[编程入门]自由下落的距离计算-题解(C语言代码) -----数学建模简化代码

#include<stdio.h>

int main()
{
    double M,N,h,l,s;
    scanf("%lf %lf",&M,&N);
    l=M;
    h=M;
    s=M;
    int i;
    for(i=1; i<=N; i++)
    {
    h=(1.0/2.0)*h;
    }
    
    for(i=2; i<=N; i++)
    {
        l=(1.0/2.0)*l;
        s=s+2*l;
    }
    
    printf("%.2lf %.2lf",h,s);
    return 0;
}

//递归
#include<stdio.h>
double digui(double M,int N);
int main()
{
    double M;
    int N;
    scanf("%lf%d",&M,&N);
    printf(" %.2lf",digui(M,N));
	return 0;
}
double digui(double M,int N){
    if(N==1){
        printf("%.2lf",M/2.0);
        return M;
    }
    return M+M/2.0+digui(M/2.0,N-1);
}

#include <stdio.h>
#include <math.h>

int main(){
	
    int j;
    double M = 0,N = 0,length = 0,high = 0,i = 0;
    scanf("%lf%lf",&M,&N);
    
	high = M * pow(0.50,N);

        for(j = 1; j < N; j++){        	
        	i = pow(0.5,j);
		    length += M * i * 2;		
            }          
        
        printf("%.2f %.2f",high,length + M);
          
    return 0;
}

#include<stdio.h>
int main()
{
    double m, zgm = 0.0, n, i = 0.0;
	scanf("%lf%lf", &m,&n);
	while (i < n)
	{
		zgm = zgm + m + m / 2.0;
		m = m / 2.0;
		i++;
	}
	printf("%.2lf %.2lf", m, zgm-m);
	return 0;
}

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>

int main(){
	float m, n, h,cnt=0;
	scanf("%f %f", &m, &n);

	for (;n > 0;n--) {

		h =(float) m + m/ 2.0;
		cnt = cnt + h;
		m = (float)m / 2.0;
	}
	printf("%.2f %.2f", m, cnt-m);
	
}我这么简单的代码就搞定了

#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>

int main(){
	float m, n, h,cnt=0;
	scanf("%f %f", &m, &n);

	for (;n > 0;n--) {

		h =(float) m + m/ 2.0;
		cnt = cnt + h;
		m = (float)m / 2.0;
	}
	printf("%.2f %.2f", m, cnt-m);
	
}

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
int main()
{
    int n,i,j,s,b[1000];
   scanf("%d",&n);
   for(i=3;i<=n;i++)
   {
       int sum=1,x=0;
       for(j=2;j<i;j++)
       {
        if(i%j==0)
          {sum=j+sum;
          b[x]=j;
          x++;}
       }
       if(sum==i)
         {printf("%d its factors are 1",i);
         for(s=0;s<x;s++)
            printf(" %d",b[s]);
         printf("\n");
         memset(b,0,sizeof(b));
         }
   }
   return 0;
}

感觉这种解法属于本末倒置了，本来写代码就是为了简化计算，现在为了简化代码而自己算出求和的结果，这都增加了自己的计算量，那这个代码写了也没有什么意义了。而且这个简化代码也看出有什么简化。