Luo Gu - some interesting questions

P5731 【 Deep base 5. xi 6】 Serpentine matrix - Output format

  1  2  3  4
 12 13 14  5
 11 16 15  6
 10  9  8  7

  It can be said to be a water problem , But all WA,debug For a long time , It turns out that the output format is toxic

AC：

#include<bits/stdc++.h>
using namespace std;

int a[10][10];
int main()
{
	int n;
	cin>>n;
	int k=0,i,j;
	int x=1,y=0;    // from a[1][1] Start , Convenient operation 
	
	while(k<n*n)
	{
		while(y<n&&a[x][y+1]==0) // From left to right 
		a[x][++y]=++k;
		
		while(x<n&&a[x+1][y]==0) // From top to bottom 
		a[++x][y]=++k;
		
		while(y>1&&a[x][y-1]==0) // From right to left 
		a[x][--y]=++k;
		
		while(x>1&&a[x-1][y]==0) // From bottom to top 
		a[--x][y]=++k;
	}
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=n;j++)
		printf("%3d",a[i][j]);
		printf("\n");
		//cout<<a[i][j]<<' ';
		//cout<<endl;
	}
	return 0;
}

P5732 【 Deep base 5. xi 7】 Yang hui triangle - Classical mathematics

 

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

  Ideas ： So let's define one Two dimensional array ：a[N][N], Slightly larger than the number of lines to print . Let the number on both sides be 1, That is, when the first and last number of each row are 1.a[i][0]=a[i][i-1]=1,n Is the number of rows . Except for the numbers on both sides , Any number is the sum of the upper two vertices , namely a[i][j] = a[i-1][j-1] + a[i-1][j].

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
typedef long long LL;
const int N=1e5+10;
using namespace std;
 
 int a[21][21];
 int n;
 
int main()
{
 	cin>>n;
 	for(int i=1;i<=n;i++)    // from 1 Start good operation 
 	a[i][1]=a[i][i]=1;    // Easy to see , The first column and diagonal are 1
 	
 	for(int i=1;i<=n;i++)
 	{
 		for(int j=2;j<i;j++)
 		{
 			a[i][j]=a[i-1][j]+a[i-1][j-1]; // The core formula 
		}
	}
	
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=i;j++)
		cout<<a[i][j]<<' ';
		cout<<endl;
	}
		
	return 0;
}

P1957 Oral arithmetic exercises

 

4
a 64 46
275 125
c 11 99
b 46 64

Little knowledge ：

sprintf Function usage

sprintf Format of function ：
int sprintf( char *buffer, const char *format [, argument,…] );

1、 Can control the accuracy , Can also “ rounding ”

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int main()
{
	char str1[200];
	char str2[200];
	double f1=14.305100;
	double f2=14.304900;
	//double f2=14.305000;
	sprintf(str1,"%20.2f\n",f1);
	sprintf(str2,"%20.2f\n",f2);
	cout<<str1<<str2;
	return 0;
}

 

 2、 You can connect multiple numerical data

char str[20];
int a=20984,b=48090;
sprintf(str,”%3d%6d”,a,b);
str[]=”20984 48090”

3、 You can connect multiple strings into a string

char str[20];
	char s1[5]={'A','B','C'};
	char s2[5]={'x','y','z'};
	sprintf(str,"%.3s%.3s",s1,s2);
	str="ABCxyz";

 %m.n In the output of the string ,m Width , The number of columns shared by strings ;n Represents the actual number of characters .%m.n In floating point numbers ,m Also means width ;n Represents the number of decimal places .

4、

The subsequent operations are basically the same as printf equally , nothing but sprintf Function to print to a string （ Note that the length of the string should be enough to accommodate the printed content , Otherwise, there will be a memory overflow ）, and printf Function prints out to the screen .

understand sprintf After usage , This is a direct second kill ：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
typedef long long LL;
const int N=1e5+10;
using namespace std;

int main()
{
	char ch;
	int n,a,b;
	char s[10010],x[2];
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>x;
		if(x[0]>='a'&&x[0]<='c')// See whether the first character is a number or a letter 
		{
			ch=x[0];
			cin>>a>>b;// It is letters that directly output numbers 
		}
		else
		{
			sscanf(x,"%d",&a);// It is the conversion of characters and numbers into numbers 
			cin>>b;
		}
		
		if(ch=='a')
			sprintf(s,"%d+%d=%d",a,b,a+b);// Direct output , Very easy to use 
		else if(ch=='b')
			sprintf(s,"%d-%d=%d",a,b,a-b);
		else if(ch=='c')
			sprintf(s,"%d*%d=%d",a,b,a*b);
		cout<<s<<endl<<strlen(s)<<endl;
	}
	return 0;
}