USACO2.3.3--Zero Sum

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
题解：先对每个数字之间添加'+'或者'-'或者空格。这个可以用深搜实现。然后再对表达式求值，如果表达式的和为零，那么就是符合要求的，输出即可。
丑陋的代码o(╯□╰)o

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  1 /*

  2 ID:spcjv51

  3 PROG:zerosum

  4 LANG:C

  5 */

  6 #include<stdio.h>

  7 #include<string.h>

  8 int n;

  9 char ss[50];

 10 void print()

 11 {

 12     char s[50];

 13     int i,len,lr,j,ans;

 14     ans=0;

 15     strcpy(s,ss);

 16     len=strlen(s);

 17     i=1;

 18     lr=1;

 19     if(s[i]==' ')

 20         {

 21             lr=lr*10+s[i+1]-'0';

 22             j=i+2;

 23             while(s[j]==' ')

 24             {

 25                 lr=lr*10+s[j+1]-'0';

 26                 j+=2;

 27 

 28             }

 29             i=j;

 30         }

 31     ans+=lr;

 32 

 33     while(i<len)

 34     {

 35 

 36 

 37         if(s[i]=='-')

 38         {

 39             j=i+2;

 40             lr=s[i+1]-'0';

 41             while(s[j]==' ')

 42             {

 43                 lr=lr*10+s[j+1]-'0';

 44                 j+=2;

 45 

 46             }

 47             i=j;

 48             ans-=lr;

 49 

 50         }

 51         if(s[i]=='+')

 52         {

 53             j=i+2;

 54             lr=s[i+1]-'0';

 55             while(s[j]==' ')

 56             {

 57                 lr=lr*10+s[j+1]-'0';

 58                 j+=2;

 59 

 60             }

 61             i=j;

 62             ans+=lr;

 63         }

 64 

 65     }

 66     if(ans==0)

 67     printf("%s\n",ss);

 68 }

 69 void dfs(int step,int last)

 70 {

 71     char str[3],ssr[50];

 72     if(step==n)

 73     {

 74         print();

 75         return;

 76     }

 77     sprintf(str,"%d",step+1);

 78     last=last*10+step+1;

 79     strcpy(ssr,ss);

 80     strcat(ss," ");

 81     strcat(ss,str);

 82     dfs(step+1,last);

 83     strcpy(ss,ssr);

 84     strcat(ss,"+");

 85     strcat(ss,str);

 86     last=step+1;

 87     dfs(step+1,last);

 88     strcpy(ss,ssr);

 89     strcat(ss,"-");

 90     strcat(ss,str);

 91     last=-(step+1);

 92     dfs(step+1,last);

 93     strcpy(ss,ssr);

 94 }

 95 

 96 int main(void)

 97 {

 98     freopen("zerosum.in","r",stdin);

 99     freopen("zerosum.out","w",stdout);

100     scanf("%d",&n);

101     strcpy(ss,"1");

102     dfs(1,1);

103     return 0;

104 }