HDU1228 字符串处理

题意：有a+b计算结果

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 1 #include<stdio.h>
 2 #include<string>
 3 #include<stdlib.h>
 4 int mp( char a[] ){
 5     if( strcmp(a,"zero")==0 )
 6         return 0;
 7     if( strcmp(a,"one")==0 )
 8         return 1;
 9     if( strcmp(a,"two")==0 )
10         return 2;
11     if( strcmp(a,"three")==0 )
12         return 3;
13     if( strcmp(a,"four")==0 )
14         return 4;
15     if( strcmp(a,"five")==0 )
16         return 5;
17     if( strcmp(a,"six")==0 )
18         return 6;
19     if( strcmp(a,"seven")==0 )
20         return 7;
21     if( strcmp(a,"eight")==0 )
22         return 8;
23     if( strcmp(a,"nine")==0 )
24         return 9;
25 }
26 int main(){
27     /*
28     mp["zero"]=0;
29     mp["one"]=1;
30     mp["two"]=2;
31     mp["three"]=3;
32     mp["four"]=4;
33     mp["five"]=5;
34     mp["six"]=6;
35     mp["seven"]=7;
36     mp["eight"]=8;
37     mp["nine"]=9;
38     */
39     char ss[8];
40     int num[3];
41     char a[1005];
42     while( gets(a)!=NULL ){
43         num[1]=num[2]=0;
44         int flag=1;
45         for( int i=0;a[i]!='=';i++ ){
46             if( a[i]==' ' )
47                 continue;
48             if( a[ i ]=='+' ){
49                 flag++;
50                 continue;
51             }
52             int cnt=0;
53             //ss.clear();
54             for( int j=i;;j++ ){
55                 if( a[j]<'a'||a[j]>'z' ){
56                     j--;
57                     i=j;
58                     //ss.at(cnt)='\0';
59                     ss[cnt]='\0';
60                     break;
61                 }
62                 //ss.at(cnt)=a[j];
63                 ss[ cnt ]=a[j];
64                 cnt++;
65             }
66             if( num[flag]==0 ) num[flag]+=mp(ss);
67             else {
68                 num[flag]*=10;
69                 num[flag]+=mp(ss);
70             }
71         }
72         if( num[1]+num[2]==0 )
73             break;
74         printf("%d\n",num[1]+num[2]);
75     }
76     return 0;
77 }

 

posted @ 2013-02-17 16:45  xxx0624  阅读(255)  评论(0)  编辑  收藏  举报