#include<bits/stdc++.h>
using namespace std;
int sum1=0,sum2=0;
void read(string s)
{
sum1=0,sum2=0;
//sum1存未知数前面的系数,sum2存已知数的总和。
int i=0,sign=1,ceof=0;
while(s[i])
{
sign=1,ceof=0;
if(s[i]=='-'){
sign=-1;i++;
}
else if(s[i]=='+'){
sign=1;i++;
}
if(s[i]>='0'&&s[i]<='9'){
while(s[i]>='0'&&s[i]<='9'){
ceof=ceof*10+s[i]-'0';
i++;
}
if(s[i]>='a'&&s[i]<='z'){
sum1+=(sign*ceof);
i++;
}
else{
sum2+=(sign*ceof);
}
}
else if(s[i]>='a'&&s[i]<='z'){
ceof=1;
sum1+=(sign*ceof);
i++;
}
}
}
int main()
{
string s,s1,s2;
cin>>s;char ch;
for(int i=0;i<s.size();i++){
if(s[i]>='a'&&s[i]<='z'){
ch=s[i];break;
}
}
s1=s.substr(0,s.find("="));
s2=s.substr(s.find("=")+1,s.size()-s.find("=")-1);
read(s1);
int ceof_unknown_left=sum1;
int ceof_known_left=sum2;
read(s2);
int ceof_unknown_right=sum1;
int ceof_known_right=sum2;
int aa=ceof_unknown_left-ceof_unknown_right;
int bb=ceof_known_right-ceof_known_left;
double x=bb*1.0/aa*1.0;
printf("%c=%.3lf",ch,x);
}
最坑的就是直接带公式x=0输出-0.000……