Nyist oj Part 1 and question (DFS search), ojdfs

Nyist oj Part 1 and question (DFS search), ojdfs
Partial and problem time limit: 1000 MS | memory limit: 65535 KB difficulty: 2

Description

Given integers a1, a2,... an, judge whether several numbers can be selected from them to make their sum exactly K.

Input

First, n and k, n represent the number of numbers, and k represent the sum of numbers.
Next, the number of n rows.
(1 <= n <= 20, which must not exceed the int range)

Output

If the sum can be k, "YES" is output, and the sum of which numbers is output in order of input. Otherwise, "NO" is output"

Sample Input

4 131 2 4 7

Sample output

YES2 4 7

Source

Typical questions

Uploaded

TC _ Yang zhuangliang

When searching simple questions, you must pay attention to the ending condition when recursion, that is, the critical condition to reduce the overhead of recursion; The following code is used:

# Include <cstdio> # include <cstring> using namespace std; int a [30], n, k, sum; bool visit [30], flag; void dfs (int pos) {if (flag = true) return; if (sum> = k) {if (sum = k) {flag = true; printf ("YES \ n "); for (int I = 0; I <n; I ++) if (visit [I]) // mark printf ("% d", a [I]);} return; // here is the condition for Recursive termination. Adding an end condition here reduces the time consumption.} for (int I = pos; I <n; I ++) // The search process {sum + = a [I]; visit [I] = 1; dfs (I + 1); sum-= a [I]; visit [I] = 0 ;}} int main () {int I; while (scanf ("% d", & n, & k )! = EOF) {for (I = 0; I <n; I ++) scanf ("% d", & a [I]); memset (visit, 0, sizeof (visit); flag = false; dfs (0); if (! Flag) printf ("NO \ n");} return 0 ;}