Instruction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 163 Accepted Submission(s): 46

Problem Description

Nowadays, Jim Green has produced a kind of computer called JG. In his computer, the instruction is represented by binary code. However when we code in this computer, we use some mnemonic symbols. For example, ADD R1, R2 means to add the number in register R1 and R2, then store the result to R1. But this instruction cannot be execute directly by computer, before this instruction is executed, it must be changed to binary code which can be executed by computer. Each instruction corresponds to a 16-bit binary code. The higher 6 bits indicates the operation code, the middle 5 bits indicates the destination operator, and the lower 5 bits indicates the source operator. You can see Form 1 for more details.

15

instruction ADD Ra,Rb SUB Ra,Rb DIV Ra,Rb MUL Ra,Rb MOVE Ra,Rb SET Ra function Add the number in register Ra and Rb, then store the result to Ra. Subtract the number in register Ra to Rb, then store the result to Ra. Divide the number in register Ra by Rb, then store the result to Ra. Mulplicate the number in register Ra and Rb, then store the result to Ra. Move the number in register Rb to Ra. Set 0 to Ra.

Operation

Input

Multi test cases (about 50000), every case contains two lines.

First line contains a type sign, ‘0’ or ‘1’.

‘1’ means you should transfer an instruction into a 16-bit binary code;

‘0’ means you should transfer a 16-bit binary code into an instruction.

For the second line.

If the type sign is ‘1’, an instruction will appear in the standard form which will be given in technical specification;

Otherwise, a 16-bit binary code will appear instead.

Please process to the end of file.

[Technical Specification]

The standard form of instructions is

ADD Ra,Rb

SUB Ra,Rb

DIV Ra,Rb

MUL Ra,Rb

MOVE Ra,Rb

SET Ra

which are also listed in the Form 2.

Output

For type ‘0’，if the 16-bit binary code cannot be transferred into a instruction according to the description output “Error!” (without quote), otherwise transfer the 16-bit binary code into instruction and output the instruction in the standard form in a single line.

For type ‘1’, transfer the instruction into 16-bit binary code and output it in a single line.

Sample Input

1 ADD R1,R2 0 0000010000100010 0 1111111111111111

Sample Output

0000010000100010 ADD R1,R2 Error!

题意：将二进制编码转化成字符命令，或将字符命令转化成二进制，共有六种命令，每种命令对应一个二进制编码，见题目表格；然后操作数有31种，分别是R1~R31，对应分别二进制00000~11111；其中要注意SET命令只有一个操作数，它的二进制编码最后5位全部写为0，若SET操作输入的二进制编码后五位不全是0，则输出Error！；不满足要求的命令也输出Error！，另外操作数不能是00000.

蛋疼的一题，考虑不周，比赛没做出来。。。。代码比较乱啊啊啊啊啊

代码：

1 #include 
       
          2 #include 
        
           3 #include 
         
            4 #include 
          
             5 #include 
           
              6 #include 
            
              7 #include 
              8 #include 
              
                9 #include 
               
                 10 #include 
                
                  11 #include 
                 
                   12 #pragma comment (linker,"/STACK:102400000,102400000") 13 #define maxn 1005 14 #define MAXN 2005 15 #define mod 1000000009 16 #define INF 0x3f3f3f3f 17 #define pi acos(-1.0) 18 #define eps 1e-6 19 typedef long long ll; 20 using namespace std; 21 22 char ope[6][5]={ { "ADD"},{ "SUB"},{ "DIV"},{ "MUL"},{ "MOVE"},{ "SET"}}; 23 char code[6][7]={ { "000001"},{ "000010"},{ "000011"},{ "000100"},{ "000101"},{ "000110"}}; 24 char co[31][6]={ { "00001"} , { "00010"} , { "00011"} , { "00100"} , { "00101"} , { "00110"} , { "00111"} , { "01000"} , { "01001"} , { "01010"} 25 , { "01011"} , { "01100"} , { "01101"} , { "01110"} , { "01111"} , { "10000"} , { "10001"} , { "10010"} , { "10011"} , { "10100"} 26 , { "10101"} , { "10110"} , { "10111"} , { "11000"} , { "11001"} , { "11010"} , { "11011"} , { "11100"} , { "11101"} , { "11110"} 27 , { "11111"}}; 28 char s1[5],s2[10]; 29 30 int OP(char str[]) 31 { 32 for (int i=0;i<6;i++) 33 { 34 if (strcmp(str,ope[i])==0) 35 return i; 36 } 37 } 38 39 int main() 40 { 41 int n; 42 while (~scanf("%d",&n)) 43 { 44 if (n==1) 45 { 46 scanf("%s%s",s1,s2); 47 int x=OP(s1); 48 int dd=x; 49 printf("%s",code[x]); 50 x=s2[1]-'0'; 51 int ok=0; 52 if (isdigit(s2[2])) 53 { 54 ok=1; 55 x=x*10; 56 x=x+s2[2]-'0'; 57 } 58 x--; 59 printf("%s",co[x]); 60 if (dd==5) 61 { 62 printf("00000\n"); 63 continue; 64 } 65 if (ok) 66 { 67 x=s2[5]-'0'; 68 if (isdigit(s2[6])) 69 { 70 x=x*10; 71 x=x+s2[6]-'0'; 72 } 73 } 74 else 75 { 76 x=s2[4]-'0'; 77 if (isdigit(s2[5])) 78 { 79 x=x*10; 80 x=x+s2[5]-'0'; 81 } 82 } 83 x--; 84 printf("%s\n",co[x]); 85 } 86 else if (n==0) 87 { 88 scanf("%s",s1); 89 int len=strlen(s1); 90 int t=1; 91 int sum=0; 92 for (int i=5;i>=0;i--) 93 { 94 int x=s1[i]-'0'; 95 sum+=t*x; 96 t=t*2; 97 } 98 if (sum>6||sum==0) 99 {100 printf("Error!\n");101 continue;102 }103 sum--;104 int o=sum;105 int flag=sum;106 // printf("%s",ope[sum]);107 t=1;108 sum=0;109 for (int i=10;i>=6;i--)110 {111 int x=s1[i]-'0';112 sum+=t*x;113 t=t*2;114 }115 if (sum>31||sum==0)116 {117 printf("Error!\n");118 continue;119 }120 int aa=sum;121 // printf(" R%d",sum);122 if (flag==5)123 {124 t=1;125 sum=0;126 for (int i=15;i>=11;i--)127 {128 int x=s1[i]-'0';129 // printf("x=%d\n",x);130 sum+=t*x;131 t=t*2;132 }133 if (sum==0)134 {135 printf("%s",ope[o]);136 printf(" R%d",aa);137 printf("\n");138 continue;139 }140 else141 {142 printf("Error!\n");143 continue;144 }145 }146 t=1;147 sum=0;148 for (int i=15;i>=11;i--)149 {150 int x=s1[i]-'0';151 // printf("x=%d\n",x);152 sum+=t*x;153 t=t*2;154 }155 if (sum>31||sum==0)156 {157 printf("Error!\n");158 continue;159 }160 printf("%s",ope[o]);161 printf(" R%d",aa);162 printf(",R%d\n",sum);163 }164 }165 return 0;166 }

View Code

转载于:https://www.cnblogs.com/i8888/p/4051055.html

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