Zero Sum（DFS）

Zero Sum

Consider the sequence of digits from 1 through N (where N=9) in increasing order: 1 2 3 ... N.

Now insert either a `+' for addition or a `-' for subtraction or a ` ' [blank] to run the digits together between each pair of digits (not in front of the first digit). Calculate the result that of the expression and see if you get zero.

Write a program that will find all sequences of length N that produce a zero sum.

PROGRAM NAME: zerosum

INPUT FORMAT

A single line with the integer N (3 <= N <= 9).

SAMPLE INPUT (file zerosum.in)

7

OUTPUT FORMAT

In ASCII order, show each sequence that can create 0 sum with a `+', `-', or ` ' between each pair of numbers.

SAMPLE OUTPUT (file zerosum.out)

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

 

    题意：

    给出 N（3 ~ 9），代表有N个数，有3种操作符，分别是 + 加法，- 减法，” “ 连续（空格说明相邻的这两个数合在一起，比如 2  3 代表23，1 2 3代表123），找到适合的等式使结果和为0。最后按字典序输出。

 

    思路：

    DFS。先把符号存起来当为 N - 1的时候再继续判断等式时候和为0，注意处理空格符号。

  

    AC：

/*
TASK:zerosum
LANG:C++
ID:sum-g1
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
char way[5],fin[10];
int n;

void dfs(int ans,char ope)
{
    fin[ans] = ope;
    if(ans == n - 1)
    {
        int num[10],sum = 0;
        for(int i = 1;i <= n;i++)   num[i] = i;
        for(int i = 1;i <= n - 1;i++)
        {
            int k = i,t = num[i];
            while(fin[k] == ' ')
            {
                t = t * 10 + num[k + 1];
                k++;
            }
            if(t != num[i]) num[i] = t;
        }
        sum += num[1];
        for(int i = 1;i <= n - 1;i++)
        {
            if(fin[i] == ' ') continue;
            if(fin[i] == '+') sum += num[i + 1];
            if(fin[i] == '-') sum -= num[i + 1];
        }
        if(!sum)
        {
            printf("%d",1);
            for(int i = 1;i <= n - 1;i++)
                printf("%c%d",fin[i],i + 1);
            printf("\n");
        }
        return;
    }
    for(int i = 1;i <= 3;i++)
        dfs(ans + 1,way[i]);
}

int main()
{
    freopen("zerosum.in","r",stdin);
    freopen("zerosum.out","w",stdout);
    way[1] = ' ';way[2] = '+';way[3] = '-';
    scanf("%d",&n);
    for(int i = 1;i <= 3;i++)   dfs(1,way[i]);
    return 0;
}