2019.10.12 班级练习赛_C

概述https://www.luogu.org/problem/T103143 https://www.luogu.org/problem/T103188 https://www.luogu.org/problem/T103328 c++版 Max And Min 1 #include <stdio.h> 2 #include <math.h> 3 #include <algorithm>

https://www.luogu.org/problem/T103143

https://www.luogu.org/problem/T103188

https://www.luogu.org/problem/T103328

c++版

Max And Min

 1 #include <stdio.h> 2 #include <math.h> 3 #include <algorithm> 4  5 using namespace std; 6  7 int main() 8 { 9     int a,b,c,d;10     scanf("%d%d%d%d",&a,&b,&c,&d);11     int ma; ma=max(max(max(a,b),c),d);12     int mi; mi=min(min(min(a,d);13     printf("%d\n%d\n",ma,mi);14     return 0;15 }

解方程

 1 #include<iostream> 2 #include<cmath> 3 #include<iomanip> 4  5 using namespace std; 6  7 double pd(double a,double b,double c) 8 { 9     double m=b*b-4*a*c;10     return m;11 }12 13 int main()14 {15     double a,c;16     cin>>a>>b>>c;17     if(a==0)18     {19         cout<<1<<endl<<fixed<<setprecision(4)<<-(c/b)<<endl;20     }21     else 22     {23         double disc=pd(a,c);24         if(disc<0) cout<<"N0"<<endl;25         else 26         {27             double x=(-b+sqrt(disc))/(2*a);28             double xx=(-b-sqrt(disc))/(2*a);29             double X=max(xx,x),Y=min(xx,x);30             if(xx==x) cout<<1<<endl<<fixed<<setprecision(4)<<X<<endl;31             else cout<<2<<endl<<fixed<<setprecision(4)<<Y<<" "<<X<<endl;32         }33     }34     return 0;35 }

进制数：

 1 #include<iostream> 2  3 using namespace std; 4  5 int n,m; 6  7 int main() 8 { 9     cin>>n>>m;10     int sum=0;11     while(n)12     {13         sum+=n%m;14         n/=m;15     }16     cout<<sum<<endl;17 }

c版

min and max

 1 #include <stdio.h> 2  3 int max(int x,int y) 4 { 5     if(x<y) return y; 6     return x; 7 } 8  9 int min(int x,int y)10 {11     if(x>y) return y;12     return x;13 }14 15 int main()16 {17     int a,d;18     scanf("%d%d%d%d",&d);19     int ma; ma=max(max(max(a,d);20     int mi; mi=min(min(min(a,d);21     printf("%d\n%d\n",mi);22     return 0;23 }

解方程

 1 #include<stdio.h> 2 #include<cmath> 3  4 double max(double x,double y) 5 { 6     if(x<y) return y; 7     return x; 8 } 9 10 double min(double x,double y)11 {12     if(x>y) return y;13     return x;14 }15 16 double pd(double a,double c)17 {18     double m=b*b-4*a*c;19     return m;20 }21 22 int main()23 {24     double a,c;25     scanf("%lf%lf%lf",&c);26     if(a==0)27     {28         printf("1\n%.4lf\n",-(c/b));29     }30     else 31     {32         double disc=pd(a,c);33         if(disc<0) printf("%N0\n");34         else 35         {36             double x=(-b+sqrt(disc))/(2*a);37             double xx=(-b-sqrt(disc))/(2*a);38             double X=max(xx,x);39             if(xx==x) printf("1\n%.4lf",X);40             else printf("2\n%.4lf %.4lf",Y,X);41         }42     }43     return 0;44 }

进制数：

 1 #include<stdio.h> 2  3 int n,m; 4  5 int main() 6 { 7     scanf("%d%d",&n,&m); 8     int sum=0; 9     while(n)10     {11         sum+=n%m;12         n/=m;13     }14     printf("%d\n",sum);15     return 0;16 }