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：

//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：

//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;
}