• 高精度模板


    用类实现的高精度模板实在是太臃肿了,用过程实现一次方便记忆

      1 #include<iostream>
      2 #include<algorithm>
      3 #include<cstring>
      4 #include<iomanip>
      5 using namespace std;
      6 
      7 const int MAXN=1000;
      8 const int DLEN=4;
      9 const int WIDE=10000;
     10 class BigNum
     11 {
     12 public:
     13     int NUM[MAXN];
     14     int L;
     15     bool flag;
     16     BigNum(){memset(NUM,0,sizeof(NUM));L=1;flag=0;}
     17     BigNum(const BigNum &T){memcpy(NUM,T.NUM,sizeof(NUM));L=T.L;flag=T.flag;}
     18     BigNum(int n){memset(NUM,0,sizeof(NUM));NUM[0]=n;L=1;while(NUM[L-1]>=WIDE){NUM[L]+=NUM[L-1]/WIDE;NUM[L-1]%=WIDE;L++;}flag=0;}
     19 };
     20 
     21 void Input(string s,BigNum &T)
     22 {
     23     int k=1,num=0;
     24     memset(T.NUM,0,sizeof(int)*MAXN);T.L=0;
     25     for(int i=s.size()-1;i>=0;i--)
     26     {
     27         if(k==WIDE) T.NUM[T.L++]=num,num=0,k=1;
     28         num+=k*(s[i]-'0');
     29         k*=10;
     30     }
     31     if(num>0) T.NUM[T.L++]=num;
     32 }
     33 
     34 void Output(const BigNum T)
     35 {
     36     if(T.flag==1) cout<<'-';
     37     cout<<T.NUM[T.L-1];
     38     for(int i=T.L-2;i>=0;i--)
     39     {
     40         cout.width(DLEN);
     41         cout.fill('0');
     42         cout<<T.NUM[i];
     43     }
     44 }
     45 
     46 bool cmp(const BigNum A,const BigNum B)
     47 {
     48     if(A.L!=B.L) return A.L<B.L;
     49     for(int i=A.L-1;i>=0;i--)
     50         if(A.NUM[i]!=B.NUM[i])
     51             return A.NUM[i]<B.NUM[i];
     52     return 0;
     53 }
     54 
     55 BigNum Add(const BigNum A,const BigNum B)
     56 {
     57     BigNum C;
     58     int L=max(A.L,B.L);C.L=L;
     59     for(int i=0;i<L;i++)
     60     {
     61         C.NUM[i]+=A.NUM[i]+B.NUM[i];
     62         if(C.NUM[i]>=WIDE)
     63             C.NUM[i]-=WIDE,C.NUM[i+1]++;
     64     }
     65     if(C.NUM[L]>0) C.L++;
     66     return C;
     67 }
     68 
     69 BigNum Add(const BigNum A,int B)
     70 {
     71     BigNum C(A);
     72     C.NUM[0]+=B;
     73     for(int i=0;i<C.L;i++)
     74         if(C.NUM[i]>=WIDE)
     75             C.NUM[i+1]+=C.NUM[i]/WIDE,C.NUM[i]%=WIDE;
     76     if(C.NUM[C.L]>0) C.L++;
     77     return C;
     78 }
     79 
     80 BigNum Dec(BigNum A,BigNum B)
     81 {
     82     BigNum *X=&A,*Y=&B,C;
     83     int L=max(X->L,Y->L);
     84     C.L=L;
     85     if(cmp(*X,*Y)) swap(X,Y),C.flag=1;
     86     for(int i=0;i<L;i++)
     87     {
     88         C.NUM[i]+=X->NUM[i]-Y->NUM[i];
     89         if(C.NUM[i]<0)
     90             C.NUM[i]+=WIDE,C.NUM[i+1]--;
     91     }
     92     while(C.NUM[C.L-1]==0) C.L--;
     93     return C;
     94 }
     95 
     96 BigNum Dec(BigNum A,int B)
     97 {
     98     BigNum C(A);
     99     C.NUM[0]-=B;
    100     for(int i=0;i<C.L;i++)
    101         if(C.NUM[i]<0)
    102             C.NUM[i+1]+=C.NUM[i]/WIDE-1,C.NUM[i]=WIDE-abs(C.NUM[i]%WIDE);
    103     if(C.NUM[C.L-1]==0) C.L--;
    104     return C;
    105 }
    106 
    107 BigNum Mult(const BigNum A,const BigNum B)
    108 {
    109     BigNum C;
    110     for(int i=0;i<A.L;i++)
    111         for(int j=0;j<B.L;j++)
    112         {
    113             C.L=i+j;
    114             C.NUM[C.L]+=A.NUM[i]*B.NUM[j];
    115             if(C.NUM[C.L]>=WIDE)
    116                 C.NUM[C.L+1]+=C.NUM[C.L]/WIDE,C.NUM[C.L]%=WIDE;
    117         }
    118     C.L=A.L+B.L;
    119     if(C.NUM[C.L-1]==0) C.L--;
    120     return C;
    121 }
    122 
    123 BigNum Mult(const BigNum A,int B)
    124 {
    125     BigNum C(A);
    126     int i,tmp,k=0;
    127     for(i=0;i<C.L||k;i++)
    128     {
    129         tmp=C.NUM[i]*B+k;
    130         k=tmp/WIDE;
    131         C.NUM[i]=tmp%WIDE;
    132     }
    133     C.L=i;
    134     return C;
    135 }
    136 
    137 BigNum Div2(const BigNum A)
    138 {
    139     BigNum C(A);
    140     for(int i=C.L-1;i>=0;i--)
    141     {
    142         if(C.NUM[i]%2&&i>0)
    143             C.NUM[i-1]+=WIDE;
    144         C.NUM[i]/=2;
    145     }
    146     if(C.NUM[C.L-1]==0) C.L--;
    147     return C;
    148 }
    149 
    150 BigNum Div(const BigNum A,const BigNum B)
    151 {
    152     BigNum L(1),R(A),T;
    153     while(cmp(Add(L,1),R))
    154     {
    155         T=Div2(Add(L,R));
    156         if(cmp(A,Mult(T,B)))
    157             R=T;
    158         else 
    159             L=T;
    160     }
    161     return L;
    162 }
    163 
    164 BigNum Div(const BigNum A,int B)
    165 {
    166     BigNum C(A);
    167     int k=0;
    168     for(int i=C.L-1;i>=0;i--)
    169     {
    170         k=k*WIDE+C.NUM[i];
    171         C.NUM[i]=k/B;
    172         k%=B;
    173     }
    174     while(C.NUM[C.L-1]==0) C.L--;
    175     return C;
    176 }
    177 
    178 BigNum A,B,C;
    179 
    180 int main()
    181 {
    182     string s1,s2;
    183     cin>>s1>>s2;
    184     Input(s1,A);
    185     Input(s2,B);
    186     C=Add(A,B);
    187     Output(C);
    188     return 0;
    189 }
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  • 原文地址:https://www.cnblogs.com/InWILL/p/5924982.html
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