△UVA10106

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

 1 #include<cstdio>
 2 #include<string.h>
 3 
 4 #define maxn 2005
 5 
 6 int main()
 7 {
 8     char a[maxn],b[maxn];
 9     int ans[maxn];
10     int i,j,k;
11     int la,lb;
12     while(scanf("%s%s",a,b) != EOF)
13     {
14         la=strlen(a);
15         lb=strlen(b);
16         memset(ans,0,sizeof(ans));
17         for(i=la-1;i>=0;i--)
18         {
19             for(j=lb-1,k=la-1-i;j>=0;j--)//大数相乘模拟手算，关键在于错位，即k=la-1-i
20             {
21                 ans[k++] += (a[i]-'0')*(b[j]-'0');//ans[0]存的是个位，ans[1]存的是十位，ans[2]存的是百位……
22             }
23         }
24         for(i=0;i<maxn-1;i++)
25         {
26             if(ans[i]>9)//进位
27             {
28                 ans[i+1]+=ans[i]/10;//因为是乘法，不再像加法一样是进1还是不进位了
29                 ans[i]%=10;//只留一位
30             }
31         }
32         for(i=maxn-1;i>=0;i--)
33         {
34             if(ans[i])//忽略前导0
35                 break;
36         }
37         if(i>=0)
38         {
39             for(;i>=0;i--)
40             {
41                 printf("%d",ans[i]);
42             }
43         }
44         else
45             printf("0");
46         printf("
");
47     }
48     return 0;
49 }