1 /*A Famous Music Composer  2 时间限制：1000 ms  |  内存限制：65535 KB  3 难度：1  4 描述  5 Mr. B is a famous music composer. One of his most famous work was his set of preludes. These 24 pieces span the 24 musical keys   6 (there are musically distinct 12 scale notes, and each may use major or minor tonality). The 12 distinct scale notes are:   7  A        A#=Bb     B           C          C#=Db    D          D#=Eb     E          F           F#=Gb     G          G#=Ab  8   9 Five of the notes have two alternate names, as is indicated above with equals sign. Thus, there are 17 possible names of scale notes, but 10  only 12 musically distinct notes. When using one of these as the keynote for a musical key, we can further distinguish between major and 11   minor tonalities. This gives 34 possible keys, of which 24 are musically distinct.  12 In naming his preludes, Mr. B used all the keys except the following 10, which were named instead by their alternate names:  13  Ab minor     A# major    A# minor     C# major     Db minor 14  D# major     D# minor    Gb major     Gb minor     G# major  15 Write a program that, given the name of a key, give an alternate name if it has one, or report the key name is unique.  16 输入 17 Each test case is described by one line having the format "note tonality", where "note" is one of the 17 names for the scale notes given above, 18  and "tonality" is either "major" or "minor" (quotes for clarify). 19 输出 20 For each case output the required answer, following the format of the sample. 21 样例输入 22 Ab minor 23 D# major 24 G minor 25 样例输出 26 Case 1: G# minor 27 Case 2: Eb major 28 Case 3: UNIQUE 29 来源 30 hdu 31 上传者 32 李如兵 33 */  34 #include
     
       35 #include
      
        36 int main() 37 { 38     char s[10], c[10],s1[]="A",s2[]="B",s3[]="C",s4[]="D",s5[]="E",s6[]="F",s7[]="G",c1[]="A#",c2[]="Bb",c3[]="C#",c4[]="Db",c5[]="D#",c6[]="Eb",c7[]="F#",c8[]="Gb",c9[]="G#",c10[]="Ab"; 39     int i=1; 40     memset(s,'\0',10); 41     memset(c,'\0',10); 42     while(scanf("%s%s",s,c) != EOF ) 43     { 44         getchar(); 45         if( strcmp(s,s1) == 0 || strcmp(s,s2) == 0 || strcmp(s,s3) == 0 || strcmp(s,s4) == 0 || strcmp(s,s5) == 0 || strcmp(s,s6) == 0 || strcmp(s,s7) == 0) 46         { 47             printf("Case %d: UNIQUE\n",i++); 48             continue; 49         } 50         else if( strcmp(s,c1) == 0 ) 51         { 52             strcpy(s,c2); 53             printf("Case %d: %s %s\n",i++,s,c); 54             continue; 55         } 56         else if( strcmp(s,c2) == 0 )  57         { 58             strcpy(s,c1); 59             printf("Case %d: %s %s\n",i++,s,c); 60             continue; 61         } 62         else if( strcmp(s,c3) == 0 ) 63         { 64             strcpy(s,c4); 65             printf("Case %d: %s %s\n",i++,s,c); 66             continue; 67         } 68         else if( strcmp(s,c4) == 0 ) 69         { 70             strcpy(s,c3); 71             printf("Case %d: %s %s\n",i++,s,c); 72             continue; 73         } 74         else if( strcmp(s,c5) == 0 ) 75         { 76             strcpy(s,c6); 77             printf("Case %d: %s %s\n",i++,s,c); 78             continue; 79         } 80         else if( strcmp(s,c6) == 0 ) 81         { 82             strcpy(s,c5); 83             printf("Case %d: %s %s\n",i++,s,c); 84             continue; 85         } 86         else if( strcmp(s,c7) == 0 ) 87         { 88             strcpy(s,c8); 89             printf("Case %d: %s %s\n",i++,s,c); 90             continue; 91         } 92         else if( strcmp(s,c8) == 0 ) 93         { 94             strcpy(s,c7); 95             printf("Case %d: %s %s\n",i++,s,c); 96             continue; 97         } 98         else if( strcmp(s,c9) == 0 ) 99         {100             strcpy(s,c10);101             printf("Case %d: %s %s\n",i++,s,c);102             continue;103         }104         else if( strcmp(s,c10) == 0 )105         {106             strcpy(s,c9);107             printf("Case %d: %s %s\n",i++,s,c);108             continue;109         }110     }111     return 0;112 }