Introduction to Luogu 3 [circular structure] problem list solution

Catalog

Find the minimum

Classification average  

A foot's weight

Digital right triangle

Sum of factorials

High precision

Counting problem

Sum of series

Gold coin

Prime pocket

Prime Palindromes

I. Judge the palindrome number

II. Judge the prime number

Xiaoyu is swimming

Number reversal

When the moon falls and the crow dies, the money is

  Longest hyphen

Prime factor decomposition

Find triangle

Davor

Jinjin's savings plan

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【 introduction 3】 Loop structure - List of questions - Luogu | New ecology of computer science education (luogu.com.cn)

Find the minimum

• Pay attention to finding the minimum value , The minimum value should be initialized to a larger number ;
• Find the maximum , The maximum value should be initialized to a smaller number .

int n,a[100],min=1000; // Initialize to a larger number .
	cin >> n;

	for (int i = 0; i < n; i++)
		cin >> a[i];
	for (int i = 0; i < n; i++)
	{
		if (a[i] < min)
			min = a[i];
	}
	cout << min;

Classification average  

int n, k,sum1=0,sum2=0;
	double ave1=0, ave2=0;
	cin >> n >> k;
	for (int i = 1; i <= n; i++)
	{
		if (i%k == 0)
		{
			ave1 += i;
			sum1++;
		}
		if (i%k != 0)
		{
			ave2 += i;
			sum2++;
		}
	}
	ave1 /= sum1;
	ave2 /= sum2;

	printf("%.1lf %.1lf", ave1, ave2);

A foot's weight

	int a,sum=1;
	cin >> a;
	while (1)
	{
		if (a == 1)
			break;
		a /= 2;
		sum++;
	}
	cout << sum;

Digital right triangle

• Leading 0 It can be used %02d

#include<iostream>
#include<algorithm>
#include<math.h>

using namespace std;

int main() {
	int n, sum = 0, k;
	cin >> n;
	k = n;// Set up k  Or the following i<=(n The expression of ） there n It changes with it .
// Pay attention to this intermediate setting , There will be many questions related to this original value change in the future , But later operations also use the original value , At this time , Be sure to start with a variable equal to the original value .
	for (int i = 1; i <= (n+1)*n/2; i++)
	{
		
		printf("%02d", i);
		sum++;
		if (sum == k)
		{
			sum = 0;// Be careful sum=0 The location of 
			cout << endl;
			k--;
		}
	}


	return 0;
}

Sum of factorials

High precision

#include<iostream>
#include<iomanip>
#include<math.h>
#include<algorithm>
#include<string>
#include<cstring>
#include<stdio.h>
using namespace std;
int a[101] = { 0 }, b[101] = { 0 };

void mul(int x)// ride 
{
	int y = 0;
	for (int i = 100; i >= 0; i--)
	{
		a[i] = a[i] * x + y;      //a[101] except a[100] Others are initialized to 0, After that, each carry must only be based on the previous digit, that is y carry . In addition to the first to use x.
		y = a[i] / 10;//y It's carry 
		a[i] = a[i] % 10;
	}
}

void add( )// Add 
{
	int y = 0;
	for (int i = 100; i >= 0; i--)
	{
		b[i] = b[i] +a[i] + y;
		y = b[i] / 10;     //y It's carry 
		b[i] = b[i] % 10;
	}

}

int main()
{
	int n,i,w;
	cin >> n;
	a[100] = b[100] = 1;
	for (i = 2; i <= n; i++)
	{
		mul(i);   // Run from small to large , Add   2!   The next operation starts from 2!*3  namely 3！ To calculate .
		add();
	}
	int m = 0;
	while (b[m] == 0) m++;// Remove the end 0
	
	for (int i = m; i <= 100; i++)
		cout << b[i];

	return 0;
}

Counting problem

int sum = 0;
	int n, x,a;
	cin >> n >> x;
	for (int i = 1; i <=n; i++)
	{
		a = i;
		while (a)// If you don't set a=i, Back i Arithmetic , change i The value of the ,i The next time for Cycle from 0 Start .
		{
			if (a % 10 == x)// Single digit ,
				sum++;
			a /= 10;// After using single digits , Change ten digits into single digits , Then judge why the number of single digits .
		}// Determine what number each bit of this number is   We must know this while loop . 
	}
	cout << sum;

Sum of series

int k;
	double n = 0;
	double i = 1;
	cin >> k;
	while(i++)
	{
		n += 1 / i;
	    if (n > k)
		{
			break;
		}
	}

	cout << ceil(n);

Gold coin

#include<iostream>
using namespace std;
int main()
{
	int i = 1,m=1,k,sum=0;
	cin >> k;
	while (1)
	{
		while (m--) // Every time the m God ,m by 0 when , For the next group of days, press m=i, namely m+1 To calculate .
		{
			sum += i; //m individual i Add up    namely n God *n Gold coin 
			k--;
			if (k == 0)
				break;
		}
		i++;  // The number of gold coins , Receive... Every day i Gold coins 
		m = i;// Days , continuity m God 
		if (k == 0)
		{
	     	break;	
		}
	}
	cout << sum;
	return 0;
}

Prime pocket

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
using namespace std;
int main()
{
	int L,sum=1,num=2;
	cin >> L;
	if (L == 1)
		cout << "0";
	else
	{
		for (int i = 2; num <= L; i++)
		{
			if (i == 2)
				cout << i << endl;
			bool prime = true;// Note the location of the initialization 
			for (int j = 2; j <= i / 2 + 1; j++)
			{
				if (i % j == 0)// Remainder  
				{
					prime = false;
					break;
				}
			}
			if (prime == true)
			{
				num += i;
				if (num > L)
					break;
				cout << i << endl;
				sum++;
			}
		}
		cout << sum;
	}
	return 0;
}

Prime Palindromes

I. Judge the palindrome number

bool isPalindrome(int x)
{
	if (x < 0 || (x % 10 == 0 && x != 0)) // Not 0 outside 10 The multiple of is definitely not palindromes . 
	{
		return false;
	}
	int re = 0;
	while (x > re)
	{
		re = re * 10 + x % 10;
		x /= 10;
	}
	return x == re || x == re / 10;
}

  Figure source force buckle

II. Judge the prime number

bool isprime(int y)
{
	if (y != 2 && y % 2 == 0) // except 2 Even numbers other than are definitely not prime numbers 
		return false;
	int n = y,sum=0;

	if (isPalindrome(y) && y != 11) // except 11 Palindromes other than even digits are definitely not prime numbers , There are 11 It's a factor 
	{
		while (y)
		{
			y /= 10;
			sum++;
		}
		if (sum % 2 == 0)
			return false;
	}
	for (int i = 2; i <= sqrt(n); i++)
	{
		if (n%i == 0)
			return false;
	}
	return true;
}

#include<iostream>
#include<math.h>
#include<algorithm>

using namespace std;

bool isPalindrome(int x)
{
	if (x < 0 || (x % 10 == 0 && x != 0))
	{
		return false;
	}
	int re = 0;
	while (x > re)
	{
		re = re * 10 + x % 10;
		x /= 10;
	}
	return x == re || x == re / 10;
}

bool isprime(int y)
{
	if (y != 2 && y % 2 == 0)
		return false;
	int n = y,sum=0;

	if (isPalindrome(y) && y != 11)
	{
		while (y)
		{
			y /= 10;
			sum++;
		}
		if (sum % 2 == 0)
			return false;
	}
	for (int i = 2; i <= sqrt(n); i++)
	{
		if (n%i == 0)
			return false;
	}
	return true;
}

int main()
{
	int a, b;
	cin >> a >> b;
	if (b > 10000000) b = 10000000; // except 11 Palindromes other than even digits are definitely not prime numbers . Run to the 10000000 Can .
	for (int i = a; i <= b; i++)
	{
		if (isPalindrome(i))
		{
			
			if (isprime(i))
			{
				cout << i << endl;	
			}
		}
	}
	return 0;
}

Xiaoyu is swimming

#include<iostream>

using namespace std;
int main()
{
	double meter=2,x;
	int step=0;
	cin >> x;
	while (x > 0)
	{
		x -= meter;
		meter*=0.98;
		step++;
	}
	cout << step;


	return 0;
}

Number reversal

Pay attention to cut off all at the end 0

#include<iostream>
using namespace std;
int main()
{
	int N,a,b,count0=1;
	cin >> N;
	if (N == 0)
		cout << 0;
	else {
		if (N < 0)
		{
			cout << "-";
			N = -N;
		}
		a = N;
		b = N;
		while (b)// Judge how many there are at the end of this number 0; Cut them all off. 
		{
			N = b % 10;
			b /= 10;
			if (N == 0)
			{
				count0 *= 10;
			}
			else
				break;
		}
		a /= count0; // Cut off the end 0
		b = a;
		while (b)
		{
			a = b;
			a %= 10;
			cout << a;//2
			b /= 10;
		}
	}
	return 0;
}

When the moon falls and the crow dies, the money is

• Don't be confused by formulas , Don't use formulas directly ！！！
• To find a pattern
• The title has already said Fibonacci sequence . The characteristic of this series is f(n-1)+f(n-2)=f(n)

• #include <stdio.h> // The header file 
  int main()
  {
      double f[50];
      int n,i;
      f[0]=0;
      f[1]=1;
      f[2]=1;   // Recursive boundary conditions 
      scanf("%d",&n);
      for (i=3;i<=n;i++)
      f[i]=f[i-1]+f[i-2];  // Start using Fibonacci series 
      printf("%0.2lf",f[n]); // Output , Keep two decimal places 
      return 0;
  }

  Longest hyphen

#include<iostream>
using namespace std;
int main()
{
	int n,serial=1,a,b,max=1;
	cin >> n;

	cin >> a;// First number 
	n--;
	while (n--)
	{
		cin >> b;// Other numbers , The next number of the previous number .
		if (a == b - 1)
		{
			serial++;
			if (serial > max)
				max = serial;
		}
		if (a != b - 1)
		{
			serial = 1;
		}
		a = b;// The last number 
	}


	cout << max;
	return 0;
}

Prime factor decomposition

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	int n,p,a;
	cin >> n;
	// The product of two prime numbers , The factor of this product can only be the sum of these two prime numbers and 1 And itself .
	// There are many iterations from big to small , Traverse from small to large, find smaller divisors, and traverse fewer , Reuse n/
	for (int i =2; i<=sqrt(n); i++)// Judge the prime number 
	{
		if (n%i == 0)// First, let's see if the product can be divisible 
		{
			p = i;
			break;
		}
	}
	cout << n/p;
	return 0;
}

Find triangle

#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
int main()
{
	int scale,row=0,a,b,newrow=0;
	cin >> scale;

	a = scale * scale;
	for (int i = 1; i <=a ; i++)
	{
		printf("%02d", i);
		row++;
		if (row % scale == 0)
			cout << endl;
	}

	cout << endl;
	cout << setw(2 * (scale - 1)) << ""; // The first row of the second square 01 The space in front of it 
	b = (scale + 1)*scale / 2;
	row = 1;
	for (int j = 1; j <= b; j++)// A number means how many numbers there are in a row . Another is that each line starts from 0 Start , Output a number ++
	{

		printf("%02d", j);
		newrow++;
		if (row == newrow && row<scale)//row<scale Otherwise, one more line will be output 
		{
			cout << endl;	
			cout << setw(2 * (scale - row-1)) << "";
			newrow=0;
			row++;//row Is a line of several numbers .
		}

	}

	return 0;
}

Davor

#include<iostream>
using namespace std;
int main()
{
	int n, x, k;
	cin >> n;
	//1092k+364x=n;
	for (int i = 1; i < n; i++)
	{//k As small as possible , from 1 Start arranging , Line up first to the smaller k
		if ((n - 1092 * i) % 364 == 0)
		{
			x = (n - 1092 * i) / 364;
			if (x > 100)
			{
				continue;
			}
			k = i;
			
			break;
		}
	}
	cout << x << endl;
	cout << k << endl;
	return 0;
}

Jinjin's savings plan

#include<iostream>
using namespace std;
int main()
{
	int a[12],sum=0,rest=0;//sum For the money saved in mom's place 
	bool is = true;
	for (int i = 0; i < 12; i++)
	{
		cin >> a[i];//a[i] It means the money spent this month . If you save money for mom , That's a flower .
	}
	for (int i = 0; i < 12; i++)
	{
		if (a[i] <= 100)//300-a[i] The rest of the money .
		{
			sum += 200;
			a[i] += 200;//a[i] It means the money spent this month . If you save money for mom , That's a flower .

		}
		else if (a[i] <= 200)
		{
			sum += 100;
			a[i] += 100;
		}
		rest += (300 - a[i]);// The money you save ,

		if (rest >=100)
		{
			//cout << "rest is >=100  "<<rest << endl;
			sum += 100;
			rest -= 100;// Note that it's not just this month , The money left over from last month , She saves until next month .
		
		}
		if (rest < 0)
		{
			//cout << "rest is <0 " << rest << endl;
			cout << "-" << i+1;
			is = false;
			break;
		}
	}
	if (is == true)
	{
		cout << sum*1.2+rest;
	}


	return 0;
}