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  • 【紫书】BigInteger 高精度类型 原书上有一个bug:A+B!=B+A

     存个代码

    struct BigInterger {
        static const int BASE = 1e8;
        static const int WIDTH = 8;
        vector<int> s;
        BigInterger(long long num = 0) { *this = num; }
        BigInterger operator =(long long num) {
            s.clear();
            do {
                s.push_back(num%BASE);
                num /= BASE;
            } while (num > 0);//num==0
            return *this;
        }
        BigInterger operator =(const string& str) {
            s.clear();
            int x, len = (str.length() - 1) / WIDTH + 1;//len==width
            for (int i = 0; i < len; i++) {
                int end = str.length() - i*WIDTH;
                int start = max(0, end - WIDTH);
                sscanf(str.substr(start, end - start).c_str(), "%d", &x);
                s.push_back(x);
            }
            return *this;
        }
    
    
        BigInterger operator +(const BigInterger& b)const {
            BigInterger c;
            c.s.clear();
            for (int i = 0, g = 0;; i++) {
                if (g == 0 && i >= max(s.size(), b.s.size()))break;//此处bug
                int x = g;
                if (i < s.size())x += s[i];
                if (i < b.s.size()) x += b.s[i];
                c.s.push_back(x%BASE);
                g = x / BASE;
            }
            return c;
        }
        BigInterger operator +=(const BigInterger& b) {
            *this = *this + b; return *this;
        }
    
        bool operator<(const BigInterger& b)const {
            if (s.size() != b.s.size()) return s.size() < b.s.size();
            for (int i = s.size() - 1; i >= 0; i--) {
                if (s[i] != b.s[i]) return s[i] < b.s[i];//Width 个数字一起比
            }
            return false;//==
        }
        bool operator>(const BigInterger &b)const { return b < *this; }
        bool operator<=(const BigInterger &b)const { return !(b < *this); }
        bool operator>=(const BigInterger &b)const { return !(*this < b); }
        bool operator!=(const BigInterger &b)const { return b < *this || *this<b; }
        bool operator==(const BigInterger &b)const { return!(b < *this) && !(*this<b); }
    };
    ostream& operator <<(ostream &out, const BigInterger& x) {
        out << x.s.back();
        for (int i = x.s.size() - 2; i >= 0; i--) {
            char buf[20];
            sprintf(buf, "%08d", x.s[i]);
            for (int j = 0; j < strlen(buf); j++)cout << buf[j];
        }
        return out;
    }
    istream&operator>>(istream &in, BigInterger&x) {
        string s;
        if (!(in >> s))return in;
        x = s;
        return in;
    }
     
    成功的路并不拥挤,因为大部分人都在颓(笑)
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  • 原文地址:https://www.cnblogs.com/SuuT/p/8733148.html
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