ACM题目————The partial sum problem

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!

 

题目大意就是，给你n个数， 再给一个sum，能不能用这n个数，加起来等于sum！

 

主要难点在减少循环。

 

//时间超限代码：

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//Asimple #include <stdio.h> #include <iostream> #include <string.h> using namespace std; #define CLS(a) memset(a,0,sizeof (a)) const int maxn = 25; int n, T, num, cnt, point, line, x, y; int vis[maxn];//标记数组，标记是否走过 int a[maxn]; bool flag;   bool check() {     for(int i=0; i<n; i++)         if( !vis[i] )             return false ;     return true; }   void DFS(int cnt) {     if( check() ) return ;     if( cnt == num )     {         flag = true ;         return ;     }     for(int i=0; i<n; i++)     {         if( !vis[i] && cnt + a[i] <= num )         {             vis[i] = 1 ;             DFS(cnt + a[i] ) ;             vis[i] = 0 ;         }     } }   int main() {     while( cin >> n )     {         cnt = 0 ;         CLS(vis);         flag = false ;         for(int i=0; i<n; i++)             cin >> a[i] ;         cin >> num ;         DFS(cnt);         if( flag ) cout << "Of course,I can!" << endl ;         else cout << "Sorry,I can't!" << endl ;     }       return 0; }

 

减少循环后的代码：

 

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//Asimple #include <iostream> using namespace std; const int maxn = 25; int n, num, cnt; int a[maxn]; bool flag;   void DFS(int x) {     if( cnt > num ) return ;     if( cnt == num )     {         flag = true ;         return ;     }     for(int i=x; i<n; i++)     {         cnt += a[i] ;         DFS(i+1);         cnt -= a[i] ;     } }   int main() {     while( cin >> n )     {         cnt = 0 ;         flag = false ;         for(int i=0; i<n; i++)             cin >> a[i] ;         cin >> num ;         DFS(0);         if( flag ) cout << "Of course,I can!" << endl ;         else cout << "Sorry,I can't!" << endl ;     }       return 0; }