3093 - 2001年秋浙江省计算机等级考试二级C 编程题(2)

10
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0

10.00 1.00

Solution C

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

int main(){
	int i,n;
	scanf ("%d",&n);
	float a[100];
	for (i=0;i<n;i++)
		scanf ("%f",&a[i]);
	float max=a[0],min=a[0];
	for (i=0;i<n;i++)
	{
		max=max>a[i]?max:a[i];
		min=min>a[i]?a[i]:min;
	}
	printf ("%.2f %.2f\n",max,min);














	return 0;
}

Solution C++

#include<iostream>
#include<cstdio>
using namespace std;
float a[10000];
float fmax(float a[],int n)	//查找最大值
{
	float max=a[1];
	for (int i=2; i<=n; i++)
		if (max<a[i]) max=a[i];
	return max;
}
float fmin(float a[],int n)	//查找最小值
{
	float min=a[1];
	for (int i=2; i<=n; i++)
		if (min>a[i]) min=a[i];
	return min;
}
int main()
{
	int n;
	cin>>n;
	for (int i=1; i<=n; i++) cin>>a[i];
	printf("%.2f %.2f\n",fmax(a,n),fmin(a,n));
	return 0;
}