UVa 10106 Product

高精度乘法问题，WA了两次是因为没有考虑结果为0的情况。

Product

 

 

The Problem

The problem is to multiply two integers X, Y. (0<=X,Y<10²⁵⁰)

The Input

The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer.

The Output

For each input pair of lines the output line should consist one integer the product.

Sample Input

 

12

12

2

222222222222222222222222

 

Sample Output

 

144

444444444444444444444444

AC代码：

 1 //#define LOCAL

 2 #include <iostream>

 3 #include <cstdio>

 4 #include <cstring>

 5 using namespace std;

 6 

 7 const int maxn = 265;

 8 char a[maxn], b[maxn];

 9 int x[maxn], y[maxn], mul[maxn * 2 + 10];

10 void Reverse(char s[], int l);

11 

12 int main(void)

13 {

14     #ifdef LOCAL

15         freopen("10106in.txt", "r", stdin);

16     #endif

17 

18     while(gets(a))

19     {

20         gets(b);

21         int la = strlen(a);

22         int lb = strlen(b);

23         memset(mul, 0, sizeof(mul));

24         memset(x, 0, sizeof(x));

25         memset(y, 0, sizeof(y));

26         Reverse(a, la);

27         Reverse(b, lb);

28         int i, j;

29         for(i = 0; i < la; ++i)

30             x[i] = a[i] - '0';

31         for(i = 0; i < lb; ++i)

32             y[i] = b[i] - '0';

33 

34         for(i = 0; i < lb; ++i)

35         {

36             int s = 0, c = 0;

37             for(j = 0; j < maxn; ++j)

38             {

39                 s = y[i] * x[j] + c + mul[i + j];

40                 mul[i + j] = s % 10;

41                 c = s / 10;

42             }

43         }

44 

45         for(i = maxn * 2 + 9; i >= 0; --i)

46             if(mul[i] != 0)

47                 break;

48         if(i < 0)

49             cout << 0;

50         else

51         {

52             for(; i >=0; --i)

53                 cout << mul[i];

54         }

55         cout << endl;

56     }

57     return 0;

58 }

59 //用来反转数组

60 void Reverse(char s[], int l)

61 {

62     int i;

63     char t;

64     for(i = 0; i < l / 2; ++i)

65     {

66         t = s[i];

67         s[i] = s[l - i -1];

68         s[l - i -1] = t;

69     }

70 }

代码君