#include <iostream> #include <cstring> #include <string> #include <cstdio>using namespace std;/ /500 digits at moststruct num{int num[1000],len; Num () {memset (num,0,sizeof (num)); Len=1; } Num (const string &s) {len=s.size (); for (int i=0;i<len;++i) {num[i]=s[len-1-i]-' 0 '; }} num& operator= (const num& right) {copy (right.num,right.num+1000,this->num); Len=right.len; return *this; } friend ostream& operator<< (ostream &os,const Num &output) {for (int i=output.len-1;i>=0;-- i) os<<output.num[i]; return OS; } Friend Num operator+ (num &x,num &y) {num ans; Ans.len=max (X.len,y.len); for (int i=0;i<ans.len;++i) {ans.num[i]=x.num[i]+y.num[i]; } for (int i=0;i<ans.len;++i) {ANS.NUM[I+1]+=ANS.NUM[I]/10; ans.num[i]%=10; } if (Ans.num[ans.len]) ++ans.len; return ans; } Friend Num operator-(const Num &left,const num& right) {num ans; Ans.len=max (Left.len,right.len); for (int i=0;i<ans.len;++i) ans.num[i]=left.num[i]-right.num[i]; for (int i=0;i<ans.len;++i) {if (ans.num[i]<0) {--ans.num[i+1]; ans.num[i]+=10; }} while (Ans.len>=1 &&!ans.num[ans.len-1])--ans.len; return ans; } Friend Num mul (const num &x,const num &y) {if (x.len==1 && y.len==1) return to_string (x.num[0]* Y.num[0]); Num A,b,c,d,ans; int Maxlen=max (X.len,y.len),len1=maxlen>>1; Copy (X.num,x.num+len1,b.num); B.len=len1; Copy (X.num+len1,x.num+maxlen,a.num); A.len=maxlen-len1; Copy (Y.num,y.num+len1,d.num); D.len=len1; Copy (Y.num+len1,y.num+maxlen,c.num); C.len=maxlen-len1; Num Tem_ac=mul (a,c),Tem_bd=mul (b,d), Ac,bd,fin_cdab; Num Cdab=mul (c+d,a+b)-tem_ac-tem_bd; Copy (tem_ac.num,tem_ac.num+tem_ac.len,ac.num+len1*2); Ac.len=tem_ac.len+2*len1; Copy (CDAB.NUM,CDAB.NUM+CDAB.LEN,FIN_CDAB.NUM+LEN1); Fin_cdab.len=cdab.len+len1; Ans=ac+fin_cdab; ANS=ANS+TEM_BD; while (ans.len>1 &&!ans.num[ans.len-1])--ans.len; return ans; } Friend num operator* (const num &x,const num &y) {return mul (x, y); }};int Main () {string A, B; while (cin>>a>>b) {Num x=a,y=b; cout<< "***************************************************" <<endl; cout<<a<<endl<< "*" <<endl<<b<<endl<< "=" <<endl<<x*y< <endl; cout<< "***************************************************" <<endl; } return 0;}
2 power-Number large integer divide-and-conquer algorithm