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Interpreter 본문
A certain computer has ten registers and 1,000 words of RAM. Each register or RAM location holds a three-digit integer between 0 and 999. Instructions are encoded as three-digit integers and stored in RAM. The encodings are as follows:
100 means halt
2dn means set register d to n (between 0 and 9)
3dn means add n to register d
4dn means multiply register d by n
5ds means set register d to the value of register s
6ds means add the value of register s to register d
7ds means multiply register d by the value of register s
8da means set register d to the value in RAM whose address is in register a
9sa means set the value in RAM whose address is in register a to the value of register s
0ds means goto the location in register d unless register s contains 0
All registers initially contain 000. The initial content of the RAM is read from standard input. The first instruction to be executed is at RAM address 0. All results are reduced modulo 1,000.
Input
The input begins with a single positive integer on a line by itself indicating the number of cases, each described as below. This is followed by a blank line, and there will be a blank line between each two consecutive inputs.
Each input case consists of up to 1,000 three-digit unsigned integers, representing the contents of consecutive RAM locations starting at 0. Unspecified RAM locations are initialized to 000.
Output
The output of each test case is a single integer: the number of instructions executed up to and including the halt instruction. You may assume that the program does halt. Separate the output of two consecutive cases by a blank line.
Sample Input
1
299
492
495
399
492
495
399
283
279
689
078
100
000
000
000
Sample Output
16
- time limit exceed 발생 ㅜ.ㅜ
- scanf를 이용해 공백 라인을 읽는 방법 참고
- time limit exceed의 원인은 switch()의 default:에서 iPC++을 안 넣어줘서
- iNumOfCases의 앞에 빈 printf("\n");는 output 형식을 맞추기 위함 ->없으면 presentation error
100 means halt
2dn means set register d to n (between 0 and 9)
3dn means add n to register d
4dn means multiply register d by n
5ds means set register d to the value of register s
6ds means add the value of register s to register d
7ds means multiply register d by the value of register s
8da means set register d to the value in RAM whose address is in register a
9sa means set the value in RAM whose address is in register a to the value of register s
0ds means goto the location in register d unless register s contains 0
All registers initially contain 000. The initial content of the RAM is read from standard input. The first instruction to be executed is at RAM address 0. All results are reduced modulo 1,000.
Input
The input begins with a single positive integer on a line by itself indicating the number of cases, each described as below. This is followed by a blank line, and there will be a blank line between each two consecutive inputs.
Each input case consists of up to 1,000 three-digit unsigned integers, representing the contents of consecutive RAM locations starting at 0. Unspecified RAM locations are initialized to 000.
Output
The output of each test case is a single integer: the number of instructions executed up to and including the halt instruction. You may assume that the program does halt. Separate the output of two consecutive cases by a blank line.
Sample Input
1
299
492
495
399
492
495
399
283
279
689
078
100
000
000
000
Sample Output
16
- time limit exceed 발생 ㅜ.ㅜ
- scanf를 이용해 공백 라인을 읽는 방법 참고
- time limit exceed의 원인은 switch()의 default:에서 iPC++을 안 넣어줘서
- iNumOfCases의 앞에 빈 printf("\n");는 output 형식을 맞추기 위함 ->없으면 presentation error
#include#include #define REG_SIZE 10 #define RAM_SIZE 1000 #define CMD_SIZE 4 int main() { int i, j, d, n, s, a, iRC, iNumOfCases, Reg[REG_SIZE], iRAMNum, iPC, iNumOfIns; char pcLine[CMD_SIZE], RAM[RAM_SIZE][CMD_SIZE]; iRC = scanf("%d%*[^\n]", &iNumOfCases); if ( iRC < 0 ) return 1; getchar(); // remove '\n' at the end of iNumOfCases getchar(); // remove blank line between iNumOfCases and first instruction //printf("NumOfCases: %d\n", iNumOfCases); for ( i = 0; i < iNumOfCases; ++i ) { //printf("------------------------------\n"); // initialize register & RAM memset(Reg, '\0', sizeof(int) * REG_SIZE); memset(RAM, '\0', sizeof(char) * RAM_SIZE * CMD_SIZE); iRAMNum = 0; memset(pcLine, '\0', CMD_SIZE); while ( ( iRC = scanf("%4[^\n]%*[^\n]", pcLine) ) >= 0 ) { if ( iRC >= 0 ) getchar(); // remove '\n' at the end of pcLine if ( 0 == strlen(pcLine) ) // blank line between instruction sets break; // goto next instructions set // copy instructions into RAM strncpy(RAM[iRAMNum], pcLine, sizeof(char) * CMD_SIZE); //printf("%s\n", pcLine); //printf("RAM[%d] = %s\n", iRAMNum, RAM[iRAMNum]); ++iRAMNum; memset(pcLine, '\0', CMD_SIZE); } iPC = 0; iNumOfIns = 0; while ( 1 ) { if ( 0 == strcmp(RAM[iPC], "100") ) { ++iNumOfIns; //printf("halt\n"); break; } switch ( RAM[iPC][0] ) { case '2': d = RAM[iPC][1] - '0'; n = RAM[iPC][2] - '0'; Reg[d] = n; ++iPC; break; case '3': d = RAM[iPC][1] - '0'; n = RAM[iPC][2] - '0'; Reg[d] += n; Reg[d] %= 1000; ++iPC; break; case '4': d = RAM[iPC][1] - '0'; n = RAM[iPC][2] - '0'; Reg[d] *= n; Reg[d] %= 1000; ++iPC; break; case '5': d = RAM[iPC][1] - '0'; s = RAM[iPC][2] - '0'; Reg[d] = Reg[s]; ++iPC; break; case '6': d = RAM[iPC][1] - '0'; s = RAM[iPC][2] - '0'; Reg[d] += Reg[s]; Reg[d] %= 1000; ++iPC; break; case '7': d = RAM[iPC][1] - '0'; s = RAM[iPC][2] - '0'; Reg[d] *= Reg[s]; Reg[d] %= 1000; ++iPC; break; case '8': d = RAM[iPC][1] - '0'; a = RAM[iPC][2] - '0'; Reg[d] = atoi(RAM[Reg[a]]); ++iPC; break; case '9': s = RAM[iPC][1] - '0'; a = RAM[iPC][2] - '0'; sprintf(RAM[Reg[a]], "%d", Reg[s]); ++iPC; break; case '0': d = RAM[iPC][1] - '0'; s = RAM[iPC][2] - '0'; if ( Reg[s] != 0 ) iPC = Reg[d]; else ++iPC; break; default: ++iPC; break; } ++iNumOfIns; //for ( j = 0; j < REG_SIZE; ++j ) // printf("%d|", Reg[j]); //printf("\n"); } if ( 0 < i ) printf("\n"); printf("%d\n", iNumOfIns); } return 0; }
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