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2143 | bpr | 1 | /* Le compte est bon - version récursive |
2 | * Copyright (C) 2002 Lucas Nussbaum <lucas@lucas-nussbaum.net> |
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3 | * |
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4 | * This program is free software; you can redistribute it and/or modify |
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5 | * it under the terms of the GNU General Public License as published by |
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6 | * the Free Software Foundation; either version 2 of the License, or |
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7 | * (at your option) any later version. |
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8 | * |
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9 | * This program is distributed in the hope that it will be useful, |
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10 | * but WITHOUT ANY WARRANTY; without even the implied warranty of |
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11 | * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the |
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12 | * GNU General Public License for more details. |
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13 | * |
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14 | * You should have received a copy of the GNU General Public License |
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15 | * along with this program; if not, write to the Free Software |
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16 | * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA |
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17 | */ |
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18 | |||
19 | #include <stdlib.h> |
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20 | #include <stdio.h> |
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21 | |||
22 | typedef enum {ADD, MULT, SUB, DIV, EMPTY} operation; |
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23 | |||
24 | struct result { |
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25 | unsigned int val; /* value */ |
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26 | operation op; /* operator */ |
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27 | struct result * l; /* left operand */ |
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28 | struct result * r; /* right operand */ |
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29 | }; |
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30 | |||
31 | const operation ops[4] = { ADD, MULT, SUB, DIV }; |
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32 | const char dispops[4] = { '+', '*', '-', '/' }; |
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33 | |||
34 | /* initializes a result struct */ |
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35 | struct result * resinit() |
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36 | { |
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37 | struct result * p; |
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38 | p = (struct result *)malloc(sizeof(struct result)); |
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39 | p->l = NULL; |
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40 | p->r = NULL; |
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41 | p->op = EMPTY; |
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42 | return p; |
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43 | } |
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44 | |||
45 | /* display the result in a readable form. called recursively. */ |
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46 | void dispres(struct result * p) |
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47 | { |
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48 | if (p->op!=EMPTY) |
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49 | { |
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50 | dispres(p->l); |
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51 | dispres(p->r); |
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52 | printf("%u %c %u = %u\n", p->l->val, dispops[p->op],p->r->val, p->val); |
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53 | } |
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54 | } |
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55 | |||
56 | /* calculate the result with those 2 operands and the operator provided */ |
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57 | /* /!\ MUST HAVE p->val >= q->val */ |
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58 | struct result * add(struct result * p, struct result * q, int op) |
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59 | { |
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60 | unsigned int res = 0; |
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61 | struct result *r; |
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62 | switch(ops[op]) |
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63 | { |
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64 | case ADD: |
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65 | if (!(p->val&&q->val)) return NULL; |
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66 | res = p->val + q->val; |
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67 | break; |
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68 | case MULT: |
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69 | if ((p->val==1)||(q->val==1)) return NULL; |
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70 | res = p->val * q->val; |
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71 | break; |
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72 | case SUB: |
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73 | if (!(p->val&&q->val)) return NULL; |
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74 | res = p->val - q->val; |
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75 | break; |
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76 | case DIV: |
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77 | if (!q->val) return NULL; |
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78 | if (q->val==1) return NULL; |
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79 | if (!(p->val%q->val)) |
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80 | res = p->val / q->val; |
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81 | else |
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82 | return NULL; |
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83 | break; |
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84 | case EMPTY: |
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85 | return NULL; /* prevents a compiler warning */ |
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86 | } |
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87 | r = resinit(); |
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88 | r->op = ops[op]; |
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89 | r->l = p; |
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90 | r->r = q; |
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91 | r->val = res; |
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92 | return r; |
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93 | } |
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94 | |||
95 | struct result * best = NULL; |
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96 | int min = 10000; |
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97 | int goal; |
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98 | |||
99 | /* tests if result is better */ |
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100 | void resultest(struct result * res) |
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101 | { |
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102 | int tmp; |
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103 | if ((tmp = abs(res->val - goal)) < min) |
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104 | { |
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105 | min = tmp; |
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106 | best = res; |
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107 | if (!min) |
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108 | { |
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109 | printf("FOUND\n"); |
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110 | dispres(best); |
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111 | exit(0); |
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112 | } |
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113 | } |
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114 | } |
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115 | |||
116 | /* recursively try to find an appropriate result */ |
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117 | void compute (struct result ** base, int nb) |
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118 | { |
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119 | struct result * new[6]; |
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120 | struct result * backup1, * backup2; |
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121 | int i, j, k; |
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122 | |||
123 | /* generate a brand new array ! */ |
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124 | memcpy(new, base, sizeof(struct result *) * 6); |
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125 | if (nb > 1) |
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126 | { |
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127 | for (i = 0; i < nb - 1; i++) |
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128 | { |
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129 | for (j = i + 1; j < nb; j++) |
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130 | { |
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131 | /* now try to replace the 2 values with a combined one */ |
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132 | backup1 = new[i]; |
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133 | backup2 = new[j]; |
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134 | for (k = 0; k < 4; k++) /* for each operator */ |
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135 | { |
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136 | if ((backup1->val > backup2->val) ? (new[i] = add(backup1, backup2, k)) != NULL : (new[i] = add(backup2, backup1, k)) != NULL) |
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137 | { |
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138 | resultest(new[i]); |
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139 | new[j] = new[nb - 1]; |
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140 | compute(new, nb - 1); |
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141 | new[j] = backup2; |
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142 | } |
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143 | new[i] = backup1; |
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144 | } |
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145 | } |
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146 | } |
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147 | } |
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148 | } |
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149 | |||
150 | /* main routine */ |
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151 | int main(int argc, char * argv[]) |
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152 | { |
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153 | struct result * base[6]; |
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154 | int i; |
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155 | |||
156 | if (argc!=8) |
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157 | { |
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158 | printf("There should be 7 arguments. goal, following by the 6 numbers" |
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159 | "to use\n"); |
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160 | exit(-1); |
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161 | } |
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162 | /* we read the goal */ |
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163 | goal = atoi(argv[1]); |
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164 | /* we read the 6 base numbers */ |
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165 | for (i = 0; i < 6; i++) |
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166 | { |
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167 | base[i] = resinit(); |
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168 | base[i]->val = atoi(argv[i+2]); |
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169 | } |
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170 | |||
171 | printf("Processing "); |
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172 | for (i = 0; i < 6; i++) |
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173 | printf("%d ", base[i]->val); |
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174 | printf(". Goal : %d\n", goal); |
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175 | |||
176 | for (i = 0; i < 6; i++) |
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177 | resultest(base[i]); |
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178 | |||
179 | compute(base, 6); |
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180 | |||
181 | printf("NOTFOUND %d %d\n", best->val, min); |
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182 | dispres(best); |
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183 | exit(0); |
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184 | } |