UVA10494 - If We Were a Child Again

来源：互联网发布：张国荣同性恋知乎编辑：程序博客网时间：2024/05/31 05:29
问题描述

输入两个数求余和求模的结果
思路

套用大数模板。。
代码

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int maxn=1000;struct bign{    int len,s[maxn];    bign()    {        memset(s,0,sizeof(s));        len=1;    }    bign(int num)    {        *this=num;    }    bign(const char *num)    {        *this=num;    }    bign operator = (const int num)    {        char s[maxn];        sprintf(s,"%d",num);        *this=s;        return *this;    }    bign operator = (const char *num)    {        len=strlen(num);        for(int i=0;i<len;i++)            s[i]=num[len-i-1]-'0';        return *this;    }    //输出    string str() const    {        string res="";        for(int i=0;i<len;i++)        {            res=(char)(s[i]+'0')+res;        }        if(res=="")            res="0";        return res;    }    void clean()    {        while(len>1&&!s[len-1])            len--;    }    bign operator + (const bign &b) const    {        bign c;        c.len=0;        for(int i=0,g=0;g||i<max(len,b.len);i++)        {            int x=g;            if(i<len) x+=s[i];            if(i<b.len) x+=b.s[i];            c.s[c.len++]=x%10;            g=x/10;        }        return c;    }    bign operator - (const bign& b) const    {        bign c;        c.len=0;        for(int i=0,g=0;i<len;i++)        {            int x=s[i]-g;            if(i<b.len) x-=b.s[i];            if(x>=0)                g=0;            else            {                g=1;                x+=10;            }            c.s[c.len++]=x;        }        c.clean();        return c;    }    bign operator * (const bign& b) const    {        bign c;        c.len=len+b.len;        for(int i=0;i<len;i++)        {            for(int j=0;j<b.len;j++)                c.s[i+j]+=s[i]*b.s[j];        }        for(int i=0;i<c.len-1;i++)        {            c.s[i+1]+=c.s[i]/10;            c.s[i]%=10;        }        c.clean();        return c;    }    bign operator / (const bign &b)const    {        bign ret,cur=0;        ret.len=len;        for(int i=len-1;i>=0;i--)        {            cur=cur*10;            cur.s[0]=s[i];            while(cur>=b)            {                cur-=b;                ret.s[i]++;            }        }        ret.clean();        return ret;    }    bign operator %(const bign &b) const    {        bign c=*this/b;        return *this-c*b;    }    bool operator <(const bign& b) const    {        if(len!=b.len) return len<b.len;        for(int i=len-1;i>=0;i--)            if(s[i]!=b.s[i])                return s[i]<b.s[i];        return false;    }    bool operator >(const bign& b) const    {        return b < *this;    }    bool operator <=(const bign& b) const    {        return !(b < *this);    }    bool operator >=(const bign& b) const    {        return !(*this <b);    }    bool operator ==(const bign& b) const    {        return !(b< *this)&&!(*this<b);    }    bool operator !=(const bign& b) const    {        return *this >b || *this <b;    }    bign operator +=(const bign& b)    {        *this= *this + b;        return *this;    }    bign operator -=(const bign& b)    {        *this= *this - b;        return *this;    }    bign operator *=(const bign& b)    {        *this= *this * b;        return *this;    }    bign operator /=(const bign& b)    {        *this= *this / b;        return *this;    }    bign operator %=(const bign& b)    {        *this= *this % b;        return *this;    }};istream& operator >> (istream &in, bign& x){    string s;    in >> s;    x = s.c_str();    return in;}ostream& operator << (ostream &out, const bign& x){    out << x.str();    return out;}int main(){    bign a,b;    string s;    while(cin>>a>>s>>b)    {       a.clean();   //考虑清除前导0       b.clean();        if(s=="/")            cout<<a/b<<endl;        else            cout<<a%b<<endl;    }    return 0;}
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