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| Rev | Author | Line No. | Line |
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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 | } |