poj2295

给定一个方程的字符串,要求解一元一次方程,要细心。

View Code
#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;
}

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