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UVa 10106 Product

2024-07-10 08:25:39   222人阅读

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

 

12122222222222222222222222222

 

Sample Output

 

144444444444444444444444444

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 LOCAL15         freopen("10106in.txt", "r", stdin);16     #endif17 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         else51         {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

 


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