Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 26343 Accepted Submission(s): 7884Problem 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
[1,4] [10,10] [4,8] [6,9] [9,11] [30,30]
8600
Source
问题链接:HDU2058 The sum problem。
题意简述:参见上文。
问题分析:这个问题用数学方法+暴力法来解决,先做数学推导再编写枚举程序。
因为Sn=(a1+an)*n/2=a1*n+d(n-1)*n/2=a1*n+(n-1)*n/2(其中d=1),所以a1*n = m - (n-1)*n/2。
考虑最小的a1=1,则有Sn=n+(n-1)*n/2=(n+1)n/2。
那么n<=sqrt(2Sn)=sqrt(2m),即n最大值为sqrt(2m)。
基于以上推导,对n进行逐个试探即可。
程序说明:(略)
题记:(略)
AC的C语言程序如下:
/* HDU2058 The sum problem */ #include <stdio.h> #include <math.h> int main(void) { int n, m, len, a1n, i; while(scanf("%d%d", &n, &m) != EOF && (n || m)) { len = (int)sqrt(m * 2.0); for(i=len; i>0; i--) { a1n = m - (i - 1) * i / 2; //a1*n = m-(n+1)*n/2; if(a1n % i == 0) printf("[%d,%d] ", a1n / i, a1n / i + i - 1); } putchar(' '); } return 0; }