Input length (N<=100):（此处括号、冒号为英文符号，后面无空格；length和左括号间有空格）
6
8 10 13 1 23 7
The max is 23
The min is 1（此处输出结束不换行，is后面有一个空格）
/*
Input length (N<=100):（此处括号、冒号为英文符号，后面无空格；length和左括号间有空格）
6
8 10 13 1 23 7
The max is 23
The min is 1（此处输出结束不换行，is后面有一个空格）
*/
#include <stdio.h>
int main()
{
int a[100],n,i,max,min;
printf("Input length (N<=100):\n");
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
max = min = a[0];
for(i=1;i<n;i++)
{
if(max < a[i])
max = a[i];
if(min > a[i])
min = a[i];
}
printf("The max is %d\n",max);
printf("The min is %d",min);
return 0;
}
#include <iostream>
using namespace std;
int main()
{
int N; cin >> N;
int max = -1000000;
int min = 10000000;
for (int i = 0; i < N; i++)
{
int a; cin >> a;
if (a > max)
{
max = a;
}
if (a < min)
{
min = a;
}
}
cout << "The max is " << max << endl;
cout << "The min is " << min;
}
一次遍历找出最大值最小值
#include<stdio.h>
int main()
{
printf("Input length (N<=100):");
int n,t[100],j;
scanf("%d",&n);
for(j=0; j<n; j++)
scanf("%d",&t[j]);
int max=t[0];
int min=t[0];
for(j=1; j<n; j++)
{
if(max<t[j])
max=t[j];
if(min>t[j])
min=t[j];
}
printf("The max is %d\nThe min is %d",max,min);
return 0;
}