1 /*
2 POJ2389 Bull Math
3 http://poj.org/problem?id=2389
4 高精度乘法
5 *
6 */
7 #include <cstring>
8 #include <cstdio>
9 #include <algorithm>
10 using namespace std;
11 const int Nmax=1000;
12 struct BigInt
13 {
14 int a[Nmax];
15 int n;
16 void init()
17 {
18 for(int i=0;i<Nmax;i++)
19 a[i]=0;
20 }
21 BigInt()
22 {
23 init();
24 n=0;
25 }
26 BigInt(int _a[],int _n)
27 {
28 init();
29 n=_n;
30 //for(int i=0;i<Nmax;i++)
31 //a[i]=0;
32 for(int i=1;i<=n;i++)
33 a[i]=_a[i];
34 maintain();
35 }
36 BigInt(char s[])
37 {
38 init();
39 n=strlen(s+1);
40 for(int i=1;i<=n;i++)
41 a[i]=s[n-i+1]-'0';
42 maintain();
43 }
44 BigInt(long long x)
45 {
46 init();
47 n=0;
48 while(x>0LL)
49 {
50 a[++n]=x%10LL;
51 x/=10LL;
52 }
53 }
54 BigInt(int x)
55 {
56 init();
57 n=0;
58 while(x>0)
59 {
60 a[++n]=x%10;
61 x/=10;
62 }
63 }
64 void read()
65 {
66 init();
67 n=0;
68 char c=getchar();
69 if(c==-1)
70 return;
71 while(c==' ' || c=='\n' )
72 {
73 c=getchar();
74 if(c==-1)
75 break;
76 }
77 int num[Nmax];
78 while(c!=' ' && c!='\n' && c!=-1)
79 {
80 num[++n]=c-'0';
81 c=getchar();
82 }
83 for(int i=1;i<=n;i++)
84 a[i]=num[n-i+1];
85 maintain();
86 }
87 void print()
88 {
89 if(n==0)
90 printf("0");
91 for(int i=n;i>=1;i--)
92 printf("%d",a[i]);
93 }
94 void maintain()
95 {
96 for(int i=1;i<=n;i++)
97 {
98 a[i+1]+=a[i]/10;
99 a[i]%=10;
100 }
101 int j=n+1;
102 while(a[j]!=0)
103 {
104 a[j+1]+=a[j]/10;
105 a[j]%=10;
106 j++;
107 }
108 n=j-1;
109 while(a[n]==0 && n>1)
110 n--;
111 }
112 friend BigInt operator + (BigInt a,BigInt b)
113 {
114 int len=max(a.n,b.n);
115 BigInt ans;
116 ans.n=len;
117 for(int i=1;i<=len;i++)
118 ans.a[i]=a.a[i]+b.a[i];
119 ans.maintain();
120 return ans;
121 }
122 //friend BigInt operator - (BigInt a,BigInt b)
123 //{
124 //int len=max(a.n,b.n);
125 //BigInt ans;
126 //ans.n=len;
127 //for(int i=1;i<=len;i++)
128 //ans.a[i]=a.a[i]-b.a[i];
129 //for(int i=1;i<=len;i++)
130 //{
131 //if(ans.a[i]<0)
132 //{
133 //ans.a[i]+=10;
134 //a.a[i+1]--;
135 //}
136 //ans.a[i]+=10
137 //}
138 //}
139 friend BigInt operator * (BigInt b,int a)
140 {
141 int n=b.n;
142 BigInt ans;
143 ans.n=n;
144 for(int i=1;i<=n;i++)
145 ans.a[i]=b.a[i]*a;
146 ans.maintain();
147 return ans;
148 }
149 friend BigInt operator * (int a,BigInt b)
150 {
151 int n=b.n;
152 BigInt ans;
153 ans.n=n;
154 for(int i=1;i<=n;i++)
155 ans.a[i]=b.a[i]*a;
156 ans.maintain();
157 return ans;
158 }
159 friend BigInt operator * (BigInt a,BigInt b)
160 {
161 BigInt ans;
162 ans.n=a.n+b.n+1;
163 int k;
164 for(int i=1;i<=a.n;i++)
165 for(int j=1;j<=b.n;j++)
166 ans.a[j+i-1]+=a.a[i]*b.a[j];
167 ans.maintain();
168 return ans;
169 }
170 };
171 int main()
172 {
173 char s[1002];
174 char s1[1002];
175 BigInt a,b;
176 //freopen("test.in","r",stdin);
177 a.read();
178 b.read();
179 (a*b).print();
180 printf("\n");
181 return 0;
182 }
