C++实现高精度运算-优快云博客

本文介绍了作者在C++中创建的一个高精度结构体，该结构体重新定义了常用运算符，适合处理超过long long范围的数值计算。作者对代码进行了优化，修正了除法的错误，并减少了冗余代码。

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就算是long long（或int64）还不够怎么办？用高精度算法。自己动手丰衣足食
_int128用户狂怒

Update2020.10.29 对原代码逻辑进行升级修改一处代码错误（除法）并删除大量冗余代码

闲着没事，敲了一个高精度全家桶的结构体，重定义了常用运算符，挺实用的

具体思路详见代码

MAXN可改
常数 O(能过)
代码超短 (共3396B)
本代码适用于大部分高精度程序

#include<cstdio>
#include<cstring>
#define MAXN 10000
using namespace std;
struct bignum{
    char num[MAXN+1];
    int len;
    void clean(){memset(num,0,sizeof(num));len=0;}
    void print(){
        bool flag=0;
        for(int i=MAXN;i>=0;i--)if(flag||num[i])putchar(num[i]+48),flag=1;
    }
    void read(){
        char tmp[MAXN+1];
        memset(tmp,0,sizeof(tmp));
        clean();
        scanf("%s",tmp);
        len=strlen(tmp);
        for(int i=0,j=strlen(tmp);j;i++,j--)num[i]=tmp[j-1]-48;
    }
    int size(){
        len=0;
        bool flag=0;
        for(int i=MAXN;i>=0;i--)if(flag||num[i])len++,flag=1;
        return len;
    }
    bool operator<(bignum &b){
        if(len!=b.len)return len<b.len;
        for(int i=b.len;i>=0;i--)if(num[i]!=b.num[i])return num[i]<b.num[i];
        return 0;
    }
    bool operator>(bignum &b){return b<*this;}
    bool operator!=(bignum &b){return *this>b||*this<b;}
    bool operator==(bignum &b){return!(*this!=b);}
    bool operator<=(bignum &b){return!(*this>b);}
    bool operator>=(bignum &b){return!(*this<b);}
    bignum operator+(bignum &b){
        static bignum ans;
        ans.clean();
        for(int i=0;i<MAXN+1;i++)ans.num[i]=num[i]+b.num[i];
        for(int i=0;i<MAXN+1;i++)if(ans.num[i]>9)ans.num[i]-=10,ans.num[i+1]++;
        return ans;
    }
    bignum operator-(bignum &b){
        static bignum ans;
        ans.clean();
        for(int i=0;i<MAXN+1;i++)ans.num[i]=num[i]-b.num[i];
        for(int i=0;i<MAXN+1;i++)if(ans.num[i]<0)ans.num[i]+=10,ans.num[i+1]--;
        return ans;
    }
    bignum operator*(bignum &b){
        static bignum ans;
        ans.clean();
        for(int i=0;i<len;i++)
            for(int j=0;j<b.len;j++)
                ans.num[i+j]+=num[i]*b.num[j];
        for(int i=0;i<MAXN+1;i++)ans.num[i+1]+=ans.num[i]/10,ans.num[i]%=10;
        ans.len=ans.size();
        return ans;
    }
    bignum operator/(bignum &b){
        static bignum ans,tmp,ten,rem;
        rem=*this;
        ans.clean();
        ten.clean();
        ten.num[1]=1,ten.len=2;
        while(*this>=b&&len>0){
            tmp.clean();
            tmp.num[0]=1,tmp.len=1;
            while(tmp.len+1+b.len<len)tmp=tmp*ten;
            ans=ans+tmp,tmp=tmp*b;
            for(int i=0;i<MAXN+1;i++)num[i]-=tmp.num[i];
            for(int i=0;i<MAXN+1;i++)if(num[i]<0)num[i]+=10,num[i+1]--;
            size();
        }
        *this=rem;
        return ans;
    }
    bignum operator%(bignum &b){
        static bignum ten,ans,tmp;
        ans.clean();
        ten.clean();
        ten.num[1]=1,ten.len=2;
        while(*this>b&&len>0){
            tmp.clean();
            tmp.num[0]=1,tmp.len=1;
            while(tmp.len+1+b.len<len)tmp=tmp*ten;
            tmp=tmp*b;
            for(int i=0;i<MAXN+1;i++)num[i]-=tmp.num[i];
            for(int i=0;i<MAXN+1;i++)if(num[i]<0)num[i]+=10,num[i+1]--;
            size();
        }
        for(int i=0;i<MAXN+1;i++)ans.num[i]=num[i];
        ans.len=len;
        return ans;
    }
    bignum operator+=(bignum &b){*this=*this+b;}
    bignum operator-=(bignum &b){*this=*this-b;}
    bignum operator*=(bignum &b){*this=*this*b;}
    bignum operator/=(bignum &b){*this=*this/b;}
    bignum operator%=(bignum &b){*this=*this%b;}
};
int main(){
    return 0;
}