Write a program which reads a sequence A of n elements and an integer M, and outputs "yes" if you can make M by adding elements in A, otherwise "no". You can use an element only once.
You are given the sequence A and q questions where each question contains Mi.
Input
In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers (Mi) are given.
Output
For each question Mi, print yes or no.
Constraints
-
n ≤ 20
-
q ≤ 200
-
1 ≤ elements in A ≤ 2000
-
1 ≤ Mi ≤ 2000
Sample Input 1
5
1 5 7 10 21
8
2 4 17 8 22 21 100 35
Sample Output 1
no
no
yes
yes
yes
yes
no
no
Notes
You can solve this problem by a Burte Force approach. Suppose solve(p, t) is a function which checkes whether you can make t by selecting elements after p-th element (inclusive). Then you can recursively call the following functions:
solve(0, M) solve(1, M-{sum created from elements before 1st element}) solve(2, M-{sum created from elements before 2nd element}) ...
The recursive function has two choices: you selected p-th element and not. So, you can check solve(p+1, t-A[p]) and solve(p+1, t) in solve(p, t) to check the all combinations.
For example, the following figure shows that 8 can be made by A[0] + A[2].
一种时间复杂度是O(2n)的究极暴力搜索算法
做法:判断每个元素是否可取,每次都有2种取法。
用递归的方式来解决(从m中加上w[i],用m是否等于M来判断是否这个数能用和来表示):
bool f;
void solve(int i,int m){ if(m == M){f = 1;return ;} if(i == n)return ; solve(i + 1,m) ; solve(i + 1,m + w[i]); return ; }
每次递归都有不选这个数(m)和选择这个数(m + w[i])两种选择,用f来标记是否可以用和来表示。
AC代码:
#include<bits/stdc++.h> using namespace std; const int N = 1e6 + 10; int w[N]; int n,k,M; bool f; void solve(int i,int m){ if(m == M){f = 1;return ;} if(i == n)return ; solve(i + 1,m) ; solve(i + 1,m + w[i]); return ; } int main(){ scanf("%d",&n); for(int i = 0;i < n;i ++)scanf("%d",&w[i]); scanf("%d",&k); for(int i = 0;i < k;i ++){ scanf("%d",&M); f = 0; solve(0,0); cout << (f?"yes" : "no") << endl; } return 0; }
等学了DP再来补一个DP的做法。