Description
| 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
题意：简单的2个大数的乘积；
套用刘汝佳的大数相乘模板；
要注意的是数组的大小；这里0<=X,Y<10250 ；
所以数组至少要开到500以上；
代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstdlib>
4 #include <cstring>
5 #define UNIT 10
6
7 using namespace std;
8
9 struct Bignum
10 {
11 int val[550];
12 int len;
13
14 Bignum ()
15 {
16 memset (val, 0, sizeof(val));
17 len = 1;
18 }
19
20 Bignum operator = (const int &a)
21 {
22 int t, p = a;
23 len = 0;
24 while (p >= UNIT)
25 {
26 t = p - (p/UNIT)*UNIT;
27 p = p / UNIT;
28 val[len++] = t;
29 }
30 val[len++] = p;
31 return *this;
32 }
33
34 Bignum operator + (const Bignum &a) const//大数加大数
35 {
36 Bignum x = a;
37 int L;
38 L = a.len > len ? a.len : len;
39 for (int i = 0; i < L; ++i)
40 {
41 x.val[i] += val[i];
42 if (x.val[i] >= UNIT)
43 {
44 x.val[i+1]++;
45 x.val[i] -= UNIT;
46 }
47 }
48 if (x.val[L] != 0)
49 x.len = L+1;
50 else
51 x.len = L;
52 return x;
53 }
54
55 Bignum operator - (const Bignum &a) const
56 {
57 bool flag;
58 Bignum x1, x2;
59 if (*this > a)
60 {
61 x1 = *this;
62 x2 = a;
63 flag = 0;
64 }
65 else
66 {
67 x1 = a;
68 x2 = *this;
69 flag = 1;
70 }
71 int j, L = x1.len;
72 for (int i = 0; i < L; ++i)
73 {
74 if (x1.val[i] < x2.val[i])
75 {
76 j = i+1;
77 while (x1.val[j] == 0)
78 j++;
79 x1.val[j--]--;
80 while (j > i)
81 x1.val[j--] += UNIT-1;
82 x1.val[i] += UNIT-x2.val[i];
83 }
84 else
85 x1.val[i] -= x2.val[i];
86 }
87 while (x1.val[x1.len-1] == 0 && x1.len > 1)
88 x1.len--;
89 if (flag)
90 x1.val[x1.len-1] = -x1.val[x1.len-1];
91 return x1;
92 }
93
94 Bignum operator * (const Bignum &a) const//大数乘大数
95 {
96 Bignum x;
97 int i, j, up;
98 int x1, x2;
99 for (i = 0; i < len; i++)
100 {
101 up = 0;
102 for (j = 0; j < a.len; j++)
103 {
104 x1 = val[i]*a.val[j] + x.val[i+j] + up;
105 if (x1 >= UNIT)
106 {
107 x2 = x1 - x1/UNIT*UNIT;
108 up = x1 / UNIT;
109 x.val[i+j] = x2;
110 }
111 else
112 {
113 up = 0;
114 x.val[i+j] = x1;
115 }
116 }
117 if (up != 0)
118 x.val[i+j] = up;
119 }
120 x.len = i + j;
121 while (x.val[x.len-1] == 0 && x.len > 1)
122 x.len--;
123 return x;
124 }
125
126 Bignum operator / (const int &a) const//大数除小数
127 {
128 Bignum x;
129 int down = 0;
130 for (int i = len-1; i >= 0; --i)
131 {
132 x.val[i] = (val[i]+down*UNIT) / a;
133 down = val[i] + down*UNIT - x.val[i]*a;
134 }
135 x.len = len;
136 while (x.val[x.len-1] == 0 && x.len > 1)
137 x.len--;
138 return x;
139 }
140
141 int operator % (const int &a) const//大数模小数
142 {
143 int x = 0;
144 for (int i = len-1; i >= 0; --i)
145 x = ((x*UNIT)%a+val[i]) % a;
146 return x;
147 }
148
149 Bignum operator ^ (const int &a) const
150 {
151 Bignum p, x;
152 x.val[0] = 1;
153 if(a < 0)
154 exit(-1);
155 if(a == 0)
156 return x;
157 if(a == 1)
158 return *this;
159 int n = a, i;
160 while(n > 1)
161 {
162 p = *this;
163 for(i = 1; (i<<1) <= n; i<<=1)
164 p = p * p;
165 n -= i;
166 x = x * p;
167 if(n == 1)
168 x = x * (*this);
169 }
170 return x;
171 }
172
173 bool operator > (const Bignum &a) const
174 {
175 int now;
176 if (len > a.len)
177 return true;
178 else if (len == a.len)
179 {
180 now = len - 1;
181 while (val[now] == a.val[now] && now >= 0)
182 now--;
183 if(now >= 0 && val[now] > a.val[now])
184 return true;
185 else
186 return false;
187 }
188 else
189 return false;
190 }
191 };
192
193
194 char a[555],b[555];
195 int main()
196 {
197 // freopen("ACM.txt","r",stdin);
198
199 while(cin>>a>>b)
200 {
201 Bignum x1,x2,ans;
202 int La=strlen(a);
203 int Lb=strlen(b);
204 x1.len=0;
205 x2.len=0;
206 for(int i=La-1;i>=0;i--)
207 x1.val[x1.len++]=a[i]-‘0‘;
208 for(int i=Lb-1;i>=0;i--)
209 x2.val[x2.len++]=b[i]-‘0‘;
210 ans=x1*x2;
211 for(int i=ans.len-1;i>=0;i--)
212 cout<<ans.val[i];
213 cout<<endl;
214 }
215 return 0;
216 }
时间： 2024-10-05 15:36:21