poj 1001

大数幂的题目，原来一直没有勇气写。最近终于过了 ，虽然代码很丑。

算法：模拟

#include <iostream>#include <string>using namespace std;string changeCharToString(char *p) //把char转换为string{ string str=""; size_t lenght=strlen(p); for(size_t i=0;i<lenght;i++) str+=p[i]; return str;} string itos(int b) //把int转换为string{int a[15];int i = 0,j;while(b!=0){a[i] = b%10;b = b/10;i++;}string c ;int mk = 0;char mm[15];for(j = i-1; j >=0; j--){mm[mk++] =a[j]+'0';}mm[mk] = '/0';return changeCharToString(mm);}string add(string re,string s) //两个大整数的相加{int i,j,lre,ls,jin,tmp;lre=re.length();ls=s.length();jin=0;if(lre<ls){string tmp;tmp=re;re=s;s=tmp; }lre=re.length();ls=s.length();for(i=lre-1,j=ls-1;j>=0 && i>=0;i--,j--){tmp=re[i]+s[j]-'0'-'0'+jin;re[i]=tmp%10+'0';jin=tmp/10;}for(;i>=0;i--){if(jin==0) break;tmp=re[i]+jin-'0';re[i]=tmp%10+'0';jin=tmp/10;}if(jin==1)re='1'+re;//字符串的相加，解决了那个加出来的位数比原来大的情况return re;}string bigsub(string a, int b) //一个string大整数与一个int型相乘{string re,strb;int la = a.length();int i,j=1;int temp;for( i = la -1 ; i>=0 ; i--){temp = b*(int)(a[i]-'0') *j;strb = itos(temp);re = add(strb,re);j = j*10;}return re;}string superbigsub(string a,string b) //两个大整数string的乘法{int lb = b.length();int i,j = 0;string rb;string back;string zeor;for(i = lb - 1; i>=0 ; i--){int temp = (int)(b[i]-'0');rb = bigsub(a,temp);rb+= zeor;back = add(rb,back);zeor+="0";}return back;}string doit(string a,int b) //最终的幂没有优化的{string mm = a;for(int i = 1 ; i< b; i++){mm = superbigsub(a,mm);}return mm;}string noling(string a){int i = 0;string bs;while(a[i] == '0')i++;bs = a;int j = 0;while(a[i] != '/0'){bs[j++] = a[i++];}bs[j] = '/0';return bs;}int main(){/*string a;int b;string bs;int i,j;int yidian;while(cin>>a>>b){int fid = a.find(".",0);if(fid == 1){yidian = 4;a.erase(fid,fid);cout<<a<<endl;a = noling(a);}else if(fid != -1){int la = a.length();cout<<fid<<endl;yidian = a.length() -fid;while(fid != la -1){a[fid] = a[fid+1];fid++;}a[fid] = '/0';cout<<a<<endl;}string outstr;string mm = doit(a,b);cout<<mm<<endl;}*/double a;int b;int i,j;while(cin>>a>>b){//cout<<"a: "<<a<<endl;if(a*1.0/((int)a)==1) //当输入的树为整数的时候的操作{cout<<doit(itos(a),b)<<endl;continue;}int aa = a*10000;int tuiwei = 4*b;string m = itos(aa);string mm = doit(m,b);if(a >= 1) //当输入的数如：1.2648 比0大的数{for( i = 0 ; i < mm.length() - tuiwei ; i++)cout<<mm[i];char b[100];j = 0;for( ;mm[i]!='/0'; i++){b[j++] =mm[i];}b[j] = '/0';j =strlen(b)- 1;while(b[j] == '0')j--;b[++j] = '/0';cout<<".";printf("%s/n",b);}else{cout<<".";for(i = 0 ; i < tuiwei - mm.length() ; i++)cout<<"0";i = mm.length() - 1;while( mm[i]== '0' )i--;for(j = 0 ; j <= i ; j++)cout<<mm[j];cout<<endl;}}return 0;}

额。。真的有时间 确实可以去写个完全的大整数模板呀。