用C语言编写一个程序，求n个数中的最大值和最小值，要求用函数完成。 c语言编程实现某一个数组中最大值,最小值之和。并要求最大最小...

\u6c42c\u8bed\u8a00\u7a0b\u5e8f\uff1a\u7528\u4e00\u4e2a\u51fd\u6570\u6c42N\u4e2a\u6570\u7684\u6700\u5927\u503c\u548c\u6700\u5c0f\u503c\u3002\u3002\u3002

@
741852963\u4e00
\u697c\u4e0a\u7684\u670b\u53cb\uff0c\u5bf9\u4e8e\u4f60\u7ed9\u7684\u7b54\u6848\uff0c\u6211\u60f3\u8bf4\u70b9\u6211\u81ea\u5df1\u7684\u60f3\u6cd5\uff0c\u671b\u4e0d\u8981\u89c1\u602a\u3002
1.
LZ\u8981\u7684\u662f\u4e00\u4e2a\u51fd\u6570\uff0c\u9700\u8981\u5728main
\u51fd\u6570\u4e2d\u8fdb\u884c\u8c03\u7528\uff0c\u800c\u4e0d\u662f\u8981\u5b8c\u6574\u7684main\u51fd\u6570\u3002
2.
LZ\u8981\u7684\u51fd\u6570\u529f\u80fd\u53ea\u8981\u6c42
\u6c42\u51fa\u6700\u5927\u503c\u8ddf\u6700\u5c0f\u503c\uff0c\u800c\u4e0d\u9700\u8981\u6392\u5e8f\u3002
3\u3002\u4f60\u7684\u7a0b\u5e8f\u4e2d\u7528\u7684\u662fa[10]\uff0c\u4f46n
\u662f\u53ef\u4ee5\u968f\u4fbf\u8f93\u5165\u7684\uff0c\u8fd9\u5c31\u5b58\u5728\u4e00\u4e2a\u95ee\u9898\uff1a
n
\u8981\u662f\u5927\u4e8e10\u65f6\uff0c
a[n]
\u5c31\u4f1a\u6ea2\u51fa\u3002
\u4e0b\u9762\u662f\u6211\u7f16\u5199\u7684\u4e00\u4e2a\u7a0b\u5e8f\uff0c\u5f53\u7136\u6211\u4e0d\u80fd\u4fdd\u8bc1\u5b83\u662f\u5b8c\u5168\u6b63\u786e\u7684\uff0c\u5e0c\u671b\u5927\u5bb6\u63d0\u610f\u89c1\u3002
float
fun(int
N,
float
a[
])
{
int
i;
float
max,
min;
max=min=a[0];
for
(i=1;i<N;i++)
{
if
(a[i]>=max)
max=a[i];
if
(a[i]<=min)
min=a[i];
}
prinft("max=
"
%f
",
min=
"
%f,
max,min);
return
0;
}

#include
//\u3001\u3001c\u8bed\u8a00\u7f16\u7a0b\u5b9e\u73b0\u67d0\u4e00\u4e2a\u6570\u7ec4\u4e2d\u6700\u5927\u503c,\u6700\u5c0f\u503c\u4e4b\u548c\u3002\u5e76\u8981\u6c42\u6700\u5927\u6700\u5c0f\u503c\u7528\u5b50\u51fd\u6570\u5b9e\u73b0
int Max(int a[], int n);
int Min(int a[], int n);
int Sum(int a, int b);
int main( )
{
int a[10];
int i;
printf("\u8bf7\u8f93\u516510\u4e2a\u6570:\n");
for (i = 0; i < 10; i++) {
scanf("%d", &a[i]);
}
printf("sum = %d\n",Sum(Max(a, 10), Min(a, 10)));
return 0;
}
int Max(int a[], int n)
{
int i;
int max = a[0];
for (i = 0; i < 10; i++) {
if (max < a[i]) {
max = a[i];
}
}
return max;
}
int Min(int a[], int n)
{
int i;
int min = a[0];
for (i = 0; i < 10; i++) {
if (min > a[i]) {
min = a[i];
}
}
return min;
}
int Sum(int a, int b)
{
return a + b;
}

代码如下（用冒泡排序法）：

#include <stdio.h>

void bubble(int a[], int n)
{
    int i, j, temp;
    for (j = 0; j < n - 1; j++)
        for (i = 0; i < n - 1 - j; i++)
        {
            if(a[i] > a[i + 1])
            {
                temp = a[i];
                a[i] = a[i + 1];
                a[i + 1] = temp;
            }
        }
}

int main(void)
{
    int a[10];
    for(int i = 0; i < 10; i++)
        scanf("%d", &a[i]);
    bubble(a, 10);
    printf("Max: %d, min: %d", a[9], a[0]);
    
    return 0;
}

#include<stdio.h>
#define N 1024
void fun(int *a,int n,int *max,int *min) { int i;
  *max=*min=a[0];
  for ( i=1;i<n;i++ ) if ( *max<a[i] ) *max=a[i]; else if ( *min>a[i] ) *min=a[i];
}
void main() { int i,n,max,min,a[N];
  scanf("%d",&n); if ( n>N ) n=N;
  for ( i=0;i<n;i++ ) scanf("%d",&a[i]);
  fun(a,n,&max,&min);
  printf("最大值=%d，最小值=%d。
",max,min);
}

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