高精度（较难）

struct BigInt {
 
    int n, f;  // n 是位数, f = 1 是正数, f = -1 是负数
    int a[N];
     
    BigInt(): n(0), f(1) {}
     
    BigInt(int x) {
        n = 0, f = (x < 0 ? -1 : 1);
        x = f * x;
        while (x != 0) a[n++] = x % 10, x /= 10;
    }
     
    BigInt(string& A) {
        n = 0, f = (A[0] == '-' ? -1 : 1);
        for (int i = (int) A.size() - 1; i >= (f == -1 ? 1 : 0); i--) a[n++] = A[i] - '0';
    }
     
    BigInt& operator =(int x) {
        n = 0, f = (x < 0 ? -1 : 1);
        x = f * x;
        while (x != 0) a[n++] = x % 10, x /= 10;
        return *this;
    }
     
    BigInt& operator =(const BigInt& x) {
        n = 0, f = x.f;
        for (int i = 0; i < x.n; i++) a[n++] = x.a[i];
        return *this;
    }
     
    // 判断正负, 进位/借位, 删除前导 0
    void arrange() {
        // 先判断正负, 取决于最高的非 0 位的正负
        f = 1;  // 默认是正数, 因为有可能是 0
        for (int i = n - 1; i >= 0; i++) if (a[i] != 0) {
            f = (a[i] < 0 ? -1 : 1);
            break;
        }
         
        int d = 0;  // 借位或者进位
        for (int i = 0; i < n; i++) {
            int t = a[i] * f + d;
            a[i] = t % 10;
            d = t / 10;
            if (a[i] < 0) a[i] += 10, d--;
        }
        if (d > 0) a[n++] = d;
        while (n > 1 && a[n - 1] == 0) n--; 
    }
     
    // BigInt + BigInt 
    BigInt operator +(const BigInt& x) {
        BigInt z;
        z.n = max(n, x.n);
        for (int i = 0; i < z.n; i++) z.a[i] = 0;
        for (int i = 0; i < n; i++) z.a[i] += f * a[i];
        for (int i = 0; i < x.n; i++) z.a[i] += x.f * x.a[i];
        z.arrange();
        return z;
    }
 
    // BigInt + int 
    BigInt operator +(int x) {
        BigInt w = x;
        return *this + w;
    }
     
    // BigInt - Bigint 
    BigInt operator -(const BigInt& x) {
        BigInt w = x;
        w.f *= -1;
        return *this + w;
    }
     
    // BigInt - int 
    BigInt operator -(int x) {
        BigInt w = x;
        return *this - w;
    }
 
    // BigInt * Bigint 
    BigInt operator *(const BigInt& x) {
        BigInt z;
        z.n = n + x.n - 1;
        for (int i = 0; i < z.n; i++) z.a[i] = 0;
        for (int i = 0; i < n; i++) for (int j = 0; j < x.n; j++) {
            z.a[i + j] += f * a[i] * x.f * x.a[j];
        }
        z.arrange();
        return z;
    }
     
    // BigInt * int 
    BigInt operator *(int x) {
        BigInt w = x;
        return *this * w;
    }
     
    // BigInt / int, 只支持正数
    BigInt operator /(int x) {
        int d = 0;
        BigInt z = *this;
        for (int i = n - 1; i >= 0; i--) {
            d = d * 10 + a[i];
            z.a[i] = d / x;
            d = d % x;
        }
        while (z.n > 1 && z.a[z.n - 1] == 0) z.n--;
        return z;
    }
     
    // BigInt % int, 只支持整数
    int operator %(int x) {
        int y = 0;
        for (int i = n - 1; i >= 0; i--) {
            y = y * 10 + a[i];
            y = y % x;
        }
        return y;
    }
     
    void print() {
        if (n == 0) cout << 0 << '\n';
        else {
            if (f == -1) cout << '-';
            for (int i = n - 1; i >= 0; i--) cout << a[i];
            cout << '\n';
        }
    }
};

这个语法真的很难！希望有人能看懂好吧。。