POJ Problem 1001 Solved!
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POJ Problem 1001 Solved!
1 #include <stdio.h>
2 #include <string.h>
3 #include <stdlib.h>
4
5 #define MAX_INPUT 7 低级错误出现了在这里！输入６个字节！ -邱国钦 10-5-7 下午10:19
6 #define MAX_LENGTH 150 第一个错误在这里，之前是 125 -邱国钦 10-5-7 下午10:20
7 #define RADIX 10
8
9 // function decleration ========================================
10 int *precise_multiplex (int A[], int B[], int C[]);
11 void set_value(int A[], const size_t LENGTH, int value);
12 void pretty_print (int A[], const size_t LENGTH,
13 const int num_of_fraction);
14
15 typedef struct input
16 {
17 char R[MAX_INPUT];
18 int n;
19 struct input *p_next; 为什么用 INPUT *p_next 编译不通过呢？ -邱国钦 10-5-7 下午10:29
20 } INPUT;
21
22 int main(int argc, char **argv)
23 {
24 INPUT *input_list = (INPUT *) malloc (sizeof (INPUT));
25 INPUT *p = input_list;
26
27 char R[MAX_INPUT];
28 int n;
29
30 int k;
31 while (scanf ("%s%d", R, &n) == 2)
32 {
33 p->p_next = (INPUT *) malloc (sizeof (INPUT));
34 p = p->p_next;
35
36 strcpy (p->R, R);
37 // for (k=0; k<MAX_INPUT; ++k)
38 // p->R[k] = R[k];
39 p->n = n;
40 p->p_next = NULL;
41 }
42
43 int A[MAX_LENGTH]; // A * B = C
44 int B[MAX_LENGTH];
45 int C[MAX_LENGTH];
46 int D[MAX_LENGTH];
47
48 int *m1, *m2, *product, *others; // multipliers and product
49 m1 = A;
50 m2 = B;
51 product = C;
52 others = D;
53
54 size_t length; // length of input string
55 int i, j; // loop variable
56
57 p = input_list->p_next;
58 while (p)
59 {
60 for (i=0; i<MAX_INPUT; ++i)
61 R[i] = p->R[i];
62 n = p->n;
63 p = p->p_next;
64
65 // n is zero:
66 if (!n)
67 {
68 printf ("1\n");
69 continue;
70 }
71
72 // initial *others* (=1) ====================================
73 set_value (others, MAX_LENGTH-1, 0);
74 others[MAX_LENGTH-1] = 1;
75
76 // store R into m1 in integer format ========================
77 set_value (m1, MAX_LENGTH, 0);
78 length = strlen (R);
79
80 // figure out total number of input digits ------------------
81 int start; // where the storage *start*
82 if (strchr (R, '.'))
83 {
84 // get ride of suffixing zeros
85 for (i=length-1; i>=0; --i)
86 {
87 if ('0' == R[i])
88 {
89 R[i] = '\0';
90 length -= 1;
91 }
92 else
93 break;
94 }
95 start = MAX_LENGTH - length + 1;
96 }
97 else
98 start = MAX_LENGTH - length;
99
100 int dot_pos = -1; // position of the dot
101 for (i=0, j=0; i<length; ++i)
102 {
103 // skip the dot
104 if ('.' == R[i])
105 {
106 dot_pos = i;
107 continue;
108 }
109 m1[start+j] = (int)(R[i] - '1' + 1);
110 ++j;
111 }
112
113 // figure out number of fraction
114 int number_of_fraction = 0;
115 if (-1 != dot_pos)
116 number_of_fraction = n * (length - dot_pos - 1); //
117
118 if (n == 1)
119 {
120 pretty_print (m1, MAX_LENGTH, number_of_fraction);
121 continue;
122 }
123
124 int *temp;
125
126 while (n>1)
127 {
128 if (n%2)
129 {
130 set_value (product, MAX_LENGTH, 0);
131 temp = others;
132 others = precise_multiplex (m1, others, product);
133 product = temp;
134 // copy_array (B, others, MAX_LENGTH);
135 n -= 1;
136 continue;
137 }
138
139 // copy m1 to m2; reset product
140 for (i=0; i<MAX_LENGTH; ++i)
141 {
142 m2[i] = m1[i];
143 product[i] = 0;
144 }
145
146 temp = m1;
147 m1 = precise_multiplex (m1, m2, product);
148 product = temp;
149
150 n /= 2;
151 }
152
153 // compute final result
154 set_value (product, MAX_LENGTH, 0);
155 precise_multiplex (m1, others, product);
156
157 // print the result
158 pretty_print (product, MAX_LENGTH, number_of_fraction);
159 }
160
161 return 0;
162 }
163
164 // @func: set_value
165 // set all elements in array *A* of length *LENGTH* to *value*
166 // @param A: array to be set
167 // @param LENGTH: length of *A*
168 // @param value: the desire value
169 // TODO: may exsits a better way to do this
170 void set_value(int A[], const size_t LENGTH, int value)
171 {
172 int i;
173 for (i=0; i<LENGTH; ++i)
174 {
175 A[i] = value;
176 }
177 }
178
179 // @func: precise_multiplex
180 // calculate A*Bmalloc, and store the product into C
181 // @param A, B, C: integer array storing digits of radix-10
182 // @return: the product of A*B
183 int *precise_multiplex (int A[], int B[], int C[])
184 {
185 // find the length of A and B
186 int start_A, start_B;
187 for (start_A=0; start_A<MAX_LENGTH; ++start_A)
188 {
189 if (A[start_A])
190 break;
191 }
192 for (start_B=0; start_B<MAX_LENGTH; ++start_B)
193 {
194 if (B[start_B])
195 break;
196 }
197
198 int length_A, length_B;
199 length_A = MAX_LENGTH - start_A;
200 length_B = MAX_LENGTH - start_B;
201
202 int temp, remainder, carry;
203 int i, j, k;
204
205 // multiplex every digit
206 k = 0; // 位置标志
207 for (i=MAX_LENGTH-1; i>=start_B; --i)
208 {
209 for (j=MAX_LENGTH-1; j>=start_A; --j)
210 {
211 temp = B[i] * A[j];
212 carry = temp / RADIX;
213 remainder = temp % RADIX;
214 // TODO: overflow?
215 C[j-k] += remainder;
216 C[j-1-k] += carry;
217 }
218 k++;
219 }
220
221 // 在最后处理每个数组元素
222 int start_C;
223 for (start_C=0; start_C<MAX_LENGTH; ++start_C)
224 {
225 if (C[start_C])
226 break;
227 }
228 for (i=MAX_LENGTH-1; i>=start_C; --i)
229 {
230 carry = C[i] / RADIX;
231 C[i] = C[i] % RADIX;
232 C[i-1] += carry;
233 }
234
235 return C;
236 }
237
238 void pretty_print (int A[], const size_t LENGTH,
239 const int num_of_fraction)
240 {
241 int i;
242 // find the first nonzero element
243 for (i=0; i<LENGTH; ++i)
244 if (A[i])
245 break;
246
247 if ((LENGTH-i) < num_of_fraction)
248 {
249 printf (".");
250 i = LENGTH - num_of_fraction;
251 for (i; i<LENGTH; ++i)
252 printf ("%d", A[i]);
253 }
254 else
255 {
256 int num_of_int = LENGTH - num_of_fraction;
257 for (i; i<LENGTH; ++i)
258 {
259 if (i == num_of_int)
260 printf (".");
261 printf ("%d", A[i]);
262 }
263 }
264 printf ("\n");
265 }