POJ 2389 大数问题

Bull Math

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 13438 Accepted: 6925

Description

Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers … or so they say. Farmer John wonders if their answers are correct. Help him check the bulls’ answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros). 

FJ asks that you do this yourself; don’t use a special library function for the multiplication.

Input

* Lines 1..2: Each line contains a single decimal number.

Output

* Line 1: The exact product of the two input lines

Sample Input

11111111111111
1111111111

Sample Output

12345679011110987654321

大数问题 套模板 

/*
 * cpx.cpp
 *
 *  Created on: 2015年4月9日
 *      Author: yf
 */


#include<stdlib.h>
#include<iostream>
#include<string>
#include<iomanip>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum
{
private:
    int a[500];    //可以控制大数的位数
    int len;       //大数长度
public:
    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

    friend istream& operator>>(istream&,  BigNum&);   //重载输入运算符
    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;   //重载减法运算符,两个大数之间的相减运算
    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;    //重载除法运算符,大数对一个整数进行相除运算

    BigNum operator^(const int  &) const;    //大数的n次方运算
    int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算
    bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较
    bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较

    void print();       //输出大数
};
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(const char*s)     //将一个字符串类型的变量转化为大数
{
    int t,k,index,l,i;
    memset(a,0,sizeof(a));
    l=strlen(s);
    len=l/DLEN;
    if(l%DLEN)
        len++;
    index=0;
    for(i=l-1;i>=0;i-=DLEN)
    {
        t=0;
        k=i-DLEN+1;
        if(k<0)
            k=0;
        for(int j=k;j<=i;j++)
            t=t*10+s[j]-'0';
        a[index++]=t;
    }
}
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;
}
istream& operator>>(istream & in,  BigNum & b)   //重载输入运算符
{
    char ch[MAXSIZE*4];
    int i = -1;
    in>>ch;
    int l=strlen(ch);
    int count=0,sum=0;
    for(i=l-1;i>=0;)
    {
        sum = 0;
        int t=1;
        for(int j=0;j<4&&i>=0;j++,i--,t*=10)
        {
            sum+=(ch[i]-'0')*t;
        }
        b.a[count]=sum;
        count++;
    }
    b.len =count++;
    return in;

}
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;
}

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   //两个大数之间的相减运算
{
    int i,j,big;
    bool flag;
    BigNum t1,t2;
    if(*this>T)
    {
        t1=*this;
        t2=T;
        flag=0;
    }
    else
    {
        t1=T;
        t2=*this;
        flag=1;
    }
    big=t1.len;
    for(i = 0 ; i < big ; i++)
    {
        if(t1.a[i] < t2.a[i])
        {
            j = i + 1;
            while(t1.a[j] == 0)
                j++;
            t1.a[j--]--;
            while(j > i)
                t1.a[j--] += MAXN;
            t1.a[i] += MAXN + 1 - t2.a[i];
        }
        else
            t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while(t1.a[len - 1] == 0 && t1.len > 1)
    {
        t1.len--;
        big--;
    }
    if(flag)
        t1.a[big-1]=0-t1.a[big-1];
    return t1;
}

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;
}
BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算
{
    BigNum ret;
    int i,down = 0;
    for(i = len - 1 ; i >= 0 ; i--)
    {
        ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
        down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
    }
    ret.len = len;
    while(ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算
{
    int i,d=0;
    for (i = len-1; i>=0; i--)
    {
        d = ((d * (MAXN+1))% b + a[i])% b;
    }
    return d;
}
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;
}

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 main()
{
	BigNum s1("11111111111111111111111111");
	BigNum s2("1111111111111111111111111111111111111");
	BigNum sum;
	sum=s1*s2;
	sum.print();
return 0;
}

运行结果

12345679012345679012345678888888888887654320987654320987654321

    原文作者:大整数乘法问题
    原文地址: https://blog.csdn.net/judyge/article/details/44980021
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞