Uva--424 (高精度)

2014-06-01 20:00:48

题意 &　思路：大数加，不废话。

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
using namespace std;
const int maxn = 1000;

struct bign{
    int len,s[maxn + 5];

    //初始化与赋值
    bign(){
        memset(s,0,sizeof(s));
        len = 1;//因为0的存在
    }

    bign operator = (const char *num){
        len = strlen(num);
        for(int i = 0; i < len; ++i)
            s[i] = num[len - i - 1] - '0';//倒序转化成数字
        return *this;
    }

    bign operator = (int num){
        char ts[maxn + 5];
        sprintf(ts,"%d",num);
        *this = ts;
        return *this;
    }

    bign(int num){
        *this = num;
    }

    bign(const char *num){
        *this = num;
    }
    //bign的str型转化
    string str() const{
        string res = "";
        for(int i = len - 1; i >= 0; --i)
            res +=  (char)(s[i] + '0');
        if(res == "")
            res = "0";
        return res;
    }
    //运算符重载
    //高精度加
    bign operator + (const bign & b) const{
        bign sum;
        sum.len = 0;
        for(int i = 0, g = 0; g || i < max(len,b.len); ++i){
            int x = g;//x:暂存和,g:进位
            if(i < len) x += s[i];
            if(i < b.len) x += b.s[i];
            sum.s[sum.len++] = x % 10;
            g = x / 10;
        }
        return sum;
    }

    //高精度减
    bign operator - (const bign & b) const{
        bign dif;
        int maxlen = (len > b.len) ? len : b.len;
        for(int i = 0; i < maxlen; ++i){
            dif.s[i] += s[i] - b.s[i];
            if(dif.s[i] < 0){
                dif.s[i] += 10;
                --dif.s[i + 1];
            }
        }
        dif.len = maxlen;
        while(dif.s[dif.len - 1] == 0 && dif.len > 1)
            --dif.len;
        return dif;
    }

    //高精度乘,实际上加和乘对进位的处理有所不同
    bign operator * (const bign &b) const{
        bign pro;
        pro.len = 0;
        for(int i = 0; i < len; ++i){
            for(int j = 0; j < b.len; ++j){
                pro.s[i + j] += (s[i] * b.s[j]);
                pro.s[i + j + 1] += pro.s[i + j] / 10;
                pro.s[i + j] %= 10;
            }
        }
        pro.len = len + b.len + 1;//这里注意pro.len初始值可能是题目数字范围两倍
        while(pro.s[pro.len - 1] == 0 && pro.len > 1)
            --pro.len;//最后一位不管是不是0都不能让len - 1
        if(pro.s[pro.len])
            ++pro.len;//这句有待商讨
        return pro;
    }

    //高精度乘以低精度
    bign operator * (const int num) const{
        int c = 0,t;
        bign pro;
        for(int i = 0; i < len; ++i){
            t = s[i] * num + c;
            pro.s[i] = t % 10;
            c = t / 10;
        }
        pro.len = len;
        while(c != 0){
            pro.s[pro.len++] = c % 10;
            c /= 10;
        }
        return pro;
    }

    //高精度除,模拟连减
    bign operator / (const bign & b) const{
        bign quo,f;
        for(int i = len - 1; i >= 0; --i){
            f = f * 10;
            f.s[0] = s[i];
            while(f >= b){
                f = f - b;
                ++quo.s[i];
            }
        }
        quo.len = len;
        while(quo.s[quo.len - 1] == 0 && quo.len > 1)
            --quo.len;
        return quo;
    }

    //比较运算符
    bool operator < (const bign & b) const{
        if(len != b.len) return len < b.len;
        for(int i = len - 1; i >= 0; --i)//从高位开始比较
            if(s[i] != b.s[i])
                return s[i] < b.s[i];
        //如果 本身 == b
        return false;
    }

    bool operator > (const bign &b) const{
        return b < *this;//代表 本身 > b
    }

    bool operator <= (const bign &b) const{
        return !(b < *this);//带表 !(本身 > b)
    }

    bool operator >= (const bign &b) const{
        return !(*this < b);
    }

    bool operator != (const bign &b) const{
        return *this < b || b < *this;
    }

    bool operator == (const bign &b) const{
        return !(*this < b) && !(b < *this);
    }

    friend istream & operator >> (istream & in,bign & x);
    friend ostream & operator << (ostream & out,const bign & x);
};

istream & operator >> (istream & in,bign & x){
    string ts;
    in >> ts;
    x = ts.c_str();
    return in;
}

ostream & operator << (ostream & out,const bign & x){
    out << x.str();
    return out;
}
int main(){
    char st1[maxn + 5];
    bign n1,sum;
    
    while(scanf("%s",st1) == 1){
        if(st1[0] == '0')
            break;
        n1 = st1;
        sum = sum + n1;
    }
    cout << sum << endl;

    return 0;
}