描述

One day,Tom’s girlfriend give him an array A which contains N integers and asked him:Can you choose some integers from the N integers and the sum of them is equal to K. 

输入

There are multiple test cases.
Each test case contains three lines.The first line is an integer N(1≤N≤20),represents the array contains N integers. The second line contains N integers,the ith integer represents A[i](-10^8≤A[i]≤10^8).The third line contains an integer K(-10^8≤K≤10^8).

输出

If Tom can choose some integers from the array and their them is K,printf ”Of course,I can!”; other printf ”Sorry,I can’t!”.

样例输入

4
1 2 4 7
13
4
1 2 4 7
15

样例输出

Of course,I can!
Sorry,I can‘t!

直接dfs用时长，可以剪枝

剪枝代码：用时12ms

01.#include<iostream>

02.using namespace std;

03.int n,k,a[25];

04.bool dfs(int p,int sum)

05.{

06.if(sum==k)return 1;

07.if(p==n||sum>k)return 0;

08.if(dfs(p+1,sum))return true;

09.else return dfs(p+1,sum+a[p]); 

10.}

11.int main()

12.{

13.while(cin>>n)

14.{

15.for(int i=0;i<n;i++)

16.cin>>a[i];

17.cin>>k;

18.if(dfs(0,0))

19.cout<<"Of course,I can!"<<endl;

20.else

21.cout<<"Sorry,I can‘t!"<<endl;

22.}

23.return 0;

24.}

普通dfs：用时860ms

01.#include<iostream>

02.using namespace std;

03.int n,k,a[25];

04.bool dfs(int p,int sum)

05.{

06.if(p==n)return sum==k;

07.if(dfs(p+1,sum))return true;

08.if(dfs(p+1,sum+a[p]))return true;

09.return false;  

10.}

11.int main()

12.{

13.while(cin>>n)

14.{

15.for(int i=0;i<n;i++)

16.cin>>a[i];

17.cin>>k;

18.if(dfs(0,0))

19.cout<<"Of course,I can!"<<endl;

20.else

21.cout<<"Sorry,I can‘t!"<<endl;

22.}

23.return 0;

24.}

NYOJ-927 The partial sum problem