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C++ program that uses map STL to store students' student numbers and names

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使用map STL存储学生的学号和姓名的C++程序

Suppose we have a map data structure of a student volume. The name of the volume is integer data and the name is string type data. On our standard input, we provide n queries. There must be two elements in each query (per row), or three elements for type 1 queries. The first item is the operator, the second is the volume, the third is the name, and for two-element queries, the second is the volume number. The operation is as follows -

  • Insert. This will insert the name into the map for the corresponding volume

  • removed. This will remove the corresponding roll number from the map if it exists.

  • search. This will search the map with the roll number for the name and display the name if present, otherwise display not found.

So if the input is something like n = 8, then the query = [[1,5,"Atanu"], [1,8,"Tapan"], [ 1,3,"Manish"],[2,8],[1,9,"Piyali"], [3,8],[3,3], [3,5]], then output will be [Not found, Manish, Atanu] because roll 8 does not exist and the name of the student in roll 3 is Manish and the name of the student in roll 5 is "Atanu".

To solve this problem, we will follow the following steps-

  • n := Query quantity li>
  • Define a mapping of integer type keys and string type values m.
  • When n is non-zero, reduce n in each iteration and execute:
      Get the current query type t
  • Get the volume number
  • If t Same as 1, then:
    • Get the name
    • m[roll] := name
  • ## Otherwise when t equals 2, then:
    • m[roll ] := empty string
  • otherwise
    • If m[roll] is not an empty string, then:
      • Display m[roll]
    • Otherwise
      • Display "Not Found"
##Example

Let us see the following implementation for better understanding -

#include <iostream>
#include <map>
using namespace std;
int main(){
    int n;
    cin >> n;
    map<int, string> m;
    while (n--) {
        int t;
        cin >> t;
        int roll;
        cin >> roll;
        if (t == 1) {
            string name;
            cin >> name;
            m[roll] = name;
        } else if (t == 2) {
            m[roll] = "";
        } else {
            if(m[roll] != "")
                cout << m[roll] << endl;
            else
                cout << "Not found" << endl;
        }
    }
}

Input

8
1 5 Atanu
1 8 Tapan
1 3 Manish
2 8
1 9 Piyali
3 8
3 3
3 5

Output

Not found
Manish
Atanu

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