The partial sum problem

時間限制：1000 ms  |  記憶體限制：65535 KB

輸入

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).

輸出

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!”.

樣例輸入

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

樣例輸出

Of course,I can!
Sorry,I can't!

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. 

程式碼：

#include <iostream> #include <cstdio> #include <cstring> using namespace std;

int sum,a[30],k,ok,n;//a陣列用來儲存每個數字，ok存放是否存在題目要求的組合

void dfs(int x)//深搜 {     if(sum>k)//當sum比k大的情況，就不用再往深處搜尋了         return ;     if(sum==k)//當sum==k，也就是加起來的和存在和k相等的時候，把ok賦值為1；     {         ok=1;         return ;     }     for (int i = x; i <= n; i++){         sum += a[i];//加上某一個數         dfs(i + 1);//遞迴         sum -= a[i];//把加上的這個數減去     } }

int main() {     while(~scanf("%d",&n)){          for(int i=1; i<=n; i++){             scanf("%d",&a[i]);         }         scanf("%d",&k);         ok=0,sum=0;         dfs(1);         if(ok)             cout<<"Of course,I can!"<<endl;         else             cout<<"Sorry,I can't!"<<endl;     }     return 0; }