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.
InputIn 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.
OutputFor each question Mi, print yes or no.
Constraintsn ≤ 20 q ≤ 200 1 ≤ elements in A ≤ 2000 1 ≤ Mi ≤ 2000
Sample Input 15 1 5 7 10 21 8 2 4 17 8 22 21 100 35
Sample Output 1no no yes yes yes yes no no
NotesYou 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].
设solve(i,m)为“用第i个元素后面的元素能得出m时返回true”的函数，这样一来solve(i,m)就可以分解为solve(i+1,m)和solve(i,m-A[i])这两个更小的局部问题。
函数solve(i,m)中，m==0时代表数组元素相加能够得出指定整数。相反，m>0且i>=n时表示数组元素相加得不出指定整数。
只要局部问题solve(i+1,m)和solve(i,m-A[i])之中有一个为true，原问题solve(i,m)就为true。
code/*
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*/ // Virtual_Judge —— Exhaustive Search Aizu - ALDS1_5_A.cpp created by VB_KoKing on 2019-05-04:12. /* Procedural objectives：
Variables required by the program:
Procedural thinking：
Functions required by the program:
*/ /* My dear Max said：
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first gentle breeze that passed through my ear is you,
The first star I see is also you.
The world I see is all your shadow."
FIGHTING FOR OUR FUTURE！！！
*/ #include using namespace std; int n,A[50]; //从输入值M中减去所选元素的递归函数 int solve(int i,int m) { if (m==0) return 1; if (i>=n) return 0; return solve(i+1,m)+solve(i+1,m-A[i]); } int main() { int q,M; cin>>n; for (int i = 0; i < n; i++) cin>>A[i]; cin>>q; for (int i = 0; i < q; i++) { cin>>M; if (solve(0,M)) cout<<"yes"<<endl; else cout<<"no"<<endl; } return 0; }
