Solve "sub number integer", "jump happily", "turn on the light"

A whole number of children

Title Description

For a five digit number $a_1{a}_2{a}_3{a}_4{a}_5$, It can be divided into three subnumbers ：

$su{b}_1=a_1{a}_2{a}_3$

$su{b}_2=a_2{a}_3{a}_4$

$su{b}_3=a_3{a}_4{a}_5$

for example , Five figures $20207$ Can be broken down into

$su{b}_1=202$

$su{b}_2=020(=20)$

$su{b}_3=207$

Now let's give a positive integer $K$, Ask you to program out $10000$ To $30000$ All five digits that meet the following conditions , The condition is that three of these five digits $su{b}_1,su{b}_2,su{b}_3$ Can be $K$ to be divisible by .

Input format

A positive integer K

Output format

A five digit number for each action , Require output from small to large . Do not repeat output or omit . If there is no solution , The output “No”.

Examples #1

The sample input #1

15

Sample output #1

22555
25555
28555
30000

Tips

$0<K<1000$

Answer key

// Author:PanDaoxi
#include <iostream>
using namespace std;
bool f(char s[],int k){
    
	//  Write a function to judge whether it can be divided after cutting 
	int sub[4]={
    };
	sub[0]=(s[0]-'0')*100+(s[1]-'0')*10+(s[2]-'0'),
	sub[1]=(s[1]-'0')*100+(s[2]-'0')*10+(s[3]-'0'),
	sub[2]=(s[2]-'0')*100+(s[3]-'0')*10+(s[4]-'0');
	// cout<<sub[0]<<" "<<sub[1]<<" "<<sub[2]<<endl;
	//  Return results 
	return (sub[0]%k==0&&sub[1]%k==0&&sub[2]%k==0)?true:false;
}
int main(){
    
	int k,a[30001],n=0;
	cin>>k;
	for(int i=10000;i<=30000;i++){
    
		//  It was written incorrectly , But I'm too lazy to change , Direct conversion string nested function 
		char s[7]={
    };
		s[0]=i/10000+'0',
		s[1]=(i%10000)/1000+'0';
		s[2]=(i%1000)/100+'0';
		s[3]=(i%100)/10+'0';
		s[4]=i%10+'0';
  		if(f(s,k)) a[n++]=i; //  Stored in the array 
	}
	if(n==0){
     //  unsolvable 
		cout<<"No";
		return 0;
	}
	for(int i=0;i<n;i++){
    
		cout<<a[i]<<endl;
	}
	return 0;
}

Happy jump

Title Description

One $n$ An integer array of elements , If the absolute value of the difference between two consecutive elements of the array includes $[1,n-1]$ Between all the integers , It is called consistent with “ Happy jump ”, Such as arrays $1423$ accord with “ Happy jump ”, Because the absolute values of the difference are ：$3,2,1$.

Given an array , Your task is to determine whether the array matches “ Happy jump ”.

Input format

The first row of each set of test data is represented by an integer $n(1\le n\le 1000)$ Start , Next $n$ A space separated in [$-10^8$,$10^8$] Integer between .

Output format

For each group of test data , Output a line if the array matches “ Happy jump ” The output "Jolly", Otherwise output "Not jolly".

Examples #1

The sample input #1

4 1 4 2 3

Sample output #1

Jolly

Examples #2

The sample input #2

5 1 4 2 -1 6

Sample output #2

Not jolly

Tips

$1\le n\le 1000$

Answer key

// Author:PanDaoxi
#include <iostream>
#include <cmath>
using namespace std;
int main(){
    
	int n,a[1001];
	bool flag=false;
	cin>>n;
	//  Read in the data 
	for(int i=0;i<n;i++){
    
		cin>>a[i];
	}
	for(int i=0;i<n-1;i++){
    
	    //  According to the meaning of the title , Write the formula 
		int t=abs(a[i]-a[i+1]);
		if(1<=t&&t<=n) flag=true;
		else flag=false; //  Do not conform to the 
	}
	
	if(flag==true) cout<<"Jolly";
    else cout<<"not Jolly";
	return 0;
}

turn on the light

Title Description

On an infinitely long road , There is an endless line of street lights , The number is $1,2,3,4,\dots$.

There are only two possible states for each lamp , On or off . If you press the switch of a lamp , Then the state of this lamp will change . If it turns out to be on , It's going to turn off . If it turns out to be off , It's going to turn on .

At the beginning , All the lights are off . Xiaoming can perform the following operations every time ：

Specify two numbers ,$a,t$（$a$ It's a real number ,$t$ As a positive integer ）. Will be numbered as $[a],[2\times a],[3\times a],\dots ,[t\times a]$ Once for each of the lights . among $[k]$ For real numbers $k$ The integral part of .

In Xiaoming $n$ After the first operation , Xiaoming suddenly finds out , Only one light is on at this time , Xiao Ming would like to know the number of this lamp , But this lamp is too far away from Xiaoming , Xiao Ming can't see the number .

fortunately , Xiaoming still remembers the previous $n$ operations . So Xiaoming finds you , Can you help him figure out the number of the light on ？

Input format

First line a positive integer $n$, Express $n$ operations .

Next there is $n$ That's ok , Two numbers per line ,$a_i,t_i$. among $a_i$ Is the set of real Numbers , There must be... After the decimal point $6$ position ,$t_i$ It's a positive integer. .

Output format

Just a positive integer , The number of the light on .

Examples #1

The sample input #1

3
1.618034 13
2.618034 7
1.000000 21

Sample output #1

20

Tips

remember $T=t_1+t_2+t_3+\dots +t_n$.

about $30\%$ The data of , Satisfy $T\le 1000$

about $80\%$ The data of , Satisfy $T\le 200000$

about $100\%$ The data of , Satisfy $T\le 2000000$

about $100\%$ The data of , Satisfy $n\le 5000,1\le a_i<1000,1\le t_i\le T$

Data assurance , after $n$ After the first operation , There is and only one lamp is on , There's no need to make a mistake . And for all the $i$ Come on ,$t_i\times a_i$ The maximum value of is not more than 2000000.

Answer key

// Author:PanDaoxi
#include <iostream>
using namespace std;
int main(){
    
	int n;
	//  Write a Boolean array to store data , And less space is used 
	bool flag[20000001]={
    };
	cin>>n;
	for(int i=1;i<=n;i++){
    
		//  Enter a real number a  and   Integers t
		double a;
		int t;
		cin>>a>>t;
		//  Core processing , Turn off the street lights and turn them on , Turn it on turn it off 
		for(int j=1;j<=t;j++){
    
			//  A mathematical principle 
			flag[int(j*a)]-=1;
			flag[int(j*a)]*=-1;
		}
	}
	//  Output results 
	for(int i=1;i<20000001;i++){
    
		if(flag[i]){
    
			cout<<i;
			return 0;
		}
	}
	return 0;
}

The mathematical principles mentioned above , I mean, put 1 become 0, hold 0 And then become 1 This matter . If you write directly if It will time out .
I first thought of using absolute value , Look, here is a 0, It decreases 1 Just change -1 了 ,-1 The absolute value of is 1;1 subtract 1 become 0,0 The absolute value of is still 0. So it changed .
Further simplification , Because subtract 1 In the future, non positive numbers , therefore , The absolute value of a negative number is its opposite , We can express it as -1 The opposite number of is 1, And it stipulates 0 The opposite number of is 0. therefore , We have ：

x-=1;
// x=-x;
x*=-1;