Use cin to read each pair of inputs, integer currFee and string currState, until -1 is read from input. Create a License object with currFee and currState as arguments and append the object to vector licenseList. Ex: If the input is 212 NV 331 MA 326 NM 289 FL -1, then the output is: License fee: 212, State: NV License fee: 331, State: MA License fee: 326, State: NM License fee: 289, State: FL CI OLALE, CO 29 30 31 32 33 34 35 36 37 38 39 40 42 43 44 45 46 Check /* Your code goes here while(true){ cin>>currFee; if (currFee ==-1){ break; } } cin>>currState; currLicense License (currFee, currState); licenseList.push_back (currLicense); for (i = 0; i < licenseList.size(); ++i) { currLicense = licenseList.at(i); currLicense. Print(); 4 return 0; 1 Next level 2 3 X Failed to compile main.cpp: In function 'int main()': main.cpp:37:44: error: no matching function for call to 'License:: License (int&, std::string& currLicense=License (currFee, currState); 37 | I main.cpp:5:7: note: candidate: 'License::License()' 5 class License { A¬~~~~~ 1 main.cpp:5:7: note: candidate expects o arguments, 2 provided main.cpp:5:7: note: candidate: 'License: :License (const License&)' main.cpp:5:7: note: candidate expects 1 argument, 2 provided main.cpp:5:7: note: candidate: 'License:: License (License&&)' main.cpp:5:7: note: candidate expects 1 argument, 2 provided

Q: A failsafe method of checking for integer overflow of an arithmetic operation that should result in…

A: Answer: True

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Q: Convert 3357to a numeral in base 5.       3357= blank_5

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A: Answer the above question are as follows

Q: mov r0, # 6Ah mov psw, # 00h mov a, # 22h ric a M90100564-871890336 598836 b19010056498715868 715983…

A: mov a, #22h So the content of accumulator is 0010 0010

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Array

The simplest way to start thinking about an array is as a collection, or a set of elements. In programming, these elements may be constants, variables, functions and pointers.