L1-049 ladder race seat allocation (20 points) (in-depth understanding of for loop + three-dimensional array + error analysis)

describe ：

There are a large number of players in the ladder race every year , Make sure that all members of the same school are not adjacent , The allocation of seats becomes a more troublesome thing . So we have the following strategies ： Suppose that there is N Schools compete , The first i Schools have M[i] team , Every team 10 Contestants . Line up the players of each school , The first i+1 The players of the team are in i After the team . From 1 Schools start , The first of each school 1 Members of the team were seated in sequence , And then there's the number 2 Team members …… And so on . If there is only 1 The school team has not allocated seats yet , They need to arrange their team members to sit in separate seats . This question asks you to write a program , Automatically generate the seat number of the team members for each school , from 1 Numbered starting .

Input ：

Enter the number of colleges and universities in a row N （ No more than 100 The positive integer ）; The second line gives N No more than one. 10 The positive integer , Among them the first i The number corresponds to the i The number of participating teams in Universities , Numbers are separated by spaces .

Output ：

From 1 The number of colleges and universities 1 Team start , Output the seat number of the team member in sequence . Each team has a line , Between seat numbers 1 Space separation , There must be no extra space at the beginning and end of the line . in addition , Press... On the first line of each university “#X” Output the number of the school X, from 1 Start .

The sample input

3
3 4 2

Sample output ：

#1
1 4 7 10 13 16 19 22 25 28
31 34 37 40 43 46 49 52 55 58
61 63 65 67 69 71 73 75 77 79
#2
2 5 8 11 14 17 20 23 26 29
32 35 38 41 44 47 50 53 56 59
62 64 66 68 70 72 74 76 78 80
82 84 86 88 90 92 94 96 98 100
#3
3 6 9 12 15 18 21 24 27 30
33 36 39 42 45 48 51 54 57 60

error analysis ：

There was no idea at first , It's hard to simulate , Simulated for a long time , There were only 15 branch , Test point 1,2 I can't get by , Analyzed for a long time , Found the reason for being stuck ： If there is only one school student left, there is no row ( Let's say we've reached number 80 individual ), Next to this school , If the school in the last row is still it , Then it's from 82 Start arranging , But if the last one is not this school , It's from 81 Start arranging , The back row spacing is 2;

Ideas ：

I have done row first and column first questions before , Analyze the problem , It's jumping around between different schools , We might as well The school element is set to for The innermost loop of the loop , Let the cycle jump from school to school , How to solve the interval problem , Here we introduce a sign pre Record the previous school every time , When determining the interval each time, just compare whether this school is the same as the previous school , All elements use one Three dimensional array storage

#include<bits/stdc++.h>
using namespace std;
 
typedef unsigned long long ull;
typedef long long ll;
 
const ll maxx = 1e18;
const int N = 1e6+100;
const int p = 1e4+10;
const double eps = 1e-8;

int a[101][11][11];// The first dimension is school , The second dimension is the team , The third dimension is the team position 
int max1,pre=-1,n,cnt;
int num[101];

int main()
{
    
	cin>>n;
	
	for(int i=1;i<=n;i++)
	{
    
		cin>>num[i];
		max1=max(max1,num[i]);
	}// Enter and record the maximum number of teams 
	
	for(int i=1;i<=max1;i++)// The first layer traverses the... Of each school  i  A team 
	{
    
		for(int j=1;j<=10;j++)// The second layer traverses the second layer of each team  j  A team member 
		{
    
			for(int k=1;k<=n;k++)// The last layer traverses each school 
			{
    
				if(num[k]>=i)// Make sure the school has this team 
				{
    
					if(k==pre) cnt+=2;
					else cnt+=1;// Determine the interval 
					
					a[k][i][j]=cnt;// Count 
					
					pre=k;// Write down this school 
					
				}
			}
		}
	}
	
	for(int i=1;i<=n;i++)
	{
    
		cout<<"#"<<i<<endl;
		for(int j=1;j<=num[i];j++)
		{
    
			for(int k=1;k<=10;k++)
			{
    
				if(k!=10)
				cout<<a[i][j][k]<<" ";
				else
				cout<<a[i][j][k]<<endl;
			}
		}
	}
}

reflection ：

This question is a right for The understanding of circulation , Simulation is not necessary , Through this question, I have a deeper understanding of the sequence of cycles , TIANTI race, come on ！！！