s+e）*（e-s+1）/2==M（1式），化为e*(e+1)-s*(s-1)==2*M, so , e=(int)sqrt(2*M+s*(s-1)),将得到的e再代回1式，成立则[s,e]满足条件。

但是，2*M+s*(s-1)太大……

后来参考网上的一个算法，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).

AC code:

#include <iostream>  #include <cmath>  using namespace std;  int main()  {      int N,M,i,j;      while(scanf("%d%d",&N,&M) && M+N)      {           for(j=pow(2.0*M,0.5);j>0;j--)          {              i=(2*M/j-j+1)/2;              if(j*(j+2*i-1)/2==M) printf("[%d,%d]\n",i,i+j-1);          }          printf("\n");                           }         return 0;     }

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