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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.     scanf("%s",instr);//inputから読み込み
  12.     printf("%s\n",instr);
  13.  
  14.     // 1~2のカラーコード（抵抗値）
  15.    for (i=0; i<=2; i++){
  16.    	for (j=0; j<=9; j++){
  17.    		for (k=0;k<=2; k++){
  18.     			if( !(instr[i][k]==strc[j][k])){
  19.     				goto A1;
  20.     			}
  21.    		}
  22.     		C[i]=j;
  23.     		A1:;
  24.    	}
  25.    }
  26.  
  27.    // 3のカラーコード（倍率）
  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.  
  40.    // 4のカラーコード（精度）
  41.     i=3;
  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",10*C[0]+C[1],bai[b],pre[p]);
  54.  
  55. 	return 0;
  56. }
  57.  
Success #stdin #stdout 0.01s 5276KB
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stdin
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赤,茶,黄,金
stdout
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赤,茶,黄,金
2 0 -201120672 -201120912
抵抗値=20×10^(0)[Ω]  精度=0.00[%]
https://ideone.com/dpf0ZK
language:
C (gcc 8.3)
created:
2 days ago
visibility:
public

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