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  • uva 10361 Automatic Poetry

    字符串处理,不过做的有点麻烦

    View Code
     1 #include<stdio.h>
     2 #include<string.h>
     3 int main()
     4 {
     5     int n,i,j,k,x,len1,len2,count,y;
     6     char str1[1000],str2[1000],str[1000];
     7     scanf("%d",&n);
     8     getchar();
     9     while(n--)
    10     {
    11         gets(str1);
    12         gets(str2);
    13         len1=strlen(str1);
    14         len2=strlen(str2);
    15         count=0;
    16         for(i=0;i<len1;i++)
    17         {
    18 
    19             if(str1[i]!='<'&&str1[i]!='>')
    20             {
    21                 printf("%c",str1[i]);
    22             }
    23             else
    24             {
    25                 if(count==0&&str1[i]=='<')
    26                 {
    27                     y=i;
    28                 }
    29                 count++;
    30             }
    31         }
    32         printf("\n");
    33         k=0;
    34         count=0;
    35         for(i=len1-1;i>=0;)
    36         {
    37             if(str1[i]=='>')
    38             {
    39                 count++;
    40                 if(count==1)
    41                 x=i;
    42                 while(str1[i]!='<')
    43                 {
    44                    i--;
    45                 }
    46 
    47                 j=i+1;
    48 
    49                 while(str1[j]!='>')
    50                 {
    51                     str[k]=str1[j];
    52                     k++;
    53                     j++;
    54                 }
    55                 if(i>y)
    56                 {
    57                     while(str1[i]!='>')
    58                     {
    59                         i--;
    60                     }
    61                     j=i+1;
    62                     while(str1[j]!='<')
    63                     {
    64                         str[k]=str1[j];
    65                         k++;
    66                         j++;
    67                     }
    68                 }
    69                 else
    70                 break;
    71             }
    72             else
    73             i--;
    74         }
    75         for(x+=1;x<len1;x++)
    76         {
    77             str[k]=str1[x];
    78             k++;
    79         }
    80         for(i=0;i<len2;)
    81         {
    82 
    83             if(str2[i]=='.'&&str2[i+1]=='.'&&str2[i+2]=='.')
    84             {
    85                 for(j=0;j<k;j++)
    86                 printf("%c",str[j]);
    87                 i+=3;
    88             }
    89             else{
    90             printf("%c",str2[i]);
    91             i++;
    92             }
    93 
    94         }
    95         printf("\n");
    96 
    97     }
    98     return 0;
    99 }
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  • 原文地址:https://www.cnblogs.com/wanglin2011/p/2531447.html
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