poj2295

给定一个方程的字符串，要求解一元一次方程，要细心。

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

string st;
int a, b, c, d;

int cal(string st)
{
    int a;
    sscanf(st.c_str(), "%d", &a);
    return a;
}

void make(string st, int &a, int &b)
{
    if (st == "x")
    {
        a++;
        return;
    }
    if (st == "-x")
    {
        a--;
        return;
    }
    if (st[st.length() - 1] == 'x')
        a += cal(st.substr(0, st.length() - 1));
    else
        b += cal(st);
}

int main()
{
    //freopen("t.txt", "r", stdin);
    int t;
    scanf("%d", &t);
    while (t--)
    {
        cin >> st;
        a = b = c = d = 0;
        int now = 1;
        int start = 0;
        while (1)
        {
            if (st[now] == '-' || st[now] == '+' || st[now] == '=')
            {
                make(st.substr(start, now - start), a, b);
                start = now;
            }
            if (st[now] == '=')
                break;
            now++;
        }
        now++;
        start = now;
        now++;
        while (1)
        {
            if (st[now] == '-' || st[now] == '+' || st[now] == '\0')
            {
                make(st.substr(start, now - start), c, d);
                start = now;
            }
            if (st[now] == '\0')
                break;
            now++;
        }
        a -= c;
        d -= b;
        if (a == 0 && d == 0)
            printf("IDENTITY\n");
        else if (a == 0 && d != 0)
            printf("IMPOSSIBLE\n");
        else
            printf("%d\n",(int)floor(d * 1.0 / a));
    }
    return 0;
}