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

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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 == 0 && N == 0) break;
        int len = sqrt(double(M * 2));//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 - (1 + len) / 2;
            if((a + 1 + a + len) * len / 2 == M) printf("[%d,%d]
", a + 1, a + len);
            len--;
        }
        puts("");
    }
    return 0;
}