Smart gourmet C language

Smart gourmet C Language

Title Description ：

If someone thinks that eating only needs a mouth , It would be wrong .
　　 We all know that the tongue has such a characteristic ,“ From simplicity to luxury, it's easy , It's hard to be as simple as extravagance ”（ According to the study of good people , This law also applies to many other situations ）. To be specific , If it's dessert , When the food you eat is not as sweet as what you have just eaten , It's very unpleasant .
　　 Dabao is a smart gourmet , Of course, I know this well . Once he came to a snack street , Ready to eat from one end of the street to the other . In order to eat well , He took a lot of trouble , Got all kinds of food “ Delicacy ”. He rejected unpleasant experiences , Don't go back and be refreshing （ As many times as possible ）.

Input ：

Input description :
　　 Two lines of data .
　　 The first line is an integer n, Indicates the number of snacks on the snack street
　　 Second behavior n It's an integer , respectively n Grow food “ Delicacy ”
sample input :
10
3 18 7 14 10 12 23 41 16 24

Output ：

Output description :
　　 An integer , Indicates the number of times you eat well
sample output :
6

Tips ：

HINT: The time limit ：1.0s Memory limit ：256.0MB
　　 The delicacy is 0 To 100 The integer of
　　n<1000

source ：

Blue Bridge Cup practice system ID: 320 Original link : http://lx.lanqiao.cn/problem.page?gpid=T320

#include <stdio.h>
int a[1010],b[1010],i,j,k;
int main()
{
    	int num;
	scanf("%d",&num);
	for(i=1;i<=num;i++)
		scanf("%d",&a[i]);
	for(i=1;i<=num;i++)
		b[i]=1;
	for(i=1;i<=num;i++)
		{
    
			for(j=1;j<i;j++)// Count from the first to i
				{
    
					if(a[i]>=a[j])
						{
    
							b[i]=b[i]>b[j]+1?b[i]:b[j]+1;
						}
				}
		} 
	k=1;	
	for(i=num;i>=0;i--)
		if(k<b[i])
			{
    
				k=b[i];
			}
	printf("%d",k);
	return 0;
}