The sum problem
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16197 Accepted Submission(s): 4843
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]
解题思路：
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<=(int)sqrt(2*M).
源代码：
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
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");
}
system("pause");
return 0;
}
时间： 2024-10-13 15:52:52