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  1. #include <stdio.h>
  2.  
  3. int main(void) {
  4.  
  5. int num1;//原進位制
  6. scanf("%d",&num1);
  7.  
  8. int a;//數字有幾個
  9. scanf("%d",&a);
  10.  
  11. //依序將原數字存入陣列中
  12. int arr[a];
  13. for(int i=1;i<=a;i++){
  14.     scanf("%d", &arr[i]);
  15. }
  16.  
  17.  
  18. int num2;//目標進位制
  19. scanf("%d",&num2);  
  20.  
  21. //最後數字
  22. long final;
  23. final=0;
  24.  
  25. //2轉10
  26. if((num1==2)&&(num2==10)){
  27.     for(int j =1;j<=a;j++){
  28.         final=final+arr[j]*pow(2,a-j);
  29.     }
  30. }
  31.  
  32. //8轉10
  33. if((num1==8)&&(num2==10)){
  34.     for(int j =1;j<=a;j++){
  35.         final=final+arr[j]*pow(8,a-j);
  36.     }
  37. }
  38.  
  39. //10轉8或10轉2
  40. if(((num1==10)&&(num2==8))||((num1==10)&&(num2==2))){
  41.  
  42.     //原進位制之數字(合併)
  43.     int orginal;
  44.     orginal=0;
  45.  
  46.     int temp;
  47.  
  48.     //將分散在各行的10進制的數字合併
  49.     for(int i=1;i<=a;i++){
  50.         int j=a-i;
  51.         temp=1;
  52.         while(j>0){
  53.         	temp=temp*10;
  54.         	j=j-1;
  55.         }
  56.         orginal=orginal+arr[i]*temp;
  57.     } 
  58.  
  59.  
  60. //取餘數
  61.  
  62. int count=0;//計算位數
  63.  
  64.     while(orginal !=0){
  65.         final=final+(orginal%num2)*pow(10,count);
  66.         orginal=orginal/num2;
  67.         count=count+1;
  68.         }
  69.  
  70. }    
  71.  
  72.  
  73. //8轉2(先轉10再轉2)
  74. if((num1==8)&&(num2==2)){
  75.  
  76.     int orginal;
  77.     orginal=0;
  78.  
  79.     //先轉為10進位
  80.     for(int j =1;j<=a;j++){
  81.         orginal=orginal+arr[j]*pow(8,a-j);
  82.     }
  83.  
  84.  
  85.     //取餘數
  86.     int count=0;//計算位數
  87.  
  88.     while(orginal !=0){
  89.         final=final+(orginal%num2)*pow(10,count);	
  90.         orginal=orginal/num2;
  91.         count=count+1;
  92.         }
  93.  
  94. }
  95.  
  96. //2轉8(先轉10再轉8)
  97. if((num1==2)&&(num2==8)){
  98.  
  99.     int orginal;
  100.     orginal=0;
  101.  
  102.     for(int j =1;j<=a;j++){
  103.         orginal=orginal+arr[j]*pow(2,a-j);
  104.     }
  105.  
  106.  
  107. //取餘數
  108. int count=0;//計算位數
  109.  
  110.     while(orginal !=0){
  111.         final=final+(orginal%num2)*pow(10,count);	
  112.         orginal=orginal/num2;
  113.         count=count+1;
  114.         }
  115.  
  116. }
  117.  
  118.  
  119. long m=final;
  120. int n=0;
  121. while (m!=0){
  122.     m=m/10;
  123.     n=n+1;
  124. }
  125.  
  126. int o=final;
  127.  
  128. for (int k= n-1;k>=0;k--){
  129.     o=final/pow(10,k);
  130.     printf("%d\n", o);
  131.     final=final-o*pow(10,k);
  132.  
  133. }
  134.  
  135.  
  136. return 0;
  137. }
  138.  
  139.  
Success #stdin #stdout 0s 4360KB
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stdin
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10
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stdout
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1
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https://ideone.com/3fLbgH
language:
C (gcc 8.3)
created:
5 years ago
visibility:
public

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