ALDS1_5_A: Exhaustive Search

题解

这里用的思路是：每次选择针对每个元素，每个元素都只有选与不选两种情况，所以共有2ⁿ种选择，并且这里将这些组合通过递归生成。
递归思路：这里的原问题即为solve(a, n, 0, m)，即这2ⁿ种组合中是否有和为m的，其中的子问题就是solve(a, n, i, m)，即在确定了前i个元素下其后面的元素能否组合出m。然后将solve(a, n, i, m)分治为：solve(a, n, i+1, m)和solve(a, n, i+1, m-a[i])，即每个子问题都将其分治为其下一个元素的选或不选，这样就可以递归实现了。
1. 由于子问题共有n个，所以原问题在每个子问题的分治下被分成了2ⁿ个，类似二叉树型结构
2. 这里通过i<=n即可对递归进行控制，防止其一直向下分治，当不满足条件时返回值或者不设置递归。
自己之前做这道题的思路是：每次选择针对所有元素，每次选择都要判断选择哪个元素。这样不但时间复杂度高，还需要对已选择的元素进行区别。只看到了表面的选择一组元素，没有发现选择的本质是每个元素的选与不选（类似二项分布中的组合）。

题目

Exhaustive Search
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 ≤ 2000Sample Input 1
5
1 5 7 10 21
8
2 4 17 8 22 21 100 35Sample Output 1
no
no
yes
yes
yes
yes
no
noNotes
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].

代码块

#include <iostream>
using namespace std;int solve(int *a, int n, int i, int m)
{if(m==0)//从输入值m中减去所选元素，当m==0就说明和为m，搜索成功。return 1;if(i>=n)//i代表所选元素的个数，达到n就返回0，代表不成功return 0;int res = solve(a, n, i+1, m) || solve(a, n, i+1, m-a[i]);//模拟a中第i+1个元素的选与不选：左边的代表不选，右边的代表选。如果某种情况中有1就返回1，都是0就返回0。return res;
}int main(void)
{int i, n, q;cin>>n;int a[n];for(i=0; i<n; i++)cin>>a[i];cin>>q;for(i=0; i<q; i++){int m;cin>>m;if(solve(a, n, 0, m))cout<<"yes"<<endl;elsecout<<"no"<<endl;}return 0;
}

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ALDS1_5_A: Exhaustive Search

题解

题目

代码块

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