Day 0x18 UVa10415 Eb Alto Saxophone Player

• Virtual Judge
• ZeroJudge

题意

• 输入一连串音名，输出各手指按下的次数
• 需要注意的有：
  1. 第一行输入整数 t 代表测资数
  2. 每笔测资输入歌曲的一连串音名，有可能为空
  3. 萨克斯风指法
     • 八个音名，十根手指
     • 大写代表高八度

     c: finger 2∼4, 7∼10
     d: finger 2∼4, 7∼9
     e: finger 2∼4, 7, 8
     f: finger 2∼4, 7
     g: finger 2∼4
     a: finger 2, 3
     b: finger 2
     C: finger 3
     D: finger 1∼4, 7∼9
     E: finger 1∼4, 7, 8
     F: finger 1∼4, 7
     G: finger 1∼4
     A: finger 1∼3
     B: finger 1∼2
     

  4. 若没用到就会放开，反之则持续按住

解法

• 先读入测资数，再用 while 回圈重复读入每笔测资的音名；会有 getchar() 是因为 gets() 会吃到输入测资数後的换行

  int Case;
  
  scanf("%d", &Case);
  getchar();
  
  while(Case--){
  
      char Note[201] = {0};
  
      gets(Note);
      ...
  }
  

• 宣告两字元阵列 current & next，分别代表当前有按下的手指与下个音要按下的手指，接着 for 回圈逐字元检查，观察各音名与对应的指法後可发现，手指刚好是由多到少且照顺序 (因为柱体越长音越低？)，因此透过 switch 的特性，故意不要 break，若是小写就用 if 把 1 (也就是 [0]) 放开即可

  char current[11] = "0000000000";
  
  for(i = 0; i < strlen(Note); i++){
      char next[11] = "0000000000";
      switch(Note[i]){
          case'c':
              next[9] = '1';
          case'd':
          case'D':
              next[8] = '1';
          case'e':
          case'E':
              next[7] = '1';
          case'f':
          case'F':
              next[6] = '1';
          case'g':
          case'G':
              next[3] = '1';
          case'a':
          case'A':
              next[2] = '1';
          case'b':
          case'B':
              next[1] = '1';
              if(Note[i] >= 'A' && Note[i] <= 'G'){
                 next[0] = '1';
              }
              break;
          case'C':
              next[2] = '1';
      }
      ...
  }
  

• 比较两字元阵列，计算新按下的有多少，最後更新当前按下的状态

  for(j = 0; j < 10; j++){
      if(current[j] == '0' && next[j] == '1'){
          count[j]++;
      }
  }
  strcpy(current, next);
  

• C code ver. 1

  #include<stdio.h>
  #include<string.h>
  
  int main(){
  
      int Case;
      int i, j;
  
      scanf("%d", &Case);
      getchar();
  
      while(Case--){
  
          char Note[201] = {0};
          char current[11] = "0000000000";
          int count[10] = {0};
  
          gets(Note);
  
          for(i = 0; i < strlen(Note); i++){
              char next[11] = "0000000000";
              switch(Note[i]){
                  case'c':
                      next[9] = '1';
                  case'd':
                  case'D':
                      next[8] = '1';
                  case'e':
                  case'E':
                      next[7] = '1';
                  case'f':
                  case'F':
                      next[6] = '1';
                  case'g':
                  case'G':
                      next[3] = '1';
                  case'a':
                  case'A':
                      next[2] = '1';
                  case'b':
                  case'B':
                      next[1] = '1';
                      if(Note[i] >= 'A' && Note[i] <= 'G'){
                         next[0] = '1';
                      }
                      break;
                  case'C':
                      next[2] = '1';
              }
  
              for(j = 0; j < 10; j++){
                  if(current[j] == '0' && next[j] == '1'){
                      count[j]++;
                  }
              }
              strcpy(current, next);
          }
  
          printf("%d", count[0]);
          for(j = 1; j < 10; j++){
              printf(" %d", count[j]);
          }
          printf("\n");
      }
  
      return 0;
  }
  

• 用建表法创建好各音名对应的指法

  char fingering[14][12] = {  "c0111001111", "d0111001110",
                              "e0111001100", "f0111001000",
                              "g0111000000", "a0110000000",
                              "b0100000000", "C0010000000",
                              "D1111001110", "E1111001100",
                              "F1111001000", "G1111000000",
                              "A1110000000", "B1100000000"};
  

• 一样的概念，先 for 回圈逐字扫，再用内层 for 检查是哪个音名，用 if 判断第一个字元 (音名存在 [0])，找到就更新即将按下的指法，第二个 for 回圈计算新按下的次数

  for(i = 0; i < strlen(Note); i++){
      char next[12] = {0};
      for(j = 0; j < 14; j++){
          if(Note[i] == fingering[j][0]){
              strcpy(next, fingering[j]);
              break;
          }
      }
  
      for(j = 1; j <= 10; j++){
          if(current[j] == '0' && next[j] == '1'){
              count[j - 1]++;
          }
      }
      strcpy(current, next);
  }
  

• C code ver. 2

  #include<stdio.h>
  #include<string.h>
  
  char fingering[14][12] = {  "c0111001111", "d0111001110",
                              "e0111001100", "f0111001000",
                              "g0111000000", "a0110000000",
                              "b0100000000", "C0010000000",
                              "D1111001110", "E1111001100",
                              "F1111001000", "G1111000000",
                              "A1110000000", "B1100000000"};
  
  int main(){
  
      int Case;
      int i, j;
  
      scanf("%d", &Case);
      getchar();
  
      while(Case--){
  
          char Note[201] = {0};
          char current[12] = "00000000000";
          int count[10] = {0};
  
          gets(Note);
  
          for(i = 0; i < strlen(Note); i++){
              char next[12] = {0};
              for(j = 0; j < 14; j++){
                  if(Note[i] == fingering[j][0]){
                      strcpy(next, fingering[j]);
                      break;
                  }
              }
  
              for(j = 1; j <= 10; j++){
                  if(current[j] == '0' && next[j] == '1'){
                      count[j - 1]++;
                  }
              }
              strcpy(current, next);
          }
  
          printf("%d", count[0]);
          for(j = 1; j < 10; j++){
              printf(" %d", count[j]);
          }
          printf("\n");
      }
  
      return 0;
  }