Zero Sum(DFS)

最新推荐文章于 2024-02-11 17:00:00 发布

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本文介绍了一种寻找特定长度数列中能构成零和表达式的算法实现。通过对每个数位间插入加号、减号或留空的操作，程序能够找出所有可能组合，并输出那些求和为零的表达式。

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Problem Description:

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).
Input:

The input includes several cases. For each case, input a single line with the integer N (3 <= N <= 9).

Output:

For each case, the output format is:
In ASCII order, show each sequence that can create 0 sum with a '+', '-', or ' ' between each pair of numbers.
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

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define INF 0x3f3f3f3f

int num[20];
int symbol[20];
int n;

int cou()
{
    int i,j;
    int sum=0;
    int ca;
    for(i=0;i<=10;i++)
        num[i]=i;
    for(i=1;i<n;i++)
    {
        if(symbol[i]==0)
        {
            num[i+1]+=10*num[i];
            num[i]=0;
        }
    }
    /*for(i=1;i<=n;i++)
        printf("%d ",num[i]);*/
    //printf("\n");
    sum=0;
    ca=1;
    for(i=1;i<n;i++)
    {
        if(symbol[i]==0)
            continue;
        if(symbol[i]==1)
        {
            if(ca==1)
            {
                sum+=num[i];
            }
            else
                sum-=num[i];
            ca=1;
        }
        if(symbol[i]==2)
        {
            if(ca==1)
            {
                sum+=num[i];
            }
            else
                sum-=num[i];
            ca=2;
        }
    }
    if(ca==1)
    {
        sum+=num[n];
    }
    else
        sum-=num[n];
    if(sum)
        return;
    for(i=1;i<n;i++)
    {
        printf("%d",i);
        if(symbol[i]==1)
            printf("+");
        else if(symbol[i]==2)
            printf("-");
        else
            printf(" ");
    }
    printf("%d\n",n);
}


void dfs(int a)
{
    int i;
    if(a==n)
    {
        cou();
        return;
    }
    for(i=0;i<3;i++)
    {
        symbol[a]=i;
        dfs(a+1);
    }
}

int main()
{
    int i,j,k;
    while(scanf("%d",&n)==1)
    {
        dfs(1);
    }
    return 0;
}