c++各类大数模板

悠悠 2022-08-25 09:16 142阅读 0赞

首先大数加法 两个大数相加

string sum(string s1,string s2)
    {
    	if(s1.length()<s2.length())
    	{
    		string temp=s1;
    		s1=s2;
    		s2=temp;
    	}
    	int i,j;
    	for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
    	{
    		s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));   //注意细节
    		if(s1[i]-'0'>=10)
    		{
    			s1[i]=char((s1[i]-'0')%10+'0');
    			if(i) s1[i-1]++;
    			else s1='1'+s1;
    		}
    	}
    	return s1;
    }

博客链接：

[http://blog.csdn.net/y990041769/article/details/19545179][http_blog.csdn.net_y990041769_article_details_19545179]

然后大数乘以整形数

string Multiply(string s,int x)  //大数乘以整形数
    {
        reverse(s.begin(),s.end());
        int cmp=0;
        for(int i=0;i<s.size();i++)
        {
            cmp=(s[i]-'0')*x+cmp;
            s[i]=(cmp%10+'0');
            cmp/=10;
        }
        while(cmp)
        {
            s+=(cmp%10+'0');
            cmp/=10;
        }
        reverse(s.begin(),s.end());
        return s;
    }

博客链接： [http://blog.csdn.net/y990041769/article/details/12645953][http_blog.csdn.net_y990041769_article_details_12645953]

大数除以整形数

string Except(string s,int x)  //大数除以整形数
    {
        int cmp=0,ok=0;
        string ans="";
        for(int i=0;i<s.size();i++)
        {
            cmp=(cmp*10+s[i]-'0');
            if(cmp>=x)
            {
                ok=1;
                ans+=(cmp/x+'0');
                cmp%=x;
            }
            else{
                if(ok==1)
                    ans+='0';  //注意这里啊。才找出错误
            }
        }
        return ans;
    }

大数乘法：http://poj.org/problem?id=2389

思想就是模拟乘法运算，用大数乘以另一个数的每一位然后大数相加就是ans

string sum(string s1,string s2)  //大数加法
    {
    	if(s1.length()<s2.length())
    	{
    		string temp=s1;
    		s1=s2;
    		s2=temp;
    	}
    	int i,j;
    	for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
    	{
    		s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));   //注意细节
    		if(s1[i]-'0'>=10)
    		{
    			s1[i]=char((s1[i]-'0')%10+'0');
    			if(i) s1[i-1]++;
    			else s1='1'+s1;
    		}
    	}
    	return s1;
    }
    
    string Mult(string s,int x)  //大数乘以整形数
    {
        reverse(s.begin(),s.end());
        int cmp=0;
        for(int i=0;i<s.size();i++)
        {
            cmp=(s[i]-'0')*x+cmp;
            s[i]=(cmp%10+'0');
            cmp/=10;
        }
        while(cmp)
        {
            s+=(cmp%10+'0');
            cmp/=10;
        }
        reverse(s.begin(),s.end());
        return s;
    }
    string Multfa(string x,string y)  //大数乘法
    {
        string ans;
        for(int i=y.size()-1,j=0;i>=0;i--,j++)
        {
            string tmp=Mult(x,y[i]-'0');
            for(int k=0;k<j;k++)
                tmp+='0';
            ans=sum(ans,tmp);
        }
        return ans;
    }

浮点数的n次方

string Multiply(string s,long x)  //大数乘以整形数
    {
        reverse(s.begin(),s.end());
        long cmp=0;
        for(int i=0; i<s.size(); i++)
        {
            cmp=(s[i]-'0')*x+cmp;
            s[i]=(cmp%10+'0');
            cmp/=10;
        }
        while(cmp)
        {
            s+=(cmp%10+'0');
            cmp/=10;
        }
        reverse(s.begin(),s.end());
        return s;
    }
    string Remove_later(string s)   //删除一个字符串的后倒0
    {
        int ok=1;
        for(int i=s.size()-1; i>=0; i--)
        {
            if(s[i]=='0'){
                s.erase(i);
            }
            else if(s[i]=='.')
            {
                s.erase(i);
                ok=0;
            }
            else
                ok=0;
            if(ok==0)
                break;
        }
        return s;
    }
    string factorial(string s,int n)   //浮点数的n次方
    {
        if(n==0)
            return "1";
        string cmp="",count="";
        long x=0,point=0;
        for(int i=0; i<s.size(); i++)
            if(s[i]!='.')
            {
                cmp+=s[i];
                x=x*10+(s[i]-'0');
            }
            else
                point=s.size()-1-i;
        for(int i=1; i<n; i++)
        {
            cmp=Multiply(cmp,x);
        }
        int ans_point=cmp.size()-n*point;
        if(ans_point<0)
        {
            count+='.';
            for(int i=ans_point; i!=0; i++)
                count+='0';
        }
        string::iterator it=cmp.begin();
        if(ans_point>=0&&ans_point<cmp.size())
            cmp.insert(it+ans_point,'.');
        count+=(Remove_later(cmp));
        return count;
    }

博客链接：[http://blog.csdn.net/y990041769/article/details/9262943][http_blog.csdn.net_y990041769_article_details_9262943]

字符串去除后导0函数，前导0可以先反转取后导。

string Remove_later(string s)   //删除一个字符串的后倒0
    {
        for(int i=s.size()-1; i>=0; i--)
        {
            if(s[i]=='0')
                s.erase(i);
            else
                break;
        }
        return s;
    }

题目： [http://acm.nyist.net/JudgeOnline/problem.php?pid=][http_acm.nyist.net_JudgeOnline_problem.php_pid]272

其他的正在完善中。争取写出自己的模板。

[http_blog.csdn.net_y990041769_article_details_19545179]: http://blog.csdn.net/y990041769/article/details/19545179
[http_blog.csdn.net_y990041769_article_details_12645953]: http://blog.csdn.net/y990041769/article/details/12645953
[http_blog.csdn.net_y990041769_article_details_9262943]: http://blog.csdn.net/y990041769/article/details/9262943
[http_acm.nyist.net_JudgeOnline_problem.php_pid]: http://acm.nyist.net/JudgeOnline/problem.php?pid=1