Preface

I took part in the Blue Bridge Cup competition these two days , Won the second prize in the city competition （ Average in the exam , Only more than 76% Classmate ）, Enter the provincial competition . I'm in 4 month 23 Japan （ Saturday ） The game , So do it in advance 4 month 17 Day's problem brushing work , Let's make do with it , After I finish the exam, I will share my exam experience with you .

Brush problem , Walk up

There are several questions , Let's have a look , If there is anything wrong, please point it out ！

Programming to realize ： Compare the size

Title Description ：
Given two positive integers N and M(0<N<200,0<M<200,N≠M), Compare the size of two positive integers , Then output a larger positive integer .
for example ：N=145,M=100, After comparison 145 Greater than 100, Therefore, the output 145.

// Author:PanDaoxi
#include <iostream>
using namespace std;
int main(){
    
	int n,m;
	cin>>n>>m;
	cout<<(n>m?n:m);
	return 0;
} 

Programming to realize ： Factoring integer

Title Description ：
Given a positive integer N, And then N Decompose into 3 Sum of positive integers . Calculate how many decomposition methods meet the requirements .
requirement ：
1） Disintegrated 3 Positive integers are different ;
2） Disintegrated 3 None of the positive integers contain numbers 3 and 7.
Such as ：N by 8, Decomposable into （1,1,6）、（1,2,5）、（1,3,4）、（2,2,4）、（2,3,3）, Among them, the decomposition methods that meet the requirements are 1 Kind of , by （1,2,5）.

Too lazy to write deep search （ Be lazy , Another one is a little confused ）,
Direct loop , Simulation algorithm , Others may look a little ridiculous ..

// Author:PanDaoxi
#include <iostream>
using namespace std;
int main(){
    
	int n,sum=0;
	cin>>n;
	for(int i=1;i<=n;i++){
    
		for(int j=1;j<=n;j++){
    
			for(int k=1;k<=n;k++){
    
				if(i+j+k==n){
    
					if(i!=3&&j!=3&&k!=3&&i!=7&&j!=7&&k!=7&&i!=j&&i!=k&&j!=k&&i<j&&j<k){
    
						//cout<<i<<" "<<j<<" "<<k<<endl;
						sum++;
					}
				}
			}
		}
	}
	cout<<sum;
	return 0;
}

Programming to realize ： Combine （ To think about ）

Prompt information ： Factor ： A factor is an integer a Divide by an integer b(b≠0) The quotient of is exactly an integer without a remainder , We said b yes a The factor of .
The common factor ： Given a number of integers , If there is a ( some ) Numbers are their common factor , So this ( some ) Numbers are called their common factors .
Reciprocal number ： The common factor is only 1 Two nonzero natural numbers of , It's called coprime number ; for example ：2 and 3, The common factor is only 1, Coprime number . Title Description ：
A shop packs a kind of candy into N and M Two specifications to sell （N and M Coprime number , And N and M There are countless bags ）. This way of selling will limit the number of sweets that can't be bought . Then give N and M Value , Please calculate the maximum number of candy you can't buy .
for example ：
When N=3,M=5,3 and 5 Coprime number , The number of sweets you can't buy is 1,2,4,7, The maximum amount of candy you can't buy is 7,7 Any amount of candy can be purchased through combination .

Look, it's troublesome , Ah ah ！！！
I won't gather , Not exhaustive , This ……
I made a wave on the Internet , I have referred to Blog users Aikoin The article , You can have a look at , These are the main words ：

There is a conclusion to this problem that can be used directly ： Two coprime numbers a、b The maximum number of combinations of is ab-a-b. Detailed proof is given below （ Reference resources ）：

If there is a foundation of Discrete Mathematics , We know that we can divide all nonnegative integers into a A model a Congruence equivalence class , Write it down as [0],[1],[2],…,[a-1], Respectively ak,ak+1,ak+2,…,ak+(a-1),（k∈Z）.b The multiples of must be distributed here a In an equivalent class , Again because gcd(a,b)=1, So every equivalence class has b Multiple , And evenly distributed （ I can't figure it out here. I can draw two numbers by myself , That makes sense ）. Special ,ab∈[0]. set up bKi Is an equivalence class [i] The smallest of all b Multiple , Then in the equivalence class bKi All numbers after can be expressed as ax+bKi, It must be combined . obviously , maximal bKi All subsequent integers can be combined , And to find the maximum number of combinations , We just need to consider the biggest bKi The numbers in front of you can . Put all the bKi List , Yes {0,b,2b,…,(a-1)b}. Obviously (a-1)b Is the biggest bKi, And in others a-1 In an equivalent class , There must be a number smaller than it and can be combined , These numbers are (a-1)b-1,(a-1)b-2,…,(a-1)b-(a-1). So the maximum number that cannot be combined is (a-1)b-a, namely ab-a-b.

For the sake of understanding , I take 4、7 As an example, draw a chart , Combined with the chart, it should be easy to understand the above proof process ：

Don't ask me if I understand this long passage , I don't understand either , Look at this form ？ This needs us to study . First post the code here ：

// Author:PanDaoxi
#include <iostream>
using namespace std;
int main(){
    
	int m,n;
	cin>>m>>n;
	cout<<m*n-m-n;
	return 0;
} 

Refer to the answer

The first question is

#include <iostream>
using namespace std;

int main(){
    
	int n,m;
	cin >> n >> m;
	cout << max(n,m) << endl; 
	// Mode two  if...else
	// Mode three   Three wood operators  
	return 0;
}

The second question is

#include <iostream>
using namespace std;
int n, ans;

bool isSame(int x,int y,int z){
    
	// No one can be 3,7 
	if(x==3||x==7||y==3||y==7||z==3||z==7)
		return false;
	// Two cannot be equal  
	if(x == y || x == z || y == z)
		return false;
	return true;
}
 
int main(){
    
	cin >> n;
	for(int i = 1;i < n;i++){
    
		for(int j = 1;j < n;j++){
    
			if(n-i-j > 0 && isSame(i,j,n-i-j)){
    
				ans++;
			}		
		}
	} 
	cout << ans / 6;   //3 The digits are arranged in full order  6 In this case , Need to get rid of  
	return 0;
}

Third question （ no ）