Blue bridge code error ticket

 Some secret related unit has issued some kind of bill , And take it back at the end of the year .
     Each note has a unique ID Number . Of all bills throughout the year ID The number is continuous , but ID The start number of is randomly selected .
     Because of the negligence of the staff , In the input ID There was a mistake on the , Created something ID Break the number , Another one ID Duplicate number .
     Your task is to program , Find the broken one ID And the heavy one ID.
     Suppose that the interrupt can't happen on the largest and the smallest .
 Ask the program to enter an integer first N(N<100) Indicates the number of subsequent data lines .
 Then read in N Row data .
 The data length of each row is not equal , It's a number of... Separated by spaces （ No more than 100 individual ） Positive integer （ No more than 100000）
 Each integer represents a ID Number .
 Request program output 1 That's ok , Two integers m n, Separate... With spaces .
 among ,m It's a break ID,n It's a double sign ID
 for example ：
 User input ：
2
5 6 8 11 9 
10 12 9
 Program output ：
7 9
 Another example is ：
 User input ：
6
164 178 108 109 180 155 141 159 104 182 179 118 137 184 115 124 125 129 168 196
172 189 127 107 112 192 103 131 133 169 158 
128 102 110 148 139 157 140 195 197
185 152 135 106 123 173 122 136 174 191 145 116 151 143 175 120 161 134 162 190
149 138 142 146 199 126 165 156 153 193 144 166 170 121 171 132 101 194 187 188
113 130 176 154 177 120 117 150 114 183 186 181 100 163 160 167 147 198 111 119
 Program output ：
105 120

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Their thinking

Know from the question , This ID It was originally continuous and the punctuation point was not at the head and tail , that ID The broken sign must be [ head , tail ]. So I want to correspond this continuous number to the array a[idx] The subscript idx. Such as ：1,2,4. It must be missing 3. How does the code judge ？ We just have to let a[1]=1,a[2]=2,a[4]=4 after , After traversing the array once , Found a[3]=0（ Here, assume that the array is initialized to 0）, That means 3 The number is broken .

For how to find duplicate elements , Just be right at the beginning a[idx] When setting out , Let the corresponding vis[idx]++ To record the number of times this value appears .

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Attach code

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int a[100005]={0};			// Put the number corresponding to the subscript , Such as ：200 It must be on a[200] in  
int vis[100005]={0};		// Record the number of occurrences of each number  
int main()
{
	int n;
	cin>>n;
	int min=100000;			// Record the subscript of the minimum value  
	int max=-1;				// Record the maximum subscript 
	getchar();				// Read out a carriage return  
	while(n>0)
	{
		string s;
		getline(cin,s,'\n');// Read a line of string with spaces  
		int idx=0;
		int len=s.length();
		for(int i=0;i<len;i++)
		{
			if(s[i]!=' ')	// Complete the reading of multi bit numbers  
			{
				idx=idx*10+s[i]-'0';
			}
			if(s[i]==' '||i==len-1)
			{
				a[idx]=idx;
				vis[idx]++;
				if(min>idx)
					min=idx;
				if(max<idx)
					max=idx;
				idx=0;
			}
		}
		n--;
	}
	for(int i=min;i<=max;i++)// Find the missing number  
	{
		if(a[i]==0)
		{
			cout<<i<<" "; 
			break; 
		}	
	}
	for(int i=min;i<=max;i++)// Look for duplicate numbers  
	{
		if(vis[i]>1)
		{
			cout<<i<<endl;
			break;
		}	
	}
	return 0;
}