如何在biginteger类中减去这个问题 [英] How to solve this problem in subtract in biginteger class

问题描述

class BigInteger
{
    public:
    void Subtraction()
    {
        string string1,string2;
        char x=0;
        cout<<"Enter first number :";
        cin>>string1;
        cout<<"Enter second number :";
        cin>>string2;
        size_t n=max(string1.size(),string2.size());
        if(n>string1.size())
            string1=string(n-string1.size(),'0')+string1;
        if(n>string2.size())
            string2=string(n-string2.size(),'0')+string2;
        string final(n,'0');
        string::const_reverse_iterator s1 = string1.rbegin(), e = string1.rend(), s2 = string2.rbegin();
        string::reverse_iterator f = final.rbegin();
        while (s1!=e)
        { 
            int ss1=*s1-'0';
            if(*s1-'0'<*s2-'0')
            {
                int m=ss1+10;
                ++s1;
                ss1--;
                --s1;
                x = m-(*s2-'0');
            }
            else 
            {
                x=ss1-(*s2-'0');	
            }
            *f = x + '0';
            ++s1;
            ++s2;
            ++f;     
        }
                       
        cout<< "final = " <<final << endl;
    }
};

当我输入数字1 = 35

和数字2 = 6

结果= 39

这一行

ss1--;

问题是当从第二位减去一位没有变化发生在基本字符串

when I enter number 1= 35
and number 2=6
the result = 39
in this line
ss1--;
the problem is when subtract one from second digit no change happened in the basic string

推荐答案

对于数学，您应该将输入字符串转换为数字 ParseInt64 将为真正的大数字做这件事。

For mathematics you should convert the input string to a number ParseInt64 will do it for really big numbers.

long number1 = Int64.Parse(string1);
long number2 = Int64.Parse(string1);
//and now do the math
long result = number1 - number2;

如果您需要更大的数字，那么必须将数字从右侧。

if you need bigger numbers, so must split the numbers from the right side.

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