The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 23723    Accepted Submission(s): 7052 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]

设这个子序列为a+1,a+2........a+d，d为这个子序列的长度，起始数字为a+1，M = a*d + (d + 1) * d / 2；可得2 * M > d * d，所以只需要枚举出所有的d即可！

#include <stdio.h> #include <math.h> int main(){ int n,m,i,j; while(~scanf("%d%d",&n,&m)){ if(!n && !m) break; int b,d; for(d = sqrt(2 * m); d > 0; d --){ b = m - (d + d * d) / 2; if(b % d == 0) printf("[%d,%d]\n",(b/d)+1,(b/d)+d); } putchar('\n'); } return 0; }