POJ2389 Bull Math 

  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 }

 

posted @ 2017-03-27 14:54  BBBob  阅读(150)  评论(0)    收藏  举报