c++的相关问题，关于方程的问题

关于一元二次方程的求解求解，怎么样给一元二次方程定义，给给思路，用c++的方法，
以及关于识别码，两个时间的计算方法，累加法，会的看看，给给思路

一元二次方程根：

#include <iostream>
#include <math.h>
#include <iomanip>
using namespace std;
int main()
{
    double a,b,c;
    cin >> a>>b>>c;
    double q = b*b - 4*a*c;
    if(q>0)
    {
        double s1 = (-b+sqrt(q))/(2*a);
        double s2 = (-b-sqrt(q))/(2*a);

        if(s1 == (int)s1)
            cout << (int)s1<<endl;
        else
            cout << fixed << setprecision(2) << s1<<endl;

        if(s2 == (int)s2)
            cout <<(int)s2;
        else
            cout << fixed << setprecision(2) << s2;
    }else if(q==0)
    {
        double s = -b/(2*a);
        if(s==(int)s)
            cout << (int)s;
        else
            cout <<fixed<<setprecision(2)<<s;
    }else
        cout <<"无实根";
    return 0;
}

ISBN校验码：

#include <iostream>
#include <string>
using namespace std;
int main()
{
    string str;
    char ch;
    cin >> str;
    int s = 0;
    int k = 1;
    for(int i=0;i<str.length()-1;i++)
    {
        if(str.at(i)>='0' && str.at(i)<='9')
        {
            s += (str.at(i) - '0')*k;
            k++;
        }
    }
    s = s%11; //ศกำเส
    if(s<10)
        ch = '0'+ s;
    else
        ch = 'X';

    if(str.at(str.length()-1) == ch)
        cout <<"Right";
    else
        cout << str.substr(0,str.length()-1)<<ch;
    return 0;
}

时间差：

#include <iostream>
#include <string>
using namespace std;
int main()
{
    int hourstart,minstart;
    int hourend,minend;
    cin >> hourstart >> minstart >> hourend >> minend;
    int hh,mm;
    hh = hourend - hourstart;
    mm = minend - minstart;
    if(mm <0)
    {
        hh -=1;
        mm += 60;
    }
    cout << hh<<" "<<mm;
    return 0;
}

1-n中有多少个x：

#include <iostream>
using namespace std;
//计算数字n中有多少个x
int fun(int n,int x)
{
    int count = 0;
    int t;
    while(n)
    {
        t = n%10;
        if(t==x)
            count++;
        n/=10;
    }
    return count;
}


int main()
{
    int n,x,count=0;
    cin >> n>>x;
    for (int i=1;i<=n;i++)
    {
        count += fun(i,x);
    }
    cout << count;
    return 0;
}

国王的金币：

#include <iostream>
#include <string>
using namespace std;
int main()
{
    int k;
    int ts=0;
    int dz = 1;
    long long jb = 0;
    cin >> k;
    
    for(;ts<k;ts+=dz)
    {
        jb += dz*dz;
        dz++;
    }

    jb += (k-(ts-dz)-1)*dz;
    cout << jb;

    return 0;
}

#include <iostream>
#include <cmath>
using namespace std;
 
int main() {
 
    float a, b, c, x1, x2, discriminant, realPart, imaginaryPart;
    cout << "输入 a, b 和 c: ";
    cin >> a >> b >> c;
    discriminant = b*b - 4*a*c;
    
    if (discriminant > 0) {
        x1 = (-b + sqrt(discriminant)) / (2*a);
        x2 = (-b - sqrt(discriminant)) / (2*a);
        cout << "Roots are real and different." << endl;
        cout << "x1 = " << x1 << endl;
        cout << "x2 = " << x2 << endl;
    }
    
    else if (discriminant == 0) {
        cout << "实根相同：" << endl;
        x1 = (-b + sqrt(discriminant)) / (2*a);
        cout << "x1 = x2 =" << x1 << endl;
    }
 
    else {
        realPart = -b/(2*a);
        imaginaryPart =sqrt(-discriminant)/(2*a);
        cout << "实根不同："  << endl;
        cout << "x1 = " << realPart << "+" << imaginaryPart << "i" << endl;
        cout << "x2 = " << realPart << "-" << imaginaryPart << "i" << endl;
    }
 
    return 0;
}