2018 885程序设计编程题

• 

• //定义一个函数,计算并返回如下算式的值 
  #include <stdio.h>
  #include <math.h>
  float fun(int a,int b,int c){
      float x=4.25*(a+b)+log(a+b+sqrt(a+b)+1.0/(a+b));
      float y=4.25*c+log(c+sqrt(c)+1.0/c);
      return x/y;
  }
  int main(){
      FILE *fp;
      int a,b,c,i=0;
      if((fp=fopen("data.txt","w+"))==NULL){
          printf("cannot open the file.\n");
      }
      while(i<10){
          scanf("%d%d%d",&a,&b,&c);
          printf("a=%d,b=%d,c=%d,result=%f\n",a,b,c,fun(a,b,c));
          fprintf(fp,"a=%d,b=%d,c=%d,result=%f\n",a,b,c,fun(a,b,c));
          i++;
      }
      fclose(fp);
      return 0;
  } 

   

• 

• #include <stdio.h>
  #include <math.h>
  int main(){
      int n,tem,b,i=0,k=0;
      int a[100];
      int flag=1;
      printf("please input n:\n");
      scanf("%d",&n);
      tem=n;
      while(tem>0){
          a[i++]=tem%10;
          tem/=10;
      }
      for(b=0;b<i-1;b++){
          if(a[b]>a[b+1])
              flag=0;
      }
      if(flag==1)
          printf("%d是降序数",n);
      else
          printf("%d是非降序数",n);
      return 0;
  }

   

• #include <stdio.h>
  #include <math.h>
  double f(double x){
      return pow(x,41)+pow(x,3)+1;
  }
  double fd(double x){
      return 41*pow(x,40)+3*pow(x,2);
  }
  int main(){
      float x2=-1,x1=-1,f1,f2;
      do{
           x1=x2;
           f1=f(x1);
           f2=fd(x1);
           x2=x1-f1/f2;
      }while(fabs(x2-x1)>1e-6);
      printf("%f",x2);
      return 0;
  }