用C语言解决输入 m 个数，求其最大数、最小数和平均值。_c语言输入m个数,求最大值最小值

程序代码：法一#include <stdio.h>

#include <stdlib.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[])

 {

float a=0.0,avr,max,min;

int i,m,n[m];

printf("input %d numbers:",m);

scanf("%d",&m);

    for(i=0;i<=m-1;i++){

scanf("%d",&n[i]);

a=a+n[i];

max=n[0];

min=n[0];

    if(max<n[i])

    max=n[i];else

if(min>n[i])

    min=n[i];

   }

 avr=a/m;

 printf("%.2f,%.2f,%.2f\n",avr,max,min);

return 0;

}

法二

#include <stdio.h>

#include <stdlib.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[]) {

float ave,sum=0;

int m,i,a[m],z,j,buf;

scanf("%d",&m);

for(i=0;i<m;i++)

{

scanf("%d",&a[i]);

sum=sum+a[i];

}

for (z=0;z<m-1;++z)

{

for (j=0;j<m-1-z;++j)

{

if(a[j]<a[j+1])

{

buf = a[j];

a[j] = a[j+1];

a[j+1] = buf;

}

}

}

ave=sum/m;

printf("max=%d min=%d ave=%.2f\n",a[0],a[m-1],ave);

return 0;

}运行结果：①