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  1. #include <stdio.h>
  2.  
  3. int main(void) {
  4.     char strc[12][3] = {"黒", "茶", "赤", "橙", "黄", "緑", "青", "紫", "灰", "白", "金", "銀"};
  5.     char instr[10][3];  
  6.     int bai[12] = {0, 1, 2, 3, 4, 5, 6, 7, 0, -3, -1, -2};  
  7.     double pre[12] = {0, 1, 2, 0.05, 0, 0.5, 0.25, 0.1, 0, 0, 5, 10}; 
  8.     int i, j, k, b, p;
  9.     int C[4];  
  10.  
  11.     strcpy(instr[0], "赤");  
  12.     strcpy(instr[1], "茶");  
  13.     strcpy(instr[2], "黄");  
  14.     strcpy(instr[3], "金");  
  15.  
  16.     for (i = 0; i <= 2; i++) {
  17.         for (j = 0; j <= 11; j++) {
  18.             for (k = 0; k <= 2; k++) {
  19.                 if (!(instr[i][k] == strc[j][k])) {
  20.                     goto A1;
  21.                 }
  22.             }
  23.             C[i] = j;
  24.             A1:;
  25.         }
  26.     }
  27.  
  28.     i = 2;
  29.     for (j = 0; j <= 11; j++) {
  30.         for (k = 0; k <= 2; k++) {
  31.             if (!(instr[i][k] == strc[j][k])) {
  32.                 goto A2;
  33.             }
  34.         }
  35.         b = j;
  36.         A2:;
  37.     }
  38.  
  39.     i = 3;
  40.     for (j = 0; j <= 11; j++) {
  41.         for (k = 0; k <= 2; k++) {
  42.             if (!(instr[i][k] == strc[j][k])) {
  43.                 goto A3;
  44.             }
  45.         }
  46.         p = j;
  47.         A3:;
  48.     }
  49.  
  50.     printf("%d %d %d %d\n", C[0], C[1], b, p);
  51.     printf("抵抗値=%d×10^(%d)[Ω]  精度=%.2f[%%]\n", 10 * C[0] + C[1], bai[b], pre[p]);
  52.  
  53.     return 0;
  54. }
Success #stdin #stdout 0s 5280KB
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2 1 4 10
抵抗値=21×10^(4)[Ω]  精度=5.00[%]
https://ideone.com/VEuJpa
language:
C (gcc 8.3)
created:
4 days ago
visibility:
public

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