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Dessert

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5430		Accepted: 2029

Description

FJ has a new rule about the cows lining up for dinner. Not only must the N (3 <= N <= 15) cows line up for dinner in order, but they must place a napkin between each pair of cows with a "+", "-", or "." on it. In order to earn their dessert, the cow numbers and the napkins must form a numerical expression that evaluates to 0. The napkin with a "." enables the cows to build bigger numbers. Consider this equation for seven cows:

      1 - 2 . 3 - 4 . 5 + 6 . 7

This means 1-23-45+67, which evaluates to 0. You job is to assist
the cows in getting dessert. (Note: "... 10 . 11 ...") will use the
number 1011 in its calculation.)

Input

One line with a single integer, N

Output

One
line of output for each of the first 20 possible expressions -- then a
line with a single integer that is the total number of possible answers.
Each expression line has the general format of number, space, napkin,
space, number, space, napkin, etc. etc. The output order is
lexicographic, with "+" coming before "-" coming before ".". If fewer
than 20 expressions can be formed, print all of the expressions.

Sample Input

7

Sample Output

1 + 2 - 3 + 4 - 5 - 6 + 7
1 + 2 - 3 - 4 + 5 + 6 - 7
1 - 2 + 3 + 4 - 5 + 6 - 7
1 - 2 - 3 - 4 - 5 + 6 + 7
1 - 2 . 3 + 4 + 5 + 6 + 7
1 - 2 . 3 - 4 . 5 + 6 . 7
6

题意:往 1 - n里面添加符号 + - . 问怎样使得结果为 0 ..输出方案数以及方案。。如果方案数>20，那么输出前20个方案。
题解:DFS太弱了。。不知道怎么处理点号。。参考别人的。多加练习深搜！

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iostream>
using namespace std;
int cnt,n;
int vis[];
char expe[];
void dfs(int deep,int ans,int pre)
{
    if(deep==n)
    {
        if(ans==)
        {
            cnt++;
            if(cnt<=)
            {
                for(int i=; i<n; i++)
                {
                    printf("%d %c ",i,expe[i]);
                }
                printf("%d\n",n);
            }
        }
        return;
    }
    else
    {
        int now,next,k;
        expe[deep]='+';
        dfs(deep+,ans+deep+,deep+); ///当前数为deep+1 ,由于是"+",得到的下一个数为ans+(deep+1)
        expe[deep]='-';
        dfs(deep+,ans-(deep+),deep+);
        expe[deep]='.';
        if(deep+>)
        {
            now = *pre+(deep+);
        }
        else now = *pre+(deep+);
        int j = deep-;
        while(expe[j]=='.'&&j>=) j--;
        if(expe[j]=='+')
            dfs(deep+,(ans-pre)+now,now); ///这里的话开始硬是没写出来
        else dfs(deep+,(ans+pre)-now,now);
    }
    return;
}
int main()
{
    int t = ;
    while(scanf("%d",&n)!=EOF&&n)
    {
        cnt = ;
        memset(vis,,sizeof(vis));
        expe[] = '+';  ///0+????
        dfs(,,);
        printf("%d\n",cnt);
    }
    return ;
}