uva621

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http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=562

621 - Secret Research

Time limit: 3.000 seconds

Secret Research

At a certain laboratory results of secret research arethoroughly encrypted. A result of a single experiment is stored asan information of its completion:

`positive result', `negative result', `experiment failed' or`experiment not completed'

The encrypted result constitutes a string of digits S, which maytake one of the following forms:

positive result                  S = 1 or S = 4 or S = 78 negative result                  S = S35 experiment failed                S = 9S4 experiment not completed                 S = 190S

(A sample result S35 means that if we add digits 35 from theright hand side to a digit sequence then we shall get the digitsequence corresponding to a failed experiment)

You are to write a program which decrypts given sequences ofdigits.

Input

A integer n stating the number of encrypted results andthen consecutive n lines, each containing a sequence ofdigits given as ASCII strings.

Output

For each analysed sequence of digits the following lines shouldbe sent to output (in separate lines):

                 +               for a positive result                 -               for a negative result                 *               for a failed experiment                 ?               for a not completed experiment

In case the analysed string does not determine the experimentresult, a first match from the above list should be outputted.

SampleInput

478783519078944

SampleOutput

+-?*

题意：就按照题目中的要求顺序判断，符合之后不再判断后面的条件。贡献了好几次WA啊。。其中有一种方法挺好的就是把所有的情况都保存在数组里面，输入的时候直接比较，不过WA了。。

WA的代码1：

#include<stdio.h>

#include<string.h>
int s1[4]={1,4,78};
int s2[4]={135,435,7835};
int s3[4]={914,944,9784};
int s4[4]={1901,1904,19078};
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
  int m,i;
  scanf("%d",&m);
  for(i=0;i<=3;i++)
  {
   if(m==s1[i])
   {
    printf("+\n");
    break;
   }
   elseif(m==s2[i])
   {
    printf("-\n");
    break;
   }
   elseif(m==s3[i])
   {
    printf("?\n");
    break;
   }
   elseif(m==s4[i])
   {
    printf("*\n");
    break;
   }
  }
}
return 0;
}

WA的代码2

#include<stdio.h>
#include<string.h>
char s1[5][10]={"1","4","78"};
char s2[5][10]={"135","435","7835"};
char s3[5][10]={"914","944","9784"};
char s4[5][10]={"1901","1904","19078"};
int main()
{
int n;
scanf("%d",&n);
getchar();
while(n--)
{
  char s[1010];
  int i;
  scanf("%s",s);
  for(i=0;i<3;i++)
  {
   if(strcmp(s,s1[i])==0)
   {
    printf("+\n");
    break;

   }
   if(strcmp(s,s2[i])==0)
   {
    printf("-\n");
    break;
   }
   if(strcmp(s,s3[i])==0)
   {
    printf("*\n");
    break;
   }
   if(strcmp(s,s4[i])==0)
   {
    printf("?\n");
    break;
   }
  }
}
return 0;
}

AC代码：
#include<stdio.h>
#include<string.h>
char s[1010];
int main()
{
 int n;
 scanf("%d",&n);
 getchar();
 while(n--)
 {
  scanf("%s",s);
  int len=strlen(s);
  if(!strcmp(s,"1")||!strcmp(s,"4")||!strcmp(s,"78"))
  {
   printf("+\n");
   continue;
  }
  else if(s[len-2]=='3'&&s[len-1]=='5')
  {
   printf("-\n");
   continue;
  }
  else if(s[0]=='9'&&s[len-1]=='4')
  {
   printf("*\n");
   continue;
  }
  else if((s[0]=='1')&&(s[1]=='9')&&(s[2]=='0'))
  {
   printf("?\n");
   continue;
  }
 }
 return 0;
}