题解：P1001 A+B Problem

· · 题解

该给我过题解了

AC记录

这是一道特别困难的题目，难度应该有紫题。

我们可以发现，a 和 b 的数据范围特别大，居然超过了 short！我们只好拿出秘密武器：高精度。我们可以发现 (a+b)^2=a^2+b^2+2ab。所以可以使用高精度乘法算出来 a^2,b^2,2ab，再使用高精度开根算出绝对值 a+b。这时候我们就需要特判。若 a 为正数且 b 为正数，则 a+b 为正数。若 a 为负数且 b 为负数，则 a+b 为负数。若 a 与 b 一正一负，则要比较 a 与 b 的绝对值，用到高精度比较。

高精度开根难度为紫，高精度乘法难度为绿，高精度加法难度为橙，高精度比较难度为橙，为什么这题是红题！

上代码

#include <bits/stdc++.h>
using namespace std;
string qf (string a)
{
    if (a[0] == '-') return a.substr (1 , a.size () - 1);
    return "-" + a;
}
string operator - (string a) {return qf (a);}
string gjc (string a , string b)
{
    if (a[0] == '-' && b[0] == '-') return gjc (- a , - b);
    else if (a[0] == '-') return - gjc (- a , b);
    else if (b[0] == '-') return - gjc (a , - b);
    int ans[a.size () + b.size ()] = {};
    reverse (a.begin () , a.end ());
    reverse (b.begin () , b.end ());
    for (int i = 0;i < a.size ();i ++)
        for (int j = 0;j < b.size ();j ++) ans[i + j] += (a[i] - '0') * (b[j] - '0');
    for (int i = 0;i < a.size () + b.size ();i ++)
        if (ans[i] > 9)
        {
            int x = ans[i] % 10 , y = ans[i] / 10;
            ans[i] = x;
            ans[i + 1] += y;
        }
    string ans2 = "";
    for (int i = 1;i <= a.size () + b.size ();i ++) ans2 += (char) (ans[i - 1] + '0');
    if (ans2[ans2.size () - 1] == '0') ans2 = ans2.substr (0 , ans2.size () - 1);
    reverse (ans2.begin () , ans2.end ());
    return ans2;
}
string max (string a , string b)
{
    if (a[0] == '-' && b[0] == '-')
    {
        if (- max (- a , - b) == a) return b;
        return a;
    }
    else if (a[0] == '-') return b;
    else if (b[0] == '-') return a;
    else if (a.size () > b.size ()) return a;
    else if (a.size () < b.size ()) return b;
    else
    {
        for (int i = 0;i < a.size ();i ++)
            if (a[i] > b[i]) return a;
            else if (a[i] < b[i]) return b;
        return a;
    }
}
string min (string a , string b)
{
    if (max (a , b) == a) return b;
    return a;
}
string gjj (string a , string b);
string cut (string a , string b)
{
    if (a[0] == '-' && b[0] == '-') return cut (- b , - a);
    else if (a[0] == '-') return - gjj (- a , b);
    else if (b[0] == '-') return gjj (b , - a);
    else if (max (a , b) != a) return - cut (b , a);
    reverse (a.begin () , a.end ());
    reverse (b.begin () , b.end ());
    string ans = "";
    for (int i = 1;i <= a.size () + 1;i ++) ans += "0";
    for (int i = 0;i < b.size ();i ++) ans[i] += a[i] - b[i];
    for (int i = b.size ();i < a.size ();i ++) ans[i] = a[i];
    for (int i = 0;i < ans.size ();i ++)
        if (ans[i] < '0')
        {
            ans[i] += 10;
            ans[i + 1] -= 1;
        }
    while (ans.size () > 1 && ans[ans.size () - 1] == '0') ans = ans.substr (0 , ans.size () - 1);
    reverse (ans.begin () , ans.end ());
    return ans;
}
string gjj (string a , string b)
{
    if (a[0] == '-' && b[0] == '-') return - gjc (- a , - b);
    else if (a[0] == '-') return cut (b , - a);
    else if (b[0] == '-') return cut (a , - b);
    if (a.size () < b.size ()) swap (a , b);
    reverse (a.begin () , a.end ());
    reverse (b.begin () , b.end ());
    string ans = "";
    for (int i = 1;i <= a.size () + 1;i ++) ans += "0";
    for (int i = 0;i < b.size ();i ++) ans[i] += a[i] - '0' + b[i] - '0';
    for (int i = b.size ();i < a.size ();i ++) ans[i] = a[i];
    for (int i = 0;i < ans.size ();i ++)
        if (ans[i] > '9')
        {
            ans[i] -= 10;
            ans[i + 1] += 1;
        }
    if (ans[ans.size () - 1] == '0') ans = ans.substr (0 , ans.size () - 1);
    reverse (ans.begin () , ans.end ());
    return ans;
}
string dev (string b , int a)
{
    if (a < 0 && b[0] == '-') return dev (- b , - a);
    else if (a < 0) return - dev (b , - a);
    else if (b[0] == '-') return - dev (- b , a);
    while (a % 10 == 0 && b[b.size () - 1] == '0')
    {
        a /= 10;
        b = b.substr (0 , b.size () - 1);
    }
    string ans = "";
    int yu = 0;
    for (int i = 0;i < b.size ();i ++)
    {
        yu = yu * 10 + b[i] - '0';
        ans += yu / a + '0';
        yu %= a;
    }
    while (ans.size () > 1 && ans[0] == '0') ans = ans.substr (1);
    return ans;
}
string operator / (string a , int b) {return dev (a , b);}
string operator * (string a , string b) {return gjc (a , b);}
string operator + (string a , string b) {return gjj (a , b);}
string operator - (string a , string b) {return cut (a , b);}
string sqrt (string s)
{
    if (s == "0") return "0";
    if (s == "1") return "1";
    string l = "1" , r = "1" , ans , dan = "1";
    for (int i = 1;i < s.size () / 2;i ++) l += "0";
    for (int i = 1;i <= s.size () / 2 + 1;i ++) r += "0";
    while (max (l , r) == r)
    {
        string mid = (l + r) / 2;
        if (max (mid * mid , s) == s)
        {
            ans = mid;
            l = gjj (mid , dan);
        }
        else r = mid - dan;
    }
    return ans;
}
signed main ()
{
    string a , b , c = "2";
    cin >> a >> b;
    if (a[0] == '-' && b[0] == '-') cout << - sqrt (a * a + b * b + c * a * b);
    else if (a[0] == '-')
    {
        if (max (- a , b) == b) cout << sqrt (a * a + b * b + c * a * b);
        else cout << - sqrt (a * a + b * b + c * a * b);
    }
    else if (b[0] == '-')
    {
        if (max (- b , a) == a) cout << sqrt (a * a + b * b + c * a * b);
        else cout << - sqrt (a * a + b * b + c * a * b);
    }
    else cout << sqrt (a * a + b * b + c * a * b);
}