The third phase of the school training competition

1. The disappearing tail

The meaning of the title: The title has already said it very clearly, so I won't repeat it.

Idea: Determine whether ((a*100)+i)%b is equal to 0, and if it is equal to 0, the counter will be incremented by 1.

Code:

#include<stdio.h>
int main(void)
{
	int a,b;
	int i;
	while(~scanf("%d%d",&a,&b))
	{
		if(a==0&&b==0)
			break;
		int count = 0;
		for(i = 0; i <= 99; i++)
		{
			if(((a*100)+i)%b==0)
				count++;
		}
		printf("%d\n",count);
	}
}

2. OCD 2 in WC

The meaning of the question: in a row of rice balls, each time you choose a rice ball larger than the front, how many can you choose at most?

Idea: Longest Ascending Subsequence Template Question

Code:

#include<stdio.h>
const int MAX=1001;
intmain()
{
	int i,j;
	int a[MAX];
	int dp[MAX];
    int n;
    while(~scanf("%d",&n))
    {
    	for(i=0;i<n;i++)
        	scanf("%d",&a[i]);
    	for(i=0;i<n;i++)
    	{
        	dp[i]=1;
        	for(j=0;j<i;j++)
       		{
            if(a[j]<a[i]&&dp[j]+1>dp[i])
                   dp[i]=dp[j]+1;
        	}
   		}
   		int ans=0;
    	for(i=0;i<n;i++)
        	if(ans<dp[i])
        		ans=dp[i];
        printf("%d\n",ans);
    }
    return 0;
}

3. A string of inverse zeros

Question meaning: give 0, output 1; give 1, output 0;

Idea: give 0, output 1; give 1, output 0

Code:

#include<stdio.h>
#include<string.h>
int main(void)
{
	int i;
	char s[201];
	while(gets(s)!=NULL)
	{
		int len ​​= strlen(s);
		for(i = 0; i < len; i++)
			{
				if(s[i]=='1')
					s[i] = '0';
				else
					s[i] = '1';
			}
		puts(s);
	}
	return 0;
}

4. Another simple question

Title: It has been given in the title, so I won't repeat it

Idea: Four digits, each digit is modified to judge whether it is a perfect square number

Code:

#include<stdio.h>
#include<math.h>
bool f(int n)
{
	int m;
	m = sqrt(n);
	if(m*m == n)
		return true;
	else
		return false;
}
int main(void)
{	
	int i;	
	int t;
	scanf("%d",&t);
	int years = 1;
	while(t--)
	{
		int count = 0;
		int n;
		scanf("%d",&n);
		int a = n%10;
		int b = n/10%10;
		int c = n/100%10;
		int d = n/1000;
		for(i = -a; i < 10-a; i++) // try the ones
			if(f(n+i))
			{
				count++;
			}
		for(i = -b*10; i < 100-10*b; i= i+10) //尝试十位
			if(f(n+i))
			{
				count++;
			}
		for(i = -c*100; i < 1000-100*c; i=i+100) //尝试百位
			if(f(n+i))
			{
				count++;
			}
		for(i = 1000-d*1000; i < 10000-1000*d; i=i+1000)//尝试千位
			if(f(n+i))
			{
				count++;
			}
		if(f(n))
			printf("Case %d: %d\n",ans++,count-4); //If it is a perfect square, it needs -4, because the first four judgments are repeated	
		else
			printf("Case %d: %d\n",ans++,count);  
	}
	return 0;
}