HDU2058

The sum problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20726    Accepted Submission(s): 6100

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

Recommend

linle   |   We have carefully selected several similar problems for you:  20592062206020722061

Statistic | Submit | Discuss | Note

这道题显然不可能直接暴力做出来。技巧在于所有可能数列的长度有最大值。设数列为a+1，a+2 ，…… a+len，则它的长度为len。

根据公式M=[(a+1)+(a+len)]*len/2化简得出M*2=len*len+(2a+1)*len。所以len一定小于sqrt（2*M）接下来就可以枚举lenAC这道题。

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

 int main()
 {
     int N, M;
     while(~scanf("%d%d", &N, &M)) {
         if(M ==  && N == ) break;
         int len = sqrt(double(M * ));//如果不加double会出现编译错误，sqrt的参数是浮点型的。
         while(len) {
             int a = M / len - ( + len) / ;
             if((a +  + a + len) * len /  == M) printf("[%d,%d]\n", a + , a + len);
             len--;
         }
         puts("");
     }
     return ;
 }