FOJ 1024 Simple Computer

搞了一个早上，代码有点搓。。。。

由于判断while退出的条件是HLT，而我却画蛇添足加了个PC>32的条件。

导致了几次的Wrong Answer

 

其实题目很简单，就是实现一些简单的运算罢了，主要还是对题目的理解。

 

下面是指令的翻译

 

000xxxxx   STA x   把运算器的值存入内存x store the value of the accu into memory byte x
001xxxxx   LDA x   从内存地址x载入值到运算器load the value of memory byte x into the accu
010xxxxx   BEQ x   如果运算器的值为0，那么载入内存地址x到运算器if the value of the accu is 0 load the value x into the pc
011-----   NOP     空操作 no operation
100-----   DEC     运算器减1 subtract 1 from the accu
101-----   INC     运算器加1 add 1 to the accu
110xxxxx   JMP x   载入内存地址到 pc load the value x into the pc
111-----   HLT     结束程序 terminate program

 

学过组成原理的同学都知道,x86机器内存中数据的存放是小顶机模型，也就是lower-to-higher

但是这道题目不同，是higher-to-lower.这个要注意。

 

#include <stdio.h>#include <string.h>#include <math.h>void getinstructions(char a[], char b[]){b[0] = a[0];b[1] = a[1];b[2] = a[2];b[3] = '/0';}void getoperand(char a[], char b[]){b[0] = a[3];b[1] = a[4];b[2] = a[5];b[3] = a[6];b[4] = a[7];b[5] = '/0';}void convto10(char a[],int lena,int &b,int jinzhi){int i;b = 0;int c;for (i=0;i<lena;i++){c = pow(jinzhi, lena-i-1);b += (a[i] - '0')*c;}}void tomem(char a[], int len, char mem[]){int i,j;for (i=0;i<8-len;i++)mem[i] = '0';j=0;for (;i<8;i++,j++)mem[i] = a[j];mem[i] = '/0';}void conv(int m, int n, char ret[]) { int k,l,i; int arr[50]; int count; count = 0; k=m; while (1) { l = k%n; if (k == 0) break; if (l>=10) arr[count] = 'A'+l-10; else arr[count] = l + '0'; k /= n; count++; } for (i=0;i<count;i++) ret[i] = arr[count-i-1]; ret[i] = '/0'; return; }int main(){char oper[20];char inst[20];int jinzhi10;int i;int tmp;char ret[20];char ret2[20];char mem[32][20];int pc;int add;while (scanf("%s", mem[0])!=EOF){pc = 0;add = 0;for (i=1;i<32;i++)scanf("%s", mem[i]);while (1){getoperand(mem[pc], oper);getinstructions(mem[pc], inst);//操作数的10进制形式convto10(oper,5, jinzhi10, 2);//printf("%d %d ", pc, add);if (strncmp(inst, "000", 3) == 0){//把运算器的值转换为2进制conv(add, 2, ret);//把二进制的值写入内存tomem(ret, strlen(ret), mem[jinzhi10]);/*printf("STA/n");*/} else if (strncmp(inst, "001", 3) == 0){//把内存地址x的值转换为10进制convto10(mem[jinzhi10], 8, tmp, 2);//载入运算器add = tmp;/*printf("LDA/n");*/} else if (strncmp(inst, "010", 3) == 0){//如果运算器的值为0if (add == 0){//载入运算器pc = jinzhi10;/*printf("BEQ/n");*/continue;}/*printf("BEQ/n");*/} else if (strncmp(inst, "011", 3) == 0){/*printf("NOP/n");*///NOP} else if (strncmp(inst, "100", 3) == 0){if (add == 0)add = 0xFF;else--add;/*printf("DEC/n");*/} else if (strncmp(inst, "101", 3) == 0){add ++;if (add > 255)add = 0;/*printf("INC/n");*/} else if (strncmp(inst, "110", 3) == 0){pc = jinzhi10;/*printf("JMP/n");*/continue;} else if (strncmp(inst, "111", 3) == 0){/*printf("HLT/n");*/break;}pc++;pc %=32;}conv(add, 2, ret);tomem(ret, strlen(ret), ret2);printf("%s/n", ret2);}return 0;}

 

另外加个比较good一些的代码。

这代码主要通过移位运算来做。比我那个字符串处理的简单多了。。。

 

#define memsize 32 #define EA 0x1F #define opcode 0xE0 #include <stdio.h> typedef unsigned char byte; byte memory[memsize]; byte accu,pc; int main() { int k; while(1) { int j,i; for (j=0;j<memsize;j++) { byte unit=0; char c[9]; if (scanf("%s",c)==EOF) return 0; for (i=0;i<8;i++) { unit*=2; unit+=c[i]-'0'; } memory[j]=unit; } pc=accu=0; while(1) { byte instruction=memory[pc]; pc=(pc+1) % memsize; byte code=(instruction & opcode)>>5; switch (code) { case 0: memory[instruction & EA]=accu; break; case 1: accu=memory[instruction & EA]; break; case 2: if (accu==0) pc=instruction & EA; break; case 3: break; case 4: accu--; break; case 5: accu++; break; case 6: pc=instruction & EA; break; case 7: goto Exit; } } Exit: for (k=0;k<8;k++) { putchar(((accu & 0x80)>>7)+'0'); accu<<=1; } putchar('/n'); } }