Exhaustive Search Aizu - ALDS1_5_A
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].
思路
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 <iostream>
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;
}
