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  1. #include <stdio.h>
  2. #include <string.h>
  3.  
  4. int main(void) {
  5.     char strc[12][3] = {"黒", "茶", "赤", "橙", "黄", "緑", "青", "紫", "灰", "白", "金", "銀"};
  6.     char instr[10][3];  
  7.     int bai[12] = {0, 1, 2, 3, 4, 5, 6, 7, 0, -3, -1, -2};
  8.     double pre[12] = {0, 1, 2, 0.05, 0, 0.5, 0.25, 0.1, 0, 0, 5, 10}; 
  9.     int i, j, k, b, p;
  10.     int C[5];  
  11.  
  12.     strcpy(instr[0], "紫");  
  13.     strcpy(instr[1], "黄");  
  14.     strcpy(instr[2], "橙");  
  15.     strcpy(instr[3], "赤");  
  16.     strcpy(instr[4], "茶");  
  17.  
  18.     for (i = 0; i <= 2; i++) {
  19.         for (j = 0; j <= 11; j++) {
  20.             for (k = 0; k <= 2; k++) {
  21.                 if (!(instr[i][k] == strc[j][k])) {
  22.                     goto A1;
  23.                 }
  24.             }
  25.             C[i] = j;
  26.             A1:;
  27.         }
  28.     }
  29.  
  30.     i = 3;
  31.     for (j = 0; j <= 11; j++) {
  32.         for (k = 0; k <= 2; k++) {
  33.             if (!(instr[i][k] == strc[j][k])) {
  34.                 goto A2;
  35.             }
  36.         }
  37.         b = j;
  38.         A2:;
  39.     }
  40.  
  41.     i = 4;
  42.     for (j = 0; j <= 11; j++) {
  43.         for (k = 0; k <= 2; k++) {
  44.             if (!(instr[i][k] == strc[j][k])) {
  45.                 goto A3;
  46.             }
  47.         }
  48.         p = j;
  49.         A3:;
  50.     }
  51.  
  52.     printf("%d %d %d %d\n", C[0], C[1], b, p);
  53.     printf("抵抗値=%d×10^(%d)[Ω]  精度=%.2f[%%]\n", 100 * C[0] + 10 * C[1] + C[2], bai[b], pre[p]);
  54.  
  55.     return 0;
  56. }
  57.  
Success #stdin #stdout 0s 5264KB
comments (?)
stdin 
Standard input is empty
stdout 
7 4 2 1
抵抗値=743×10^(2)[Ω]  精度=1.00[%]
https://ideone.com/9fayQ7
language:
C (gcc 8.3)
created:
3 days ago
visibility:
public

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