2845. 符号方程求解

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Kevin_K
6 年，3 月前

突然想了一段中二代码，贴出来供大佬们一笑：

Python Code
s=input()
p=0
while 1:
    if s[p]!='+' and s[p]!='-' and s[p]!='=' and not s[p].isdigit():
        a=s[p]
        break
    p+=1
s=s.replace(a,'x')
p=1
while p!=len(s):
    if s[p-1].isdigit() and s[p]=='x':
        s=s[:p]+'*'+s[p:]
    p+=1
p=s.find('=')
s=s[:p]+'-('+s[p+1:]+')'
x=0
t=eval(s)#真·核弹炸水题。emmm~
x=1
print(a+('=%.3f'%(t/(t-eval(s))) if t!=0 else '=0.000'))
#0为解巨坑！虽然可以四舍五入但小负数四舍五入后会变成-0.000！（那啥，精度问题……）
#看了大佬提示还过了很久才发现是真的菜……

ps:蟒蛇NB！
pps:C&C++什么的忘了吧！以后是玩蛇的时代！人生苦短，__。
ppps:西神这学期又变帅(qiang)了，hello新学期！
pppps:以上彩蛋只有聪明人能看见。（逃~）
ppppps:彩蛋内容本条除外，其余均大雾。（题外话：这好像是我第一次评论区贴完整代码？纪念一下。以及：空了这么多行你们是不是都发现彩蛋了。233~）

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Andrew-Malcom
4 年，7 月前 
#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);
}
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BillChen2000
5 年，9 月前

这是一段过了几天自己也注定看不太懂的代码，大概的思路就是记录下所有符号和开头结尾的位置，一个一个拆分开来判断。比较麻烦但对于x=0这种情况不用单独考虑。

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
int main(){
    char varied = '%';
    vector<int> position;
    int leftnum = 0, rightnum = 0;
    string input, temp;
    getline(cin, input);
    int equalPos = input.find('=');
    int pos = input.find('+', 0);
    while (input.find('+', pos) != string::npos){
        position.push_back(input.find('+', pos));
        pos = input.find('+', pos + 1);
    }
    pos = input.find('-', 0);
    while (input.find('-', pos) != string::npos){
        position.push_back(input.find('-', pos));
        pos = input.find('-', pos + 1);
    }
    position.push_back(equalPos);
    position.push_back(0);
    position.push_back(input.length());
    sort(position.begin(), position.end());
    for (int i = 0; i <= position.size()-1; i++){
        string temp;
        temp = input.substr(position[i], position[i + 1] - position[i]);
        if(position[i]<equalPos){
            if (temp.length() == 0)
                ;
            else if (temp[temp.length() - 1] >= 'a' && temp[temp.length() - 1] <= 'z'){
                varied = temp[temp.length() - 1];
                if (temp.length() == 1 || (temp.length()==2 && temp[0]=='+'))
                    leftnum += 1;
                else if(temp.length()==2&&temp[0]=='-')
                    leftnum -= 1;
                else
                    leftnum += atoi((temp.substr(0, temp.length() - 1)).c_str());
            }
            else{
                rightnum -= atoi(temp.c_str());
            }
        }
        else if (position[i] == equalPos){
            position[i]++;
            i--;
            continue;
        }
        else{
            if (temp.length() == 0)
                ;
            else if (temp[temp.length() - 1] >= 'a' && temp[temp.length() - 1] <= 'z'){
                varied = temp[temp.length() - 1];
                if (temp.length() == 1 || (temp.length()==2 && temp[0]=='+'))
                    leftnum -= 1;
                else if(temp.length()==2&&temp[0]=='-')
                    leftnum += 1;
                else
                    leftnum -= atoi((temp.substr(0, temp.length() - 1)).c_str());
            }
            else{
                rightnum += atoi(temp.c_str());
            }
        }
    }
    printf("%c=%.3f", varied, (double)rightnum / (double)leftnum);
    return 0;
}
回复
Li Dao
7 年，9 月前

这几天字符串的题目真是做吐了
算是我用下来比较通用的写法
EOJ 2821
EOJ 2891
EOJ 2845（本题）
这三题都可以用类似于下面的写法

include

using namespace std;
string strs;
int LA,LN,RA,RN;
char Alpha;
int main()
{
cin>>strs;
int Epos=strs.find(“=”);
int pos=0;
while(pos<Epos)
{
int symbol=1;
if(strs[pos]==’+’) pos++;
else if(strs[pos]==’-‘)
{
++pos;
symbol=-1;
}
int num=-1;
while(pos<Epos && isdigit(strs[pos]))
{
if(num==-1) num=strs[pos]-‘0’;
else num=num10+strs[pos]-‘0’;
pos++;
}
if(num==-1) num=1;
if(isalpha(strs[pos]))
{
Alpha=strs[pos];
LA+=numsymbol;
++pos;
}
else LN+=numsymbol;
}
pos++;
Epos=strs.length();
while(pos<Epos)
{
int symbol=1;
if(strs[pos]==’+’) pos++;
else if(strs[pos]==’-‘)
{
++pos;
symbol=-1;
}
int num=-1;
while(pos<Epos && isdigit(strs[pos]))
{
if(num==-1) num=strs[pos]-‘0’;
else num=num10+strs[pos]-‘0’;
pos++;
}
if(num==-1) num=1;
if(isalpha(strs[pos]))
{
Alpha=strs[pos];
RA+=numsymbol;
++pos;
}
else RN+=numsymbol;
}
printf(“%c=%.3f\n”,Alpha,(RN-LN)*1.0/(LA-RA));
return 0;
}

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Master X
6 年，9 月前

毒瘤代码毁灭世界……说多了都是泪，诶……
坑点都……不想讲了，能实现，或者说愿意做这题坑点基本踩完了。
话说我这代码几乎就是c语言的吧……
c++ string牛批！

include

using namespace std;
int main()
{
char s[81];
scanf(“%s”,s);
int l=strlen(s);
int where=0;//to judge if it’s at the left or right
int sqr=1;//to judge whether plus or submit
double sumnum=0;//the number
double suma=0;//the number of a
char c;//the ‘X’
int lin=0;//number to calculate

for(int i=0;i<l;i++)
{
     if(s[i]>='a'&&s[i]<='z')
        {c=s[i];

        if(s[i-1]<'0'||s[i-1]>'9') lin=1;
        if(lin!=0)
         suma+=lin*sqr;
         lin=0;
    }

          if((s[i]<'0'||s[i]>'9')&&lin!=0){
         sumnum=sumnum+lin*sqr;

        lin=0;
        sqr=1;
         //cout<<sumnum<<endl;
    }
    if(s[i]=='=') where=1,sumnum*=-1,suma*=-1,sqr=1;//cout<<sumnum<<ends<<suma<<endl;
    if(s[i]=='+') sqr=1;
    if(s[i]=='-') sqr=-1;



    if(s[i]>='0'&&s[i]<='9'){
       lin=lin*10+s[i]-'0';
       //cout<<sqr<<ends<<lin<<endl;
    }



}
//for(int i=0;i<l;i++)
   // cout<<s[i];

   if(lin!=0) sumnum+=lin*sqr;
   if(sumnum*-1==0) sumnum=0;
//cout<<sumnum<<ends<<suma<<endl;
suma*=-1;
   printf("%c=%.3f\n",c,sumnum/suma);
    //cout<<sumnum<<endl;
   // cout<<suma<<endl;

}

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10175102262 LarsPendragon
6 年，8 月前

最坑的就是直接带公式x=0输出-0.000……

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Cuibaby
4 年，9 月前

include

include

using namespace std;
void f(vector &v1,vector &v2,string x,char &c) {
int i=0;
while(i=‘0’&&x[j]<=‘9’) {
sum=sum10+x[j]-‘0’;
j++;
}
if(x[j]>=’a’&&x[j]<=’z’) {
c=x[j];
if(sum==0)
v1.push_back(flag);
else
v1.push_back(flagsum);
j++;
} else {
v2.push_back(sum*flag);
}
i=j;
}
}
int main() {
string s;
cin>>s;
vectorv1,v2,v3,v4;
int index=s.find(“=”);
string x=s.substr(0,index);
string y=s.substr(index+1,s.size());
char c;
f(v1,v2,x,c);
f(v3,v4,y,c);
int sum1=0;
int sum2=0;
for(int i = 0; i < v1.size(); i++)
sum1+=v1[i];
for(int i = 0; i < v3.size(); i++)
sum1-=v3[i];
for(int i = 0; i < v2.size(); i++)
sum2-=v2[i];
for(int i = 0; i < v4.size(); i++)
sum2+=v4[i];
printf(“%c=%.3f\n”,c,(float)sum2/sum1);
return 0;
}

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Hairaa
4 年，12 月前 
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;

double coefA = 0, coefB = 0;
char x;

void compute(string str,bool flag)
{
    bool tag = 0;//默认为常数
    for (int i = 0;i < str.size();i++) { if (str[i] >= 'a'&&str[i] <= 'z') { x = str[i];tag = 1; } }
    if (flag == 0)
    {
        if (tag == 1 && str.size() == 1) { coefA += 1; }
        else if (tag == 1 && str.size() == 2 && str[0] == '+') { coefA += 1; }
        else if (tag == 1 && str.size() == 2 && str[0] == '-') { coefA -= 1; }
        else if (tag == 1) { coefA += stoi(str); }
        if (tag == 0) { coefB += stoi(str); }
    }
    else
    {
        if (tag == 1 && str.size() == 1) { coefA -= 1; }
        else if (tag == 1 && str.size() == 2 && str[0] == '+') { coefA -= 1; }
        else if (tag == 1 && str.size() == 2 && str[0] == '-') { coefA += 1; }
        else if (tag == 1) { coefA -= stoi(str); }
        if (tag == 0) { coefB -= stoi(str); }
    }
}

int main()
{
    string s, str="";
    cin >> s;
    bool flag = 0;//默认为等号前端
    for (int i = 0;i < s.size();i++)
    {
        str += s[i];
        if ((s[i + 1] == '+' || s[i + 1] == '-') && flag == 0) { compute(str,flag);str = ""; }
        else if ((s[i + 1] == '+' || s[i + 1] == '-') && flag == 1) { compute(str, flag);str = ""; }
        if (s[i + 1] == '=') { compute(str, flag);flag = 1;str = "";i++; }
    }
    if (str != "") { compute(str, flag); }
    cout << x << "=";
    cout << fixed << setprecision(3)<< coefB / -coefA << endl;
}
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Hairaa
4 年，12 月前 
#include <iostream>
#include <string>
#include <iomanip>
using namespace std;

double coefA = 0, coefB = 0;
char x;

void compute(string str,bool flag)
{
    bool tag = 0;//默认为常数
    for (int i = 0;i < str.size();i++) { if (str[i] >= 'a'&&str[i] <= 'z') { x = str[i];tag = 1; } }
    if (flag == 0)
    {
        if (tag == 1 && str.size() == 1) { coefA += 1; }
        else if (tag == 1 && str.size() == 2 && str[0] == '+') { coefA += 1; }
        else if (tag == 1 && str.size() == 2 && str[0] == '-') { coefA -= 1; }
        else if (tag == 1) { coefA += stoi(str); }
        if (tag == 0) { coefB += stoi(str); }
    }
    else
    {
        if (tag == 1 && str.size() == 1) { coefA -= 1; }
        else if (tag == 1 && str.size() == 2 && str[0] == '+') { coefA -= 1; }
        else if (tag == 1 && str.size() == 2 && str[0] == '-') { coefA += 1; }
        else if (tag == 1) { coefA -= stoi(str); }
        if (tag == 0) { coefB -= stoi(str); }
    }
}

int main()
{
    string s, str="";
    cin >> s;
    bool flag = 0;//默认为等号前端
    for (int i = 0;i < s.size();i++)
    {
        str += s[i];
        if ((s[i + 1] == '+' || s[i + 1] == '-') && flag == 0) { compute(str,flag);str = ""; }
        else if ((s[i + 1] == '+' || s[i + 1] == '-') && flag == 1) { compute(str, flag);str = ""; }
        if (s[i + 1] == '=') { compute(str, flag);flag = 1;str = "";i++; }
    }
    if (str != "") { compute(str, flag); }
    cout << x << "=";
    cout << fixed << setprecision(3)<< coefB / -coefA << endl;
}
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