POJ-1001-浮点数高精度计算

来源：互联网发布：python 没有cv2.cv 编辑：程序博客网时间：2024/06/05 22:40
#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <iostream>#include <queue>#include <map>#include <set>#include <vector>using namespace std;char s[1005];int n;#define MAX 3000000#define MAXN 9999#define MAXSIZE 10#define DLEN 4class BigNum;ostream &operator<<(ostream &output,BigNum &ca); class BigNum{public:    int a[500];    //可以控制大数的位数    int len;       //大数长度void changeto_string(string &ss);    BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数    BigNum(const int);       //将一个int类型的变量转化为大数    BigNum(const char*);     //将一个字符串类型的变量转化为大数    BigNum(const BigNum &);  //拷贝构造函数    BigNum &operator=(const BigNum &);   //重载赋值运算符，大数之间进行赋值运算bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较    friend ostream& operator<<(ostream&,  BigNum&);   //重载输出运算符    BigNum operator+(const BigNum &) const;   //重载加法运算符，两个大数之间的相加运算     BigNum operator*(const BigNum &) const;   //重载乘法运算符，两个大数之间的相乘运算     BigNum operator^(const int  &) const;    //大数的n次方运算     void print();       //输出大数};BigNum BigNum::operator^(const int & n) const    //大数的n次方运算{    BigNum t,ret(1);    int i;    if(n<0)        exit(-1);    if(n==0)        return 1;    if(n==1)        return *this;    int m=n;    while(m>1)    {        t=*this;        for( i=1;i<<1<=m;i<<=1)        {            t=t*t;        }        m-=i;        ret=ret*t;        if(m==1)            ret=ret*(*this);    }    return ret;}bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较{    int ln;    if(len > T.len)        return true;    else if(len == T.len)    {        ln = len - 1;        while(a[ln] == T.a[ln] && ln >= 0)            ln--;        if(ln >= 0 && a[ln] > T.a[ln])            return true;        else            return false;    }    else        return false;}bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较{    BigNum b(t);    return *this>b;}BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数{    int i;    memset(a,0,sizeof(a));    for(i = 0 ; i < len ; i++)        a[i] = T.a[i];}BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符，大数之间进行赋值运算{    int i;    len = n.len;    memset(a,0,sizeof(a));    for(i = 0 ; i < len ; i++)        a[i] = n.a[i];    return *this;}BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数{    int c,d = b;    len = 0;    memset(a,0,sizeof(a));    while(d > MAXN)    {        c = d - (d / (MAXN + 1)) * (MAXN + 1);        d = d / (MAXN + 1);        a[len++] = c;    }    a[len++] = d;}BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算{    BigNum t(*this);    int i,big;      //位数    big = T.len > len ? T.len : len;    for(i = 0 ; i < big ; i++)    {        t.a[i] +=T.a[i];        if(t.a[i] > MAXN)        {            t.a[i + 1]++;            t.a[i] -=MAXN+1;        }    }    if(t.a[big] != 0)        t.len = big + 1;    else        t.len = big;    return t;} BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算{    BigNum ret;    int i,j,up;    int temp,temp1;    for(i = 0 ; i < len ; i++)    {        up = 0;        for(j = 0 ; j < T.len ; j++)        {            temp = a[i] * T.a[j] + ret.a[i + j] + up;            if(temp > MAXN)            {                temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);                up = temp / (MAXN + 1);                ret.a[i + j] = temp1;            }            else            {                up = 0;                ret.a[i + j] = temp;            }        }        if(up != 0)            ret.a[i + j] = up;    }    ret.len = i + j;    while(ret.a[ret.len - 1] == 0 && ret.len > 1)        ret.len--;    return ret;} ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符{    int i;    cout << b.a[b.len - 1];    for(i = b.len - 2 ; i >= 0 ; i--)    {        cout.width(DLEN);        cout.fill('0');        cout << b.a[i];    }    return out;}void BigNum::print()    //输出大数{    int i;    cout << a[len - 1];    for(i = len - 2 ; i >= 0 ; i--)    {        cout.width(DLEN);        cout.fill('0');        cout << a[i];    }    cout << endl;}int c_to_n(int b){int len=strlen(s);int ret=0; int i;for (i=0;i<len;i++){if (i==b) continue;ret=ret*10+s[i]-'0';}return ret;}void BigNum::changeto_string(string &ss){ss="";int i,j,tmp;for(i=0;i<len-1;i++)    {tmp=a[i];for (j=1;j<=4;j++){ss+=tmp%10+'0';tmp/=10;} }tmp=a[len-1];while(tmp)//防止出现前导零{ss+=tmp%10+'0';tmp/=10;}}int main(){int i; while(cin>>s>>n){int len=strlen(s);int index=-1;for(i=0;i<len;i++){if (s[i]=='.') {index=i;break;}}int little_point=len-index-1;  ///原有小数位数int final_lpoint=little_point*n;//最后答案的小数位数BigNum  s1=c_to_n(index); s1=s1^n; string ss; s1.changeto_string(ss); string ans="";int line=0;int tmp_final_lpoint=final_lpoint;for (i=0;i<ss.length();i++){if (!line&&i<tmp_final_lpoint) if (ss[i]=='0') {final_lpoint--;continue;}//去后导零  小数位也要减少,注意不要把整数的零删掉了line=1;ans+=ss[i];} ss="";line=0;for (i=ans.length()-1;i>=0;i--){if (!line) if (ans[i]=='0') continue;//去前导零 line=1; ss+=ans[i];}int int_part=ss.length()-final_lpoint;if (ss=="")//如果结果为零printf("0");elseif (final_lpoint==0)//如果结果为整数{for (i=0;i<ss.length();i++){printf("%c",ss[i]);}}elseif (int_part<0)//如果结果整数部分为零{printf(".");int_part=-int_part;for (i=0;i<int_part;i++){printf("0");}for (i=0;i<ss.length();i++){printf("%c",ss[i]);}}else//既有int又有小数的情况{for (i=0;i<int_part;i++){printf("%c",ss[i]);}printf(".");for (i=int_part;i<ss.length();i++){printf("%c",ss[i]);}}printf("\n");}return 0;}
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