1027 大数乘法

Posted 2020-08-01 clairvoyant

给出2个大整数A,B，计算A*B的结果。

Input

第1行：大数A 第2行：大数B (A,B的长度 <= 1000，A,B >= 0）

Output

输出A＊Ｂ

Input示例

123456

234567

Output示例

28958703552

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 
 5 int judge(char* a)
 6 {
 7     int i;
 8     for (i = 0; i < strlen(a); ++i) {
 9         if (a[i] != ‘0‘)
10             return 0;
11     }
12     return 1;
13 }
14 
15 void multiplicative(char* a, char* b)
16 {
17     int lena = strlen(a), lenb = strlen(b);
18     int len = lena + lenb;
19     int i, j;
20     int a1[1005]={0}, b1[10005]={0}, r[10005]={0};
21     
22     for (i = 0; i < lena; ++i)
23         a1[i] = a[lena-i-1] - ‘0‘;
24     for (i = 0; i < lenb; ++i)
25         b1[i] = b[lenb-i-1] - ‘0‘;
26     for (i = 0; i < lena; ++i) {
27         for (j = 0; j < lenb; ++j) {
28             r[i+j] += a1[i] * b1[j];
29         }
30     }
31     for (i = 0; i < len; ++i) {
32         r[i+1] += r[i] / 10;
33         r[i] = r[i] % 10;
34     }
35     while (!r[len-1])
36         len--;
37     for (i = 0; i < len; ++i)
38         a[i] = r[len-i-1] + ‘0‘;
39     if (judge(a))
40         strcpy(a, "0");
41     printf("%s", a);
42 }
43 
44 int main ()
45 {
46     char A[10005], B[10005];
47     int fa, fb;
48     scanf("%s%s", A, B);
49     multiplicative(A, B);
50     return 0;
51 }