String operations of large numbers

#include <bits / STDC ++ H.>
 #define LL Long Long
 the using  namespace STD;
 int Compare ( String str1, String str2) // less than 0 is less than, greater than 0 is greater than, equal to 0 represents 
{
     IF (str1.length () > str2.length ()) return  . 1 ;
     the else  IF (str1.length () <str2.length ())   return - . 1 ;
     the else  return str1.compare (str2); 
} 
String the Add ( String str1, String str2) 
{ 
    String str;
     int len1=str1.length();
    int len2=str2.length();
    if(len1<len2)
    {
        for(int i=1;i<=len2-len1;i++)
           str1="0"+str1;
    }
    else
    {
        for(int i=1;i<=len1-len2;i++)
           str2="0"+str2;
    }
    len1=str1.length();
    int cf=0;
    int temp;
    for(int i=len1-1;i>=0;i--)
    {
        temp=str1[i]-'0'+str2[i]-'0'+cf;
        cf=temp/10;
        temp%=10;
        str=char(temp+'0')+str;
    }
    if(cf!=0)  str=char(cf+'0')+str;
    return str;
}
string sub(string str1,string str2)//只能大的减小的
{
    string str;
    int tmp=str1.length()-str2.length();
    int cf=0;
    for(int i=str2.length()-1;i>=0;i--)
    {
        if(str1[tmp+i]<str2[i]+cf)
        {
            str=char(str1[tmp+i]-str2[i]-cf+'0'+10)+str;
            cf=1;
        }
        else
        {
            str=char(str1[tmp+i]-str2[i]-cf+'0')+str;
            cf=0;
        }
    }
    for(int i=tmp-1;i>=0;i--)
    {
        if(str1[i]-cf>='0')
        {
            str=char(str1[i]-cf)+str;
            cf=0;
        }
        else
        {
            str=char(str1[i]-cf+10)+str;
            cf=1;
        }
    }
    str.erase(0,str.find_first_not_of('0'));
    return str;
}

string mul(string str1,string str2)
{
    string str;
    int len1=str1.length();
    int len2=str2.length();
    string tempstr;
    for(int i=len2-1;i>=0;i--)
    {
        tempstr="";
        int temp=str2[i]-'0';
        int t=0;
        int cf=0;
        if(temp!=0)
        {
            for(int j=1;j<=len2-1-i;j++)
              tempstr+="0";
            for(int j=len1-1;j>=0;j--)
            {
                t=(temp*(str1[j]-'0')+cf)%10;
                cf=(temp*(str1[j]-'0')+cf)/10;
                tempstr=char(t+'0')+tempstr;
            }
            if(cf!=0) tempstr=char(cf+'0')+tempstr;
        }
        str=add(str,tempstr);
    }
    str.erase(0,str.find_first_not_of('0'));
    return str;
}
int main(){
    string s1,s2;
    cin >> s1 >> s2;
    cout << sub(s1,s2) << ' ' << add(s1,s2) << ' ' << mul(s1,s2) << endl;
    return 0;
}
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Origin www.cnblogs.com/cherish-lin/p/11918889.html