高精大水题

https://www.luogu.org/problem/show?pid=T2379

题目背景

宁静祥和的原始森林,空气清新,万物复苏。小Z和小V在林间追跑,快乐非常。

题目描述

小Z好吃懒做, 虽然很胖可是却没有小V健壮, 很是不服气。就提出要求和小V比体重,谁重谁当“头头” 。虽然知道小V的体重是 a、小Z的体重是 b,如果a 小于 b,小Z做头头,否则,小V做头头。但小V还是答应了这个要求。聪明的小V称体重前偷偷地喝了 c 重量的水。请你判断比体重后谁是头头?

如果小V是头头输出小V及他的重量,如果小Z是头头输出小Z以及他的重量。

输入输出格式

输入格式:

一行,三个数,空格分开,表示 a、b 和 c。

输出格式:

两行,分别是一个字串以及一个数,

第一行:头头是小V还是小Z。

第二行:小V的重量或小Z的重量。

输入输出样例

输入样例#1:
5
2
2
输出样例#1:
little v
height:7
输入样例#2:
52.2
2
0
输出样例#2:
little v
height:52.2

说明

Hint#1:a,b,c范围在200位以内

Hint#2:重量相同时,小V仍然是头头

Hint#3:若不是纯数字,输出:wrong

Hint#4:a,b,c可能是小数

Hint#5:a,b,c都是高精度(别傻傻的用高精+单精)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;

const int MAXN = 410;

struct bign
{
    int len, s[MAXN];
    bign ()
    {
        memset(s, 0, sizeof(s));
        len = 1;
    }
    bign (int num) { *this = num; }
    bign (const char *num) { *this = num; }
    bign operator = (const int num)
    {
        char s[MAXN];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bign operator = (const char *num)
    {
        for(int i = 0; num[i] == '0'; num++) ;  //去前导0
        len = strlen(num);
        for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
        return *this;
    }
    bign operator + (const bign &b) const //+
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; g || i < max(len, b.len); i++)
        {
            int x = g;
            if(i < len) x += s[i];
            if(i < b.len) x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;
        }
        return c;
    }
    bign operator += (const bign &b)
    {
        *this = *this + b;
        return *this;
    }
    void clean()
    {
        while(len > 1 && !s[len-1]) len--;
    }
    bign operator * (const bign &b) //*
    {
        bign c;
        c.len = len + b.len;
        for(int i = 0; i < len; i++)
        {
            for(int j = 0; j < b.len; j++)
            {
                c.s[i+j] += s[i] * b.s[j];
            }
        }
        for(int i = 0; i < c.len; i++)
        {
            c.s[i+1] += c.s[i]/10;
            c.s[i] %= 10;
        }
        c.clean();
        return c;
    }
    bign operator *= (const bign &b)
    {
        *this = *this * b;
        return *this;
    }
    bign operator - (const bign &b)
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; i < len; i++)
        {
            int x = s[i] - g;
            if(i < b.len) x -= b.s[i];
            if(x >= 0) g = 0;
            else
            {
                g = 1;
                x += 10;
            }
            c.s[c.len++] = x;
        }
        c.clean();
        return c;
    }
    bign operator -= (const bign &b)
    {
        *this = *this - b;
        return *this;
    }
    bign operator / (const bign &b)
    {
        bign c, f = 0;
        for(int i = len-1; i >= 0; i--)
        {
            f = f*10;
            f.s[0] = s[i];
            while(f >= b)
            {
                f -= b;
                c.s[i]++;
            }
        }
        c.len = len;
        c.clean();
        return c;
    }
    bign operator /= (const bign &b)
    {
        *this  = *this / b;
        return *this;
    }
    bign operator % (const bign &b)
    {
        bign r = *this / b;
        r = *this - r*b;
        return r;
    }
    bign operator %= (const bign &b)
    {
        *this = *this % b;
        return *this;
    }
    bool operator < (const bign &b)
    {
        if(len != b.len) return len < b.len;
        for(int i = len-1; i >= 0; i--)
        {
            if(s[i] != b.s[i]) return s[i] < b.s[i];
        }
        return false;
    }
    bool operator > (const bign &b)
    {
        if(len != b.len) return len > b.len;
        for(int i = len-1; i >= 0; i--)
        {
            if(s[i] != b.s[i]) return s[i] > b.s[i];
        }
        return false;
    }
    bool operator == (const bign &b)
    {
        return !(*this > b) && !(*this < b);
    }
    bool operator != (const bign &b)
    {
        return !(*this == b);
    }
    bool operator <= (const bign &b)
    {
        return *this < b || *this == b;
    }
    bool operator >= (const bign &b)
    {
        return *this > b || *this == b;
    }
    string str() const
    {
        string res = "";
        for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
        return res;
    }
};

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

ostream& operator << (ostream &out, const bign &x)
{
    out << x.str();
    return out;
}

int main()
{
    bign a, b, c;
    cin>>a>>b>>c;
    a+=c;
    if(a>=b){
        cout<<"little v"<<endl<<"height:"<<a<<endl;
    }
    else cout<<"little z"<<endl<<"height:"<<b<<endl;
    return 0;
}
原文地址:https://www.cnblogs.com/bennettz/p/6491855.html