Dessert

Dessert

Time Limit: 1000MS		Memory Limit: 30000K

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

Source

USACO 2002 February

#include<iostream>

using namespace std;

const int Max = 16;
int n,num;   // num記錄可能得情況數。
char symbol[Max];

void print()
{
    int i;
    for(i = 1; i < n; i ++)
        printf("%d %c ", i, symbol[i]);
    printf("%d\n", i);
}

void dfs(int sum, int depth, int pre)   // dfs儘量少用全局變量，如定義成dfs(int depth)。
{
    if(depth == n)    // 深搜出口。
    {
        if(sum == 0)
        {
            num ++;
            if(num <= 20)  print();
        }
    }
    else
    {
        int next;
        symbol[depth] = '+';   // 若第depth位符號爲'+'。
        next = depth + 1;
        dfs(sum + next, depth + 1, next);

        symbol[depth] = '-';    // 若第depth位符號爲'-'。
        next = depth + 1;
        dfs(sum - next, depth + 1, next);

        symbol[depth] = '.';    // 若第depth位符號爲'.'。
        if(depth < 9)
            next = pre * 10 + depth + 1;
        else
            next = pre * 100 + depth + 1;
        int i = depth - 1;
        while(symbol[i] == '.' && i >= 0)  i --;
        if(symbol[i] == '+')
            dfs(sum + next - pre, depth + 1, next);
        else if(symbol[i] == '-')
            dfs(sum - next + pre, depth + 1, next);
    }
}
int main()
{
    cin >> n;
    symbol[0] = '+';  // 將第0個符號看成'+',構造成 0 + 1 ? 2 ? 3 …簡化後面較煩的初始化。
    num = 0;
    dfs(1, 1, 1);
    cout << num << endl;
    return 0;
}