格式不好重发*2

/*
特别考验细节的模拟。
坑1：未知数系数为1或-1。 坑2：最后一个常数项漏存。 坑3：-0.000。
*/ 
#include<cstdio>
#include<cstring>
int main()
{
    char c,s[505];
    int x=0,y=0,a=0,f=1,f1=1,i;
//x为未知数系数（=左），y为常数项（=右），a为正在读入的常数值，f为a的符号，f1为a是否在等号左边。 
    double ans;
    scanf("%s",s);
    for (i=0;i<strlen(s);i++)
    {
        if (s[i]>='0'&&s[i]<='9') a=a*10+s[i]-48; 
        if (s[i]=='-')
        {
            y+=-1*f*f1*a;  //常数项保存，如果前面是未知数此时a为0。 
            f=-1;
            a=0;
        }
        if (s[i]=='+')
        {
            y+=-1*f*f1*a;  //同上。 
            f=1;
            a=0;
        }
        if (s[i]=='=')
        {
            y+=-1*f*f1*a;  //同上。 
            f1=-1;
            f=1;
            a=0;
        }
        if (s[i]>='a'&&s[i]<='z')
        {
            c=s[i];
            if (a) x+=f*a*f1; else x+=f*f1; //未知数系数保存。 
            f=1;
            a=0;
        }
    }
    y+=-1*f*f1*a;  //最后一个常数项。 
    ans=1.0*y/x;   //计算结果。 
    if (ans<=0.000&&ans>-0.0005) printf("%c=0.000",c);
    //最后一个点会卡掉a=-0.000，注意(-0.0005,0.000]的区间。 
    else printf("%c=%.3lf",c,ans);
    return 0;
}

"""
https://vijos.org/p/1344

算比较难的题目了, 当成basic calculator去做, 字符串处理建议使用Python写

维护三个变量: parameter, total, variable
处理过程分为两部分: =左边和=右边. =左边按照正常的运算符号去处理, =右边的全部取反, 作为移项操作

然后以左边方程为例, 讨论输入的字符: 

1. 首先如果输入的字符是"+"和"-", 直接加入到stack中. 

2. 然后输入的字符是0-9的数字, 首先检查stack[-1]是否是数字, 如果是, 需要num*10+digit然后再插入栈顶
插入完之后检查下一位, 如果下一位越界或者不再是数字, 说明当前数字处理完了. 把它拿出来检查stack[-1]是否是"+"和"-". 
如果是的, 把符号拿出来并更新当前数字的正负号. 

3. 然后处理字母. 字母是未知数, 分三种情况讨论. 第一种情况是stack为空, 这时候说明字母在方程的第一位, 系数是1, 累加到parameter
第二种情况是stack[-1]是"+"或"-", 把栈顶符号取出, 系数分别是1和-1, 累加到parameter
第三种情况是stack[-1]是数字, 数字本身就是方程的系数了, 取出累加到parameter
所有情况都讨论完之后, 把当前字符更新给variable表示方程字符

全部处理完之后, stack里面全是数字, 把他们一个个累加到total, 作为方程的常数

最后-(total/parameter)就是答案, 注意保留三位小数和格式输出
"""


if __name__ == "__main__":
    s = input()
    div = s.index("=")
    total = parameter = 0
    variable = ""
    stack = []
    for i in range(div):
        ch = s[i]
        if ch == " ":
            continue
        elif ch == "+" or ch == "-":
            stack.append(ch)
        elif ch.isnumeric():
            digit = int(ch)
            # append the number 
            if stack and isinstance(stack[-1], int):
                num = stack.pop() * 10 + digit
                stack.append(num)
            else:
                stack.append(digit)
            # check if the number has ended
            if i+1 >= div or not s[i+1].isnumeric():
                num = stack.pop()
                if len(stack) == 0:
                    stack.append(num)
                elif isinstance(stack[-1], str):
                    sign = stack.pop()
                    if sign == "+":
                        stack.append(num)
                    else:
                        stack.append(-num)
        else:
            variable = ch
            if len(stack) == 0:
                parameter += 1
            elif isinstance(stack[-1], str):
                sign = stack.pop()
                if sign == "+":
                    parameter += 1
                else:
                    parameter -= 1
            else:
                parameter += stack.pop()
    
    while stack:
        total += stack.pop()

    for i in range(div+1,len(s)):
        ch = s[i]
        if ch == " ":
            continue
        elif ch == "+" or ch == "-":
            stack.append(ch)
        elif ch.isnumeric():
            digit = int(ch)
            # append the number 
            if stack and isinstance(stack[-1], int):
                num = stack.pop() * 10 + digit
                stack.append(num)
            else:
                stack.append(digit)
            # check if the number has ended
            if i+1 >= len(s) or not s[i+1].isnumeric():
                num = stack.pop()
                if len(stack) == 0:
                    stack.append(num)
                elif isinstance(stack[-1], str):
                    sign = stack.pop()
                    if sign == "+":
                        stack.append(num)
                    else:
                        stack.append(-num)
        else:
            variable = ch
            if len(stack) == 0:
                parameter -= 1
            elif isinstance(stack[-1], str):
                sign = stack.pop()
                if sign == "+":
                    parameter -= 1
                else:
                    parameter += 1
            else:
                parameter -= stack.pop()
    while stack:
        total -= stack.pop()
    
    res = - (total / parameter)
    print('%s=%.3f' % (variable, res))

#include<bits/stdc++.h>
using namespace std;

char x;
long long coe,value,opp=1,cur=0,sign=1;
double ans;
bool value_gotten=false; 
int main(){
    char c=getchar();
    while(true){
        if(c>='a' && c<='z'){
            x=c;
            if(cur==0 && !value_gotten){
                coe+=opp*sign;
                value_gotten=false;
            }
            else{
                coe+=opp*sign*cur;
                cur=0;
                sign=1;
                value_gotten=false;
            }
        }
        else if(c=='-'){
            value+=-opp*sign*cur;
            cur=0;
            sign=-1;
            value_gotten=false; 
        }
        else if(c=='+'){
            value+=-opp*sign*cur;
            cur=0;
            sign=1;
            value_gotten=false;
        }
        else if(c>='0' && c<='9'){
            cur=cur*10+c-'0';
            value_gotten=true;
        }
        else if(c=='='){
            value+=-opp*sign*cur;
            cur=0;
            sign=1;
            opp=-opp;
            value_gotten=false; 
        }
        else{
            value+=-opp*sign*cur;
            break;
        }
        c=getchar();
    }
    ans=double(value)/coe;
    printf("%c=%.3lf",x,ans==0?abs(ans):ans); 
    return 0;
}

var
p,k,i,gs,wzs,sz:longint;
s:string;
fh,wz:char;
float:real;
begin
readln(s);
k:=pos('=',s);
fh:='+';
for i:=1 to k-1 do
begin
if s[i] in ['a'..'z'] then
begin
if gs=0 then
gs:=1;
if fh='+' then
inc(wzs,gs) else
dec(wzs,gs);
wz:=s[i];
gs:=0;
end;
if s[i] in ['0'..'9'] then
begin
val(s[i],p);
gs:=gs*10+p;
end;
if s[i]='+' then
begin
if fh='+' then
inc(sz,gs) else
dec(sz,gs);
gs:=0;
fh:='+';
end;
if s[i]='-' then
begin
if fh='+' then
inc(sz,gs) else
dec(sz,gs);
gs:=0;
fh:='-';
end;
end;
if fh='+' then
inc(sz,gs) else
dec(sz,gs);
gs:=0;
sz:=(-1)*sz;
fh:='+';
for i:=k+1 to length(s) do
begin
if s[i] in ['a'..'z'] then
begin
if gs=0 then
gs:=1;
if fh='-' then
inc(wzs,gs) else
dec(wzs,gs);
wz:=s[i];
gs:=0;
end;
if s[i] in ['0'..'9'] then
begin
val(s[i],p);
gs:=gs*10+p;
end;
if s[i]='+' then
begin
if fh='+' then
inc(sz,gs) else
dec(sz,gs);
gs:=0;
fh:='+';
end;
if s[i]='-' then
begin
if fh='+' then
inc(sz,gs) else
dec(sz,gs);
gs:=0;
fh:='-';
end;
end;
if fh='+' then
inc(sz,gs) else
dec(sz,gs);
float:=sz/wzs;
if float=0.00000 then
writeln(wz,'=0.000') else
writeln(wz,'=',sz/wzs:0:3);
end.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
string a;
int main()
{
cin>>a;
double x1=0,x2=0,y1=0,y2=0;
int k=a.find('=',0),lena=a.size()-1,yx=1;
int x=0,flag=1,d;
for(d=0;d<k;++d)
{
if(a[d]>='0'&&a[d]<='9')x=x*10+a[d]-'0';
if(a[d]!='+'&&a[d]!='-'&&(a[d]>'9'||a[d]<'0'))
{
if(flag){cout<<a[d]<<"=";flag=0;}
if(x==0)y1+=2*yx-1;
else y1+=(2*yx-1)*x;
x=0;
}
if(a[d]=='-'&&x!=0)x1+=(2*yx-1)*x,yx=0,x=0;
if(a[d]=='+'&&x!=0)x1+=(2*yx-1)*x,yx=1,x=0;
if(a[d]=='-')yx=0;
if(a[d]=='+')yx=1;
}
if(a[d]=='='&&x!=0)x1+=(2*yx-1)*x;
x=0;yx=1;
for(d=k+1;d<=lena;++d)
{
if(a[d]>='0'&&a[d]<='9')x=x*10+a[d]-'0';
if(a[d]!='+'&&a[d]!='-'&&(a[d]>'9'||a[d]<'0'))
{
if(flag)cout<<a[d]<<"=",flag=0;
if(x==0)y2+=2*yx-1;
else y2+=(2*yx-1)*x;
x=0;
}
if(a[d]=='-'&&x!=0)x2+=(2*yx-1)*x,yx=0,x=0;
if(a[d]=='+'&&x!=0)x2+=(2*yx-1)*x,yx=1,x=0;
if(a[d]=='-')yx=0;
if(a[d]=='+')yx=1;
}
if(d==lena+1&&x!=0)x2+=(2*yx-1)*x;
printf("%.3lf",(x1-x2)/(y2-y1));
return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
string a;
int main()
{
cin>>a;
double x1=0,x2=0,y1=0,y2=0;
int k=a.find('=',0),lena=a.size()-1,yx=1;
int x=0,flag=1,d;
for(d=0;d<k;++d)
{
if(a[d]>='0'&&a[d]<='9')x=x*10+a[d]-'0';
if(a[d]!='+'&&a[d]!='-'&&(a[d]>'9'||a[d]<'0'))
{
if(flag){cout<<a[d]<<"=";flag=0;}
if(x==0)y1+=2*yx-1;
else y1+=(2*yx-1)*x;
x=0;
}
if(a[d]=='-'&&x!=0)x1+=(2*yx-1)*x,yx=0,x=0;
if(a[d]=='+'&&x!=0)x1+=(2*yx-1)*x,yx=1,x=0;
if(a[d]=='-')yx=0;
if(a[d]=='+')yx=1;
}
if(a[d]=='='&&x!=0)x1+=(2*yx-1)*x;
x=0;yx=1;
for(d=k+1;d<=lena;++d)
{
if(a[d]>='0'&&a[d]<='9')x=x*10+a[d]-'0';
if(a[d]!='+'&&a[d]!='-'&&(a[d]>'9'||a[d]<'0'))
{
if(flag)cout<<a[d]<<"=",flag=0;
if(x==0)y2+=2*yx-1;
else y2+=(2*yx-1)*x;
x=0;
}
if(a[d]=='-'&&x!=0)x2+=(2*yx-1)*x,yx=0,x=0;
if(a[d]=='+'&&x!=0)x2+=(2*yx-1)*x,yx=1,x=0;
if(a[d]=='-')yx=0;
if(a[d]=='+')yx=1;
}
if(d==lena+1&&x!=0)x2+=(2*yx-1)*x;
printf("%.3lf",(x1-x2)/(y2-y1));
return 0;
}

很长很傻瓜的list解法
简明易懂
只是代码很长

// P1022.cpp: 定义控制台应用程序的入口点。
//


#include <iostream>
#include <cstdio>
#include <cstring>
#include <list>
#include <cmath>
#include <cctype>
#include <cstdlib>


using namespace std;
list<char> L, L1, L2;
double unk = 0, c = 0;
char unkw; int yes = 1;

void intial()
{
    if (L.front() != '+' && L.front() != '-')
    {
        L.push_front('+');
    }
    int flag = 0;
    for (list<char>::iterator it = L.begin(); it != L.end(); it++)
        if (*it >= 'a'&&*it <= 'z')
        {
            it--;
            if (*it<'0' || *it>'9')
            {
                it++;
                L.insert(it, '1');
                it--;
            }
            it++;
        }
    for (list<char>::iterator it = L.begin(); it != L.end(); it++)
    {

        if (*it == '=')
        {
            flag = 1;
            it++;
            if (*it != '+' && *it != '-')
            {
                L.insert(it, '+');
            }
            it--;
        }
        if (flag == 0)
        {
            L1.push_back(*it);
        }
        if (flag == 1)
        {
            L2.push_back(*it);
        }
    }
    L1.push_back('+');
    L2.push_back('+');
}

void debug()
{
    cout << "left:";
    for (list<char>::iterator it = L1.begin(); it != L1.end(); it++)
        cout << *it;
    cout << endl << "right:";
    for (list<char>::iterator it = L2.begin(); it != L2.end(); it++)
        cout << *it;
    cout << endl;
}

int ltoi(list<char> tlist)
{
    int ans = 0;
    int i = 1;
    int minus = 0;
    if (tlist.front() == '-')minus = 1;
    tlist.reverse();
    for (list<char>::iterator it = tlist.begin(); it != tlist.end(); it++)
    {
        if (*it >= '0'&&*it <= '9')
        {
            ans += ((*it) - '0')*i;
            i *= 10;
        }
    }
    return minus ? (-ans) : (ans);
}

void deal(list<char> tlist)
{
    if (tlist.back() >= '0'&&tlist.back() <= '9')
    {
        c += ltoi(tlist);
    }
    else
    {
        unk += ltoi(tlist);
        if (yes)
        {
            unkw = tlist.back();
            yes = 0;
        }
    }
}

void judge()
{
    int cnt = 0;
    list<char> curr;
    list<char>::iterator left, right;
    for (list<char>::iterator it = L1.begin(); it != L1.end(); it++)
    {
        if (*it == '+' || *it == '-')
        {
            cnt++;
            if (cnt == 1)
            {
                left = it;
            }
            if (cnt == 2)
            {
                right = it--; //it++;
                curr.assign(left, right);
                cnt = 0;
                deal(curr);
                curr.clear();
            }
        }
    }


}

void judge2()
{
    int cnt = 0;
    list<char> curr;
    list<char>::iterator left, right;
    for (list<char>::iterator it = L2.begin(); it != L2.end(); it++)
    {
        if (*it == '+' || *it == '-')
        {
            cnt++;
            if (cnt == 1)
            {
                left = it;
            }
            if (cnt == 2)
            {
                right = it--; //it++;
                curr.assign(left, right);
                if (curr.front() == '+')
                {
                    curr.pop_front();
                    curr.push_front('-');
                }
                else
                {
                    curr.pop_front();
                    curr.push_front('+');
                }
                cnt = 0;
                deal(curr);
                curr.clear();
            }
        }
    }
}

int main()
{
    string str;
    cin >> str;
    for (int i = 0; i < str.length(); i++)
    {
        L.push_back(str[i]);
    }
    intial();

//  debug();
    judge();
    judge2();
    double x = 0;
    x = -(c / unk);
    if (x == 0)x = abs(x);
//  cout << c << ' ' << unk << endl;
    cout << unkw << '=';
    printf("%.3f", x);

    return 0;
}

比较考细节
分别算出两边的未知数系数和常数，再求未知数

#include<iostream>
#include<cstring> 
#include<cstdio>
using namespace std;
char s[30];
int main()
{
    int i=0,n,flag,sum1=0,xi1=0,xi2=0,sum2=0,now,zi;
    double ans;
    cin>>s;
    n=strlen(s);
    while(i<n)
    {
        flag=1;
        if(s[i]=='=')
          break;
        if(s[i]=='+')
         {  
         flag=1;
         i++;
         }
        else
         if(s[i]=='-')
         {  
          flag=-1;
          i++;
         }
         if(s[i]=='=')
          break;
        now=0;
        while(s[i]>='0'&&s[i]<='9')
        {   
            now*=10; 
            now+=s[i]-'0';
            i++;
        }
        if(s[i]>='a'&&s[i]<='z')
         {  
          zi=s[i];
          xi1+=flag*now;
          i++;
         }
        else
         {
         sum1+=flag*now;
         if(s[i]=='=')
          break;
         }
        if(s[i]=='=')
          break;
    }
    i++;
    while(i<n)
    {
        flag=1;
        if(i>=n)
          break;
        if(s[i]=='+')
         {  
         flag=1;
         i++;
         }
        else
         if(s[i]=='-')
         {  
          flag=-1;
          i++;
         }
         if(i>=n)
          break;
        now=0;
        while(s[i]>='0'&&s[i]<='9')
        {   
            now*=10; 
            now+=s[i]-'0';
            i++;
        }
        if(s[i]>='a'&&s[i]<='z')
         {  
          zi=s[i];
          xi2+=flag*now;
          i++;
         }
        else
         {
         sum2+=flag*now;
         if(i>=n)
          break;
         }
        if(i>=n)
          break;
    }
    ans=double(double(sum2)-double(sum1))/double(double(xi1)-double(xi2));
    cout<<char(zi)<<"=";
    printf("%.3lf",ans);
    return 0;
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
string a;
int main()
{
cin>>a;
double x1=0,x2=0,y1=0,y2=0;
int k=a.find('=',0),lena=a.size()-1,yx=1;
int x=0,flag=1,d;
for(d=0;d<k;++d)
{
if(a[d]>='0'&&a[d]<='9')x=x*10+a[d]-'0';
if(a[d]!='+'&&a[d]!='-'&&(a[d]>'9'||a[d]<'0'))
{
if(flag){cout<<a[d]<<"=";flag=0;}
if(x==0)y1+=2*yx-1;
else y1+=(2*yx-1)*x;
x=0;
}
if(a[d]=='-'&&x!=0)x1+=(2*yx-1)*x,yx=0,x=0;
if(a[d]=='+'&&x!=0)x1+=(2*yx-1)*x,yx=1,x=0;
if(a[d]=='-')yx=0;
if(a[d]=='+')yx=1;
}
if(a[d]=='='&&x!=0)x1+=(2*yx-1)*x;
x=0;yx=1;
for(d=k+1;d<=lena;++d)
{
if(a[d]>='0'&&a[d]<='9')x=x*10+a[d]-'0';
if(a[d]!='+'&&a[d]!='-'&&(a[d]>'9'||a[d]<'0'))
{
if(flag)cout<<a[d]<<"=",flag=0;
if(x==0)y2+=2*yx-1;
else y2+=(2*yx-1)*x;
x=0;
}
if(a[d]=='-'&&x!=0)x2+=(2*yx-1)*x,yx=0,x=0;
if(a[d]=='+'&&x!=0)x2+=(2*yx-1)*x,yx=1,x=0;
if(a[d]=='-')yx=0;
if(a[d]=='+')yx=1;
}
if(d==lena+1&&x!=0)x2+=(2*yx-1)*x;
printf("%.3lf",(x1-x2)/(y2-y1));
return 0;
}

var

s,s1,s2,s3:string; c:char;

i,j,l,l1,l2,w:integer; a,b,m:real;

begin

readln(s); l:=length(s); w:=pos('=',s); a:=0; b:=0;

for i:=1 to l do

if ((s[i]'9')) and (s[i]'-')

and (s[i]'=') and (s[i]'+') and (s[i]'.') then begin c:=s[i]; break; end;

s1:=copy(s,1,w-1); s2:=copy(s,w+1,l-w);

l1:=length(s1); l2:=length(s2);

while l1>0 do

begin

i:=2;

while (i

做完难题水题后这道题就很水了

program p1344;

const biao=[ord('a')..ord('z')];

type x=array[1..2] of longint;

var a,b:x;

str,str1,str2,ch:string;

k:longint;

root:real;

procedure init;

var i:longint;

begin

readln(str);

k:=pos('=',str);

for i:=1 to length(str) do if ord(str[i]) in biao then begin ch:=str[i]; break; end;

str1:=copy(str,1,k-1);

str2:=copy(str,k+1,length(str));

if str1[1]'-' then str1:='+'+str1;

str1:=str1+'+';

str2:=str2+'+';

fillchar(a,sizeof(a),0);

fillchar(b,sizeof(b),0);

end;

procedure calstr(str:string;k:byte);

var i,fh,xs:longint;

d:0..9;

begin

fh:=1; xs:=0;

if ord(str[1]) in biao then str:='+'+str;

for i:=1 to length(str) do

begin

case str[i] of

'0'..'9':begin val(str[i],d); xs:=xs*10+d; end;

'+':begin if ord(str) in biao then begin

if str='+' then inc(a[k],1)

else if str='-' then inc(a[k],-1)

else a[k]:=a[k]+fh*xs

end

else b[k]:=b[k]+fh*xs;

fh:=1; xs:=0;

end;

'-':begin if ord(str) in biao then begin

if str='+' then inc(a[k],1)

else if str='-' then inc(a[k],-1)

else a[k]:=a[k]+fh*xs

end

else b[k]:=b[k]+fh*xs;

fh:=-1; xs:=0;

end;

end;

end;

fh:=1; xs:=0;

end;

begin

init;

calstr(str1,1); calstr(str2,2);

if a[1]-a[2]0 then begin root:=(b[2]-b[1])/(a[1]-a[2]); if root=0 then root:=0; end;

writeln(ch,'=',root:0:3);

end.

用递归的思想

把方程左右两边各分离常数项和一次项，然后整理可变问ax=b的形式，最后解得x

program ex;

var i,j,weizhi,changshu:longint;

s:string;

x:char;

ans:real;

procedure main;

var i,j,a,m:longint;

st:string;

begin

for i:=1 to length(s) do

if (s[i] in['0'..'9','+','-','='])=false then

begin

x:=s[i];

break;

end;

i:=1;

while i

var

s,a,b:ansistring;

i,n,k,ma,na,mb,nb,q,m:longint;

ans:real; ch:char;

begin

readln(s);

for i:=1 to length(s) do

if s[i]='=' then

begin k:=i; break; end;

a:=copy(s,1,k-1); b:=copy(s,k+1,length(s)-k);

a:=a+'+'; b:=b+'+';

n:=0; q:=1; i:=1;

while i

program p1344(input,output);

var

s,s1,s2:string;

a:real;

ch:char;

i,c,m,l,b,d,x1,x2,j:integer;

procedure lj(ss:string);

var

i,l:integer;

begin

l:=length(ss);

if (ss[l]>='0')and(ss[l]='0')and(ss[l]

我这个沙茶交了四遍才AC

第一遍调试信息没删干净, 0分

第二遍没处理 x=-0.000 的情况, 83分

第三遍直接在记事本里改, 结果 No Compiled 0分

可见我有多沙茶

var

s:

├ 测试数据 01：答案错误...　├ 标准行输出

　├ 错误行输出

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Unaccepted 有效得分：83 有效耗时：0ms

为什么受伤的总是我

今天第5道字符串处理....

烦死我了

字符串处理 细节题 有点诡异的是‘-0.000’，这个...- -！

program p1344;

var

sys:set of char;

s,temp:string;

can:boolean;

k:char;

top1,top2,x1,x2,num,add,del,i,j:longint;

p:real;

begin

sys:=['+','-'];

read(s);

can:=false;

for i:=1 to length(s) do

if s[i] in ['a'..'z'] then begin

k:=s[i];

break;

end;

temp:='';top1:=0;top2:=0;

for i:=1 to length(s) do

begin

if (not(s[i] in sys))and(s[i]'=')and(ilength(s)) then

begin

temp:=temp+s[i];

continue;

end;

if s[i]='=' then can:=true;

if i=length(s) then temp:=temp+s[i];

if (can=false)or(s[i]='=') then

begin

add:=pos('+',temp);

del:=pos('-',temp);

num:=0;

for j:=1 to length(temp) do

if temp[j] in ['0'..'9'] then

num:=num*10+ord(temp[j])-ord('0');

if temp[length(temp)]k then

begin

if (add=1)or((add=0)and(del=0)) then inc(top1,num)

else dec(top1,num);

end

else

begin

if num=0 then num:=1;

if (add=1)or((add=0)and(del=0)) then inc(x1,num)

else dec(x1,num);

end;

end

else

if (can=true)or(i=length(s)) then

begin

add:=pos('+',temp);

del:=pos('-',temp);

num:=0;

for j:=1 to length(temp) do

if temp[j] in ['0'..'9'] then

num:=num*10+ord(temp[j])-ord('0');

if temp[length(temp)]k then

begin

if (add=1)or((add=0)and(del=0)) then inc(top2,num)

else dec(top2,num);

end

else

begin

if num=0 then num:=1;

if (add=1)or((add=0)and(del=0)) then inc(x2,num)

else dec(x2,num);

end;

end;

temp:=s[i];

if temp='=' then temp:='';

end;

write(k,'=');

p:=(top2-top1)/(x1-x2);

if p0 then writeln(p:0:3)

else writeln('0.000');

end.

注意答案是0的时候

他竟然输出-0.000

白交了一次...