51nod 1268

1268 and a combination of K 

Base Time Limit: 1 second Space Limit: 131072 KB Score: 20Difficulty  : Level 3 Algorithm Questions

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Given an array A consisting of N positive integers, find out whether some of them can be selected so that their sum is K. If yes, output: "Yes", otherwise output "No".

Input

Line 1: 2 numbers N, K, N is the length of the array, K is the sum to be judged (2 <= N <= 20, 1 <= K <= 10^9)
Lines 2 - N + 1: 1 number per line, corresponding to array element A[i] (1 <= A[i] <= 10^6)

Output

If yes, output: "Yes", otherwise output "No".

Input example

5 13
2
4
6
8
10

Output example

No

The number is small, you can do it directly, use a set to store the current sum, and add it to the number in the set every time you traverse a number. Of course, only those smaller than k are put into the set, which can be calculated less.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int a[22];
 4 int main() {
 5     int n, k;
 6     set<int> st[2];
 7     set<int> ::iterator it;
 8     cin >> n >> k;
 9     int last = 0;
10     for(int i = 1; i <= n; i ++) cin >> a[i];
11     for(int i = 1; i <= n; i ++) {
12         last = 1 - last;
13         it = st[1-last].begin();
14         for(; it != st[1-last].end(); ++it) {
15             if((*it)+a[i] < k) {
16                 st[last].insert((*it)+a[i]);
17             } else if((*it)+a[i] == k) return 0*printf("Yes\n");
18             st[last].insert((*it));
19         }
20         st[last].insert(a[i]);
21     }
22     if(st[last].count(k)) printf("Yes\n");
23     else printf("No\n");
24     return 0;
25 }