USCAO section 2.3 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+71+2-3-4+5+6-71-2 3+4+5+6+71-2 3-4 5+6 71-2+3+4-5+6-71-2-3-4-5+6+7

解题思路：直接深搜三种状态，最多3^8不超，注意点细节就OK了，纯水

/*
ID:nealgav1
PROG:zerosum
LANG:C++
*/
#include<fstream>
#include<cstring>
using namespace std;
const int mm=10000;
char s[mm][25],num;
char _s[25];
ifstream cin("zerosum.in");
ofstream cout("zerosum.out");
void dfs(char now,int x,int end,int sum,char pas)
{
_s[2*x-3]=now;_s[2*x-2]=x+'0';
if(now==' '&&pas=='+')
{sum+=(x-1)*9+x;
if(x==end)
{ _s[2*x-3]=now;_s[2*x-2]=x+'0';
_s[0]='0'+1;
_s[2*x-1]='\0';
if(sum==0)cout<<_s<<"\n";
return;
}
dfs(' ',x+1,end,sum,now);
dfs('+',x+1,end,sum,now);
dfs('-',x+1,end,sum,now);
}
else if(now==' '&&pas=='-')
{
sum=sum-x-(x-1)*9;
if(x==end)
{ _s[2*x-3]=now;_s[2*x-2]=x+'0';
_s[0]='0'+1;
_s[2*x-1]='\0';
if(sum==0)cout<<_s<<"\n";
return;
}
dfs(' ',x+1,end,sum,now);
dfs('+',x+1,end,sum,now);
dfs('-',x+1,end,sum,now);
}
if(now=='+')
{
sum+=x;
if(x==end)
{ _s[2*x-3]=now;_s[2*x-2]=x+'0';
_s[0]='0'+1;
_s[2*x-1]='\0';
if(sum==0)cout<<_s<<"\n";
return;
}
dfs(' ',x+1,end,sum,now);
dfs('+',x+1,end,sum,now);
dfs('-',x+1,end,sum,now);
}
if(now=='-')
{sum-=x;
if(x==end)
{ _s[2*x-3]=now;_s[2*x-2]=x+'0';
_s[0]='0'+1;
_s[2*x-1]='\0';
if(sum==0)cout<<_s<<"\n";
return;
}
dfs(' ',x+1,end,sum,now);
dfs('+',x+1,end,sum,now);
dfs('-',x+1,end,sum,now);
}
}
int main()
{

int m;
cin>>m;
num=0;
dfs(' ',2,m,1,'+');
dfs('+',2,m,1,'+');
dfs('-',2,m,1,'+');
}

USER: Neal Gavin Gavin [nealgav1]
TASK: zerosum
LANG: C++

Compiling...
Compile: OK

Executing...
Test 1: TEST OK [0.000 secs, 3580 KB]
Test 2: TEST OK [0.000 secs, 3580 KB]
Test 3: TEST OK [0.000 secs, 3580 KB]
Test 4: TEST OK [0.000 secs, 3580 KB]
Test 5: TEST OK [0.000 secs, 3580 KB]
Test 6: TEST OK [0.000 secs, 3580 KB]
Test 7: TEST OK [0.000 secs, 3580 KB]

All tests OK.
YOUR PROGRAM ('zerosum') WORKED FIRST TIME!  That's fantastic
-- and a rare thing.  Please accept these special automated
congratulations.

Here are the test data inputs:

------- test 1 ----
3
------- test 2 ----
4
------- test 3 ----
5
------- test 4 ----
6
------- test 5 ----
7------- test 6 ----
8
------- test 7 ----
9

Keep up the good work!

Thanks for your submission!

Zero SumRuss Cox
We can use a simple recursive depth first search to generate all the possible strings to be had by putting in a space, plus, or minus sign between each number.

Once we've generated each string, we evaluate it as an arithmetic sum and see if we get zero. If so, we print the string.

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

FILE *fout;
int n;

/* evaluate the string s as arithmetic and return the sum */
int
eval(char *s)
{
int term, sign, sum;
char *p;

sign = 1;
term = 0;
sum = 0;
for(p=s; *p; p++) {
switch(*p){
case '+':
case '-':
sum += sign*term;
term = 0;
sign = *p == '+' ? 1 : -1;
break;
case ' ':
break;
default:    /* digit */
term = term*10 + *p-'0';
}
}
sum += sign*term;
return sum;
}

/*
* Insert + - or space after each number, and then
* test to see if we get zero.  The first k numbers have
* already been taken care of.
*/
void
search(char *s, int k)
{
char *p;

if(k == n-1) {
if(eval(s) == 0)
fprintf(fout, "%s\n", s);
return;
}

for(p=" +-"; *p; p++) {
s[2*k+1] = *p;
search(s, k+1);
}
}

void
main(void)
{
FILE *fin;
int i;
char str[30];

fin = fopen("zerosum.in", "r");
fout = fopen("zerosum.out", "w");
assert(fin != NULL && fout != NULL);

fscanf(fin, "%d", &n);

strcpy(str, "1 2 3 4 5 6 7 8 9");
str[2*n-1] = '\0';    /* chop string to only have first n numbers */

search(str, 0);

exit(0);
}