• 补交第五次作业


    #include<stdio.h>
    int main()//1.比较三个数大小(从小到大排序)
    { 
        int a,b,c,d=0;
        printf("请输入三个数
    ");
        scanf("%d%d%d",&a,&b,&c);
    if(a>b)
    {
        d = a;
        a = b;
        b = d;
    } 
    if(b>c)
    {
        d = b;
        b = c;
        c = d;
    }
    if(a>b)
    {
        d= a;
        a= b;
        b =d;
    }
    printf("%d < %d < %d",a,b,c);
    }
    #include<stdio.h>
    int main()//2.高速公路限速处罚
    {
        int x,z;
        scanf("%d %d",&x,&z);
        double b = (double)(x-z)*100/z;
        if (b>50)printf("超过限速 %.0f%%. 吊销执照",b);
        if(b>=10&&b<=50)printf("超出限速 %.0f%%. 罚款 200",b);
        if(b<10)printf("未超过限速");
    }
    
    #include <stdio>
    using namespace std;//3.出租车计费 
    
    int count(double miles);
    
    int main(){
    int x;double wait;
    cout<<"输入公里"<<endl;
    cin>>x;
    cout<<"等待时间"<<endl;
    cin>>wait;
    cout<<count(x+int(wait/5));
    return(0);
    }
    
    int count(double x){//TODO:
    if (x<0) {cout<<"不可能"<<endl; return 0;}
    double t=0;
    if (x<3) {
    t=10;
    }else{
    t+=10;
    if (x>13) {
    t+=20+3*(x-13);
    }else{
    t+=2*(x-3);
    }
    }
    
    return int(t+0.5);
    }
    
    
    
    
    #include <stdio.h>
    int main()//4.五级制成绩分布 
    {
        int i,A,B,C,D,E,n,s;
        A=B=C=D=E=0;
        printf("Enter n:");
        scanf("%d",&n);
        for(i=0;i<n;++i)
        {
            printf("Enter grade %d:",i+1);
            scanf("%d",&s);
            switch(s/10)
            {
            case 1:
            case 2:
            case 3:
            case 4:
            case 5:E++;break;
            case 6:D++;break;
            case 7:C++;break;
            case 8:B++;break;
            case 9:
            case 10:A++;break;
            }
        }
        printf("The number of A(90~100):%d
    ",A);
        printf("The number of B(80~89):%d
    ",B);
        printf("The number of C(70~79):%d
    ",C);
        printf("The number of D(60~69):%d
    ",D);
        printf("The number of E(0~59):%d
    ",E);
        return 0;
    }
    
    
    
    #include<stdio.h>
    #include<stdio.h>
     
    int main()//5.判断三角形
    { double a,b,c,d,e,f;
      scanf("%lf %lf %lf %lf %lf %lf",&a,&b,&c,&d,&e,&f );
      double AB,BC,AC,ab,bc,ac;
      ab=(a-c)*(a-c)+(b-d)*(b-d);
      bc=(c-e)*(c-e)+(d-f)*(d-f);
      ac=(a-e)*(a-e)+(b-f)*(b-f);
      AB=sqrt(ab);
      BC=sqrt(bc);
      AC=sqrt(ac);
      if((AB<BC+AC)&&(BC<AB+AC)&&(AC<AB+BC))
      {
      double l=AB+BC+AC;
      double P = l / 2;
      double s = sqrt(P*(P-AB)*(P-BC)*(P-AC));
      printf("L = %.2f, A = %.2f",l,s); }
      else
      {printf ("Impossible");
      }
      return 0;
    }
    
    
    #include<stdio.h>
    int main()//6.双循环打印三角形 
    {
        int i,j;
        for(j=1;j<=10;j++)
        {
            for(i=1;i<=11-j;i++)
            {
                printf("*");
            }
            printf("
    ");
        }
    }
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  • 原文地址:https://www.cnblogs.com/kanghui/p/6112186.html
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