基于类和对象的高精度算法封装

高精度口诀（别问我这是从哪里来的，你在网上找不到是因为这就是我写的）

:-)

1.高精度加法：相加进位再去零
2.高精度减法：判断相减再去零
3.高精度乘法：相乘预估再去零
4.高低精乘法：相乘进位再去零

#include<iostream>
#include<cstring>
using namespace std;

class BIGINT{
    public:
        int num[505];
        int length;
        bool positive;
        BIGINT(string str=""){
            memset(num, 0, sizeof(num));
            positive = true;
            if(str == ""){
                return;
            }
            else{
                length = str.length();
                for(int i = 0; i<str.length(); i++){
                    num[i] = str[str.length()-i-1]-'0';
                }
            }
        }
        void show(){
            if(positive == false){
                cout<<"-";
            }
            for(int i = length-1; i>=0; i--){
                cout<<num[i];
            }
        }
};

int compare(BIGINT &a,BIGINT &b){
    if(a.length > b.length){
        return 1;
    }
    else if(a.length < b.length){
        return -1;
    }
    else{
        int maxlen = max(a.length,b.length);
        for(int i = maxlen-1; i>=0; i--){
            if(a.num[i] > b.num[i]){
                return 1;
            }
            else if(a.num[i] < b.num[i]){
                return -1;
            }
        }
        return 0;
    }
}

BIGINT bigadd(BIGINT &a,BIGINT &b){
    BIGINT c;
    int maxlen = max(a.length,b.length);
    //相加
    for(int i = 0;i<maxlen;i++){
        c.num[i] = a.num[i] + b.num[i];
    }
    //进位
    for(int i = 0;i<maxlen;i++){
        if(c.num[i]>=10){
            c.num[i+1] += 1;
            c.num[i] = c.num[i] % 10;
        }
    }
    //去零
    if(c.num[maxlen] == 0){
        c.length = maxlen;
    }
    else{
        c.length = maxlen + 1;
    }
    return c;
}

BIGINT bigminus(BIGINT &a,BIGINT &b){
    BIGINT c;
    int maxlen = max(a.length,b.length);
    //判断
    int maxint = compare(a,b);
    if(maxint == 1){
        //相减
        for(int i = 0;i<a.length;i++){
            if(a.num[i]>=b.num[i]){
                c.num[i] = a.num[i] - b.num[i];
            }
            else{
                a.num[i+1] -= 1;
                a.num[i] += 10;
                c.num[i] = a.num[i] - b.num[i];
            }
        }
    }
    else if(maxint == -1){
        c = bigminus(b,a);
        c.positive = false;
    }
    //去零
    for(int i = maxlen+1;i>=0;i--){
        if(c.num[i] != 0){
            c.length = i+1;
            return c;
        }
    }
    c.length = 1;
    return c;
}

BIGINT bigmultiply(BIGINT &a,BIGINT &b){
    BIGINT c;
    //相乘
    for(int i = 0;i<a.length;i++){ 
        for(int j = 0;j<b.length;j++){   
            c.num[i+j] += a.num[i]*b.num[j];
            c.num[i+j+1] += c.num[i+j]/10;
            c.num[i+j] %= 10;
        }
    }
    //预估
    int len = a.length + b.length;
    //去零
    for(int i = len+1;i>=0;i--){
        if(c.num[i] != 0){
            c.length = i+1;
            return c;
        }
    }
    c.length = 1;
    return c; 
}

BIGINT bigmultiplywithsmall(BIGINT a,int b){
    //相乘
    for(int i = 0;i<a.length;i++){
        a.num[i] *= b;
    }
    //进位
    for(int i = 0;i<a.length+1;i++){
        if(a.num[i]>=10){
            a.num[i+1] += a.num[i] / 10;
            a.num[i] %= 10;
        }
    }
    //去零
    if(a.num[a.length] != 0){
        a.length += 1;
    }
    return a;
}

int main(){
    BIGINT a("111111111");
    BIGINT b("111111111");
    BIGINT tmp; 
    
    tmp = bigadd(a,b);
    cout<<"高精度加法:";
    tmp.show();
    cout<<endl; 
    
    tmp = bigminus(a,b);
    cout<<"高精度减法:";
    tmp.show();
    cout<<endl;
    
    tmp = bigmultiply(a,b);
    cout<<"高精度乘法:";
    tmp.show();
    cout<<endl;
    
    tmp = bigmultiplywithsmall(a,100);
    cout<<"高低精乘法:";
    tmp.show();
    cout<<endl;
}

高精度运算，是指参与运算的数(加数，减数，因子……）范围大大超出了标准数据类型（整型，实型）能表示的范围的运算。例如，求两个20000位的数的和。这时，就要用到高精度算法了。

来源：https://www.cnblogs.com/lyj00912/p/11366683.html