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 }