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BestCoder15 1002.Instruction(hdu 5083) 解题报告

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5083

题目意思：如果给出 instruction 就需要输出对应的 16-bit binary code，给出16-bit binary code 就需要输出对应的instruction。

由于不会截取的技巧，代码量非常可观 = =，所以说，一直很讨厌做模拟题！！！

留下这代码，纪念一个代码还是不够精简的自己！！！内存和时间还能接受，也比较容易理解，不过好多重复代码= =。以下这个代码可以忽略，改到晕= =。之后会补上简单容易理解版滴......

一...场.......噩..........梦！！！

78Ms 284K 5708B

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstdlib>
  4 #include <cstring>
  5 using namespace std;
  6 
  7 const int maxn = 20 + 5;
  8 const int N = 10 + 2;
  9 char instruct[7][N] = {"0", "ADD", "SUB", "DIV", "MUL", "MOVE", "SET"};
 10 char num_instruct[7][N] = {"0", "000001", "000010", "000011", "000100", "000101", "000110"};
 11 
 12 char R[32][N] = {"0", "R1", "R2", "R3", "R4", "R5", "R6", "R7", "R8",
 13                 "R9", "R10", "R11", "R12", "R13", "R14", "R15", "R16",
 14                 "R17", "R18", "R19", "R20", "R21", "R22", "R23", "R24",
 15                 "R25", "R26", "R27", "R28", "R29", "R30", "R31"};
 16 char num_R[32][N] = {"0", "00001", "00010", "00011", "00100", "00101", "00110",
 17                     "00111", "01000", "01001", "01010", "01011", "01100", "01101",
 18                     "01110", "01111", "10000", "10001", "10010", "10011", "10100",
 19                     "10101", "10110", "10111", "11000", "11001", "11010", "11011",
 20                     "11100", "11101", "11110", "11111"};
 21 
 22 char s1[N], s2[maxn];
 23 char s[maxn];
 24 
 25 int main()
 26 {
 27     #ifndef ONLINE_JUDGE
 28         freopen("ljy.txt", "r", stdin);
 29     #endif
 30 
 31     int ask;
 32     while (scanf("%d", &ask) != EOF)
 33     {
 34         char tmp1[N], tmp2[N];
 35         if (ask == 1)
 36         {
 37             int i;
 38             scanf("%s%s", s1, s2);
 39             for (i = 1; i < 7; i++)
 40             {
 41                 if (!strcmp(s1, instruct[i]))
 42                 {
 43                     printf("%s", num_instruct[i]);
 44                     break;
 45                 }
 46             }
 47             if (!strcmp(s1, "SET"))     // SET指令单独处理
 48             {
 49                 for (i = 0; i < 32; i++)
 50                 {
 51                     if (!strcmp(s2, R[i]))
 52                     {
 53                         printf("%s00000
", num_R[i]);
 54                         break;
 55                     }
 56                 }
 57             }
 58             else
 59             {
 60                 int l1 = 0;
 61                 int len = strlen(s2);
 62                 for (i = 0; i < len; i++)
 63                 {
 64                     if (s2[i] == ',')
 65                         break;
 66                     tmp1[l1++] = s2[i];
 67                 }
 68                 tmp1[l1] = '';      // 截取Rdestination
 69                 int r = i+1;
 70                 for (i = 0; i < 32; i++)
 71                 {
 72                     if (!strcmp(tmp1, R[i]))
 73                     {
 74                         printf("%s", num_R[i]);
 75                         break;
 76                     }
 77                 }
 78                 int l2 = 0;
 79                 for (i = r; i < len; i++)
 80                     tmp2[l2++] = s2[i];
 81                 tmp2[l2] = '';     // 截取Rsource
 82                 for (int i = 1; i < 32; i++)
 83                 {
 84                     if (!strcmp(tmp2, R[i]))
 85                     {
 86                         printf("%s
", num_R[i]);
 87                         break;
 88                     }
 89                 }
 90             }
 91         }
 92         else
 93         {
 94             scanf("%s", s);
 95             char ans[1][maxn];
 96             int len1 = strlen(s);
 97             int l1 = 0;
 98             for (int i = 0; i < 6; i++)
 99                 tmp1[l1++] = s[i];
100             tmp1[l1] = '';
101             bool flag = false;
102             for (int i = 1; i < 7; i++)
103             {
104                 if (!strcmp(tmp1, num_instruct[i]))
105                 {
106                     strcpy(ans[0], instruct[i]);   // 有可能是SET指令，这要继续往后看判断
107                     flag = true;
108                     break;
109                 }
110              }
111             if (!flag)     // 找不到指令匹配
112                 printf("Error!
");
113             else
114             {
115                 int l1 = 0;
116                 for (int i = 6; i < 11; i++)
117                     tmp1[l1++] = s[i];
118                 tmp1[l1] = '';    // Rdestination
119                 
120                 int l2 = 0;
121                 for (int i = 11; i < 16; i++)
122                     tmp2[l2++] = s[i];
123                 tmp2[l2] = '';     // Rsource
124 
125                 if (!strcmp(ans[0], "SET"))    
126                 {
127                     if (!strcmp(tmp2, "00000") && strcmp(tmp1, "00000"))  // 符合条件的形式：000110?????00000
128                     {
129                         for (int i = 1; i < 32; i++)
130                         {
131                             if (!strcmp(tmp1, num_R[i]))
132                             {
133                                 printf("SET %s
", R[i]);
134                                 break;
135                             }
136                         }
137                     }
138                     else
139                         printf("Error!
");
140                 }
141 
142                 else
143                 {
144                     if (!strcmp(tmp2, "00000") || !strcmp(tmp1, "00000"))   // 不合法的Registers
145                         printf("Error!
");
146                     else
147                     {
148                         printf("%s ", ans[0]);    // 除SET之外的其他指令
149                         for (int i = 1; i < 32; i++)
150                         {
151                             if (!strcmp(tmp1, num_R[i]))
152                             {
153                                 printf("%s,", R[i]);
154                                 break;
155                             }
156                         }
157                         for (int i = 1; i < 32; i++)
158                         {
159                             if (!strcmp(tmp2, num_R[i]))
160                             {
161                                 printf("%s
", R[i]);
162                                 break;
163                             }
164                         }
165                     }
166                 }
167             }
168         }
169     }
170     return 0;
171 }

View Code

简化版： 62Ms 292K 2787B

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

const int N = 10 + 2;
char instruct[7][N] = {"0", "ADD", "SUB", "DIV", "MUL", "MOVE", "SET"};

char s1[2*N], s2[2*N];
char ans[2*N];

inline void ins_to_binary(int end, int st, int a)
{
    for (int i = end; i >= st; i--)
    {
        ans[i] = a % 2 + '0';
        a >>= 1;
    }
}

inline void binary_to_ins(int end, int st, int &a)
{
    int k = 1;
    for (int i = end; i >= st; i--)
    {
        a += (s1[i] - '0') * k;
        k <<= 1;
    }
}

int main()
{
    int ask, a1, a2, a3;
    while (scanf("%d", &ask) != EOF)
    {
        if (ask == 1)
        {
            scanf("%s%s", s1, s2);
            for (int i = 1; i <= 6; i++)
            {
                if (strcmp(s1, instruct[i]) == 0)
                {
                    a1 = i;
                    break;
                }
            }
            if (a1 == 6)   // SET 指令
                a3 = 0;
            else
            {
                int k = 1;
                a3 = 0;
                for (int i = strlen(s2)-1; i > 0; i--)  // R2
                {
                    if (s2[i] == 'R')
                        break;
                    else
                    {
                        a3 += (s2[i] - '0') * k;
                        k *= 10;
                    }
                }
            }
            if (s2[2] == ',' || s2[2] == '')   // R1 为个位数
                a2 = s2[1] - '0';
            else
                a2 = (s2[1]-'0') * 10 + s2[2]-'0';   // R1 为十位数

            ins_to_binary(15, 11, a3);
            ins_to_binary(10, 6, a2);
            ins_to_binary(5, 0, a1);
            ans[16] = '';
            printf("%s
", ans);
        }
        else
        {
            scanf("%s", s1);
            a1 = a2 = a3 = 0;
            binary_to_ins(5, 0, a1);
            if (a1 > 6 || a1 == 0)   // 找不到合适指令
                printf("Error!
");
            else
            {
                binary_to_ins(15, 11, a3);   // SET指令R2!=0
                if (a1 == 6 && a3 > 0)
                    printf("Error!
");
                else
                {
                    binary_to_ins(10, 6, a2);
                    if (a2 == 0 || a3 == 0 && a1 != 6)     // 除SET其他指令R1!=0
                        printf("Error!
");
                    else if (a1 == 6)
                        printf("SET R%d
", a2);
                    else
                        printf("%s R%d,R%d
", instruct[a1], a2, a3);
                }
            }
        }
    }
    return 0;
}

View Code

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原文地址：https://www.cnblogs.com/windysai/p/4053254.html