The sum problem
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7877 Accepted Submission(s): 2422
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]
#include<stdio.h> #include<math.h> int main() { int n, m; while( scanf( "%d %d", &n, &m ) && (n||m) ) { int max, i, result; max = sqrt( m*2.0 ); for( i = max; i > 0; --i ) { result = m - ( i*i + i )/2; if( result % i == 0 ) { printf( "[%d,%d]\n", result/i+1, result/i+i ); } } printf( "\n" ); } }
这题早就推了公式,只是可能有些地方没有简化,一直出不来,然后今天看了白神的代码,知道了解题思路,首先,把i看作0, 则可求出i的最大值,然后暴力i。