求集合S的最大元素和最小元素的递归分治算法.

谁能用分治法帮俺做做啊??!!

#include <stdio.h>
#define N 10
void maxmin(int s[N], int max, int min);
int main(void)
{
int a[] = {1,12,21,4,57,78,95,43,62,23};

maxmin(a, 0, 0);

return 0;
}

void maxmin(int s[N], int max, int min)
{int s1[N/2],s2[N/2],i,j,max1,max2,min1,min2;
if (N == 2)
{if (s[0] > s[1]) { max=s[0]; min=s[1];}
else { max=s[1]; min=s[0];}
}
else
{
for (i=0; i< N/2; i++)
s1[i] = s[i];
for (j=N/2; j<N; j++)
s2[j] = s[j];
maxmin(s1, max1, min1);
maxmin(s2, max2, max2);

if (max1 > max2) max = max1; else max = max2;
if (min1 > min2) min = min1; else min = min2;
printf("max is :%d\n",max);
printf("min is :%d\n",min);
}
}

我帮你写了一个,你自己看着,
晚上论文估计是抄不完了!
该死的老师，电子版的不要,非得要手写版的,大家说气人不气人!!

#include "stdio.h"
#include "conio.h"
void maxmin(int a[],int n,int *max,int *min)
{
int a1[5],a2[5];
int i,j,k=0;
int max1,max2,min1,min2;
int *ma1=&max1,*mi1=&min1,*ma2=&max2,*mi2=&min2;
if(n==1)
*max=*min=a[0];
else if(n==2)
{
if(a[0]>a[1])
{
*max=a[0];
*min=a[1];
}
else
{
*max=a[1];
*min=a[0];
}
}
else
{
for(i=0;i<n/2;i++)
a1[i]=a[i];
for(j=i;j<n;j++)
a2[k++]=a[j];
maxmin(a1,n/2,ma1,mi1);
maxmin(a2,(n-n/2),ma2,mi2);
if (*ma1 > *ma2)
*max = *ma1;
else
*max = *ma2;
if (*mi1 > *mi2)
*min = *mi2;
else
*min = *mi1;
}

}
main()
{
int a[10]={1,12,21,4,57,78,95,43,62,23};
int max,min;
maxmin(a,10,&max,&min);
printf("%d %d",max,min);
getch();
}