一.题目链接:

ZOJ-3999

二.题目大意:

读入有三行

一行分子表达式 

一行分母表达式 

一行  取值

求出  

三.分析:

表达式读入有些麻烦

具体见代码.

注意:题目中没有保证指数是有序排列的!

四.代码实现:

#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define eps 1e-6
#define pi acos(-1.0)
#define ll long long int
using namespace std;

struct node
{
    ll coef;
    ll expn;
} p[2][20];

void get_(string s, ll id)
{
    ll cnt = 0;
    ll coef = 1;
    ll expn = 1;
    bool flag = 1;
    ll len = s.size();
    for(ll i = 0; i < len; ++i)
    {
        if(s[i] == '-')
        {
            coef *= -1;
        }
        else if(s[i] >= '0' && s[i] <= '9')
        {
            if(flag)
            {
                coef *= s[i] - '0';
                flag = 0;
                if(i == len - 1 || s[i + 1] == '+' || s[i + 1] == '-')
                {
                    p[id][cnt].coef = coef;
                    flag = 1;
                    expn = 1;
                    coef = 1;
                    cnt++;
                }
            }
            else
            {
                expn *= s[i] - '0';
                p[id][cnt].expn = expn;
                cnt++;
                flag = 1;
                expn = 1;
                coef = 1;
            }
        }
        else if(s[i] == 'x')
        {
            p[id][cnt].coef = coef;
            flag = 0;
            if(i == len - 1 || s[i + 1] != '^')
            {
                p[id][cnt].expn = 1;
                flag = 1;
                expn = 1;
                coef = 1;
                cnt++;
            }
        }
    }
}

ll quick(ll a, ll b)
{
    ll sum = 1;
    while(b)
    {
        if(b & 1)
            sum *= a;
        a *= a;
        b >>= 1;
    }
    return sum;
}

void cal(ll &value1, ll &value2, ll x)
{
    value1 = value2 = 0;
    for(ll i = 0; i < 15; ++i)
    {
        value1 += p[0][i].coef * quick(x, p[0][i].expn);
        value2 += p[1][i].coef * quick(x, p[1][i].expn);
    }
}

void luo()
{
    for(ll i = 0; i < 15; ++i)
    {
        p[0][i].coef *= p[0][i].expn;
        p[0][i].expn--;
        if(p[0][i].expn == -1)
            p[0][i].coef = p[0][i].expn = 0;
        p[1][i].coef *= p[1][i].expn;
        p[1][i].expn--;
        if(p[1][i].expn <= -1)
            p[1][i].coef = p[1][i].expn = 0;
    }
}

void print(ll value1, ll value2, ll x)
{
    cal(value1, value2, x);
    if(value1 && value2)
    {
        if(value1 % value2 == 0)
            cout << value1 / value2 << "\n";
        else
        {
            ll gcd = __gcd(value1, value2);
            if(value1 * value2 > 0)
                cout << value1 / gcd << "/" << value2 / gcd << "\n";
            else
                cout << "-" << (ll)abs(value1 / gcd) << "/" << (ll)abs(value2 / gcd) << "\n";
        }
    }
    else if(value1 && !value2)
        cout << "INF\n";
    else if(!value1 && value2)
        cout << "0\n";
    else
    {
        luo();
        print(value1, value2, x);
    }
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        string s1, s2;
        ll x;
        cin  >> s1 >> s2 >> x;
        memset(p,0, sizeof(p));
        get_(s1, 0);
        get_(s2, 1);
        ll value1, value2;
        print(value1, value2, x);
    }
    return 0;
}