Write a loop that subtracts 1 from each element in lowerScores. If the element was already 0 or negative, assign 0 to the element. Ex: lowerScores = {5, 0, 2, -3} becomes {4, 0, 1, 0}. code: #include using namespace std; int main() { const int SCORES_SIZE = 4; int lowerScores[SCORES_SIZE]; int i; for (i = 0; i < SCORES_SIZE; ++i) { cin >> lowerScores[i]; } /* Your solution goes here */ for (i = 0; i < SCORES_SIZE; ++i) { cout << lowerScores[i] << " "; } cout << endl; return 0; }

Write a loop that subtracts 1 from each element in lowerScores. If the element was already 0 or negative, assign 0 to the element. Ex: lowerScores = {5, 0, 2, -3} becomes {4, 0, 1, 0}. code: #include using namespace std; int main() { const int SCORES_SIZE = 4; int lowerScores[SCORES_SIZE]; int i; for (i = 0; i < SCORES_SIZE; ++i) { cin >> lowerScores[i]; } /* Your solution goes here */ for (i = 0; i < SCORES_SIZE; ++i) { cout << lowerScores[i] << " "; } cout << endl; return 0; }

EBK JAVA PROGRAMMING

9th Edition

ISBN:9781337671385

Author:FARRELL

Publisher:FARRELL

Chapter9: Advanced Array Concepts

Section: Chapter Questions

Problem 4GZ

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Write a loop that subtracts 1 from each element in lowerScores. If the element was already 0 or negative, assign 0 to the element. Ex: lowerScores = {5, 0, 2, -3} becomes {4, 0, 1, 0}. #include <iostream>using namespace std; int main() { const int SCORES_SIZE = 4; int lowerScores[SCORES_SIZE]; int i; for (i = 0; i < SCORES_SIZE; ++i) { cin >> lowerScores[i]; } /* Your solution goes here */ for (i = 0; i < SCORES_SIZE; ++i) { cout << lowerScores[i] << " "; } cout << endl; return 0;}
Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10}, print:7 9 11 10 10 11 9 7 #include <iostream>using namespace std; int main() { const int NUM_VALS = 4; int courseGrades[NUM_VALS]; int i; for (i = 0; i < NUM_VALS; ++i) { cin >> courseGrades[i]; } /* Your solution goes here */ return 0;}
For any element in keysList with a value smaller than 40, print the corresponding value in itemsList, followed by a comma (no spaces). Ex: If keysList = {32, 105, 101, 35} and itemsList = {10, 20, 30, 40}, print: 10,40, #include <iostream>#include <string.h>using namespace std; int main() { const int SIZE_LIST = 4; int keysList[SIZE_LIST]; int itemsList[SIZE_LIST]; int i; cin >> keysList[0]; cin >> keysList[1]; cin >> keysList[2]; cin >> keysList[3]; cin >> itemsList[0]; cin >> itemsList[1]; cin >> itemsList[2]; cin >> itemsList[3]; /* Your code goes here */ cout << endl; return 0;}
Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10}, print:7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = NUM_VALS - 1. Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element array (int courseGrades[4]). Also note: If the submitted code tries to access an invalid array element, such as courseGrades[9] for a 4-element array, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message. #include <iostream>using namespace std; int main() { const int NUM_VALS = 4; int courseGrades[NUM_VALS]; int i; for (i = 0; i < NUM_VALS; ++i) { cin >> courseGrades[i]; } /* Your solution goes…
For any element in keysList with a value smaller than 40, print the corresponding value in itemsList, followed by a space. Ex: If keysList = {32, 105, 101, 35} and itemsList = {10, 20, 30, 40}, print: 10 40 #include <iostream>#include <string.h>using namespace std; int main() { const int SIZE_LIST = 4; int keysList[SIZE_LIST]; int itemsList[SIZE_LIST]; int i; cin >> keysList[0]; cin >> keysList[1]; cin >> keysList[2]; cin >> keysList[3]; cin >> itemsList[0]; cin >> itemsList[1]; cin >> itemsList[2]; cin >> itemsList[3]; /* Your code goes here */ cout << endl; return 0;}
//write code that loops through the two variables returns an array ['2 is zwei', '3 is drei', '4 is vier', '5 is fünf', '6 is sechs']// example:// return: ['2 is zwei', '3 is drei', '4 is vier', '5 is fünf', '6 is sechs']// hint: use Array.forEach, Array.map or a for loopfunction germanNumbers(){ const cardinalNumbers = [2,3,4,5,6]; const germanNumbers = ['zwei', 'drei', 'vier', 'fünf', 'sechs']; } // write code that returns an array of ONLY prime numbers that are in the array numbers// example: // return [2, 3, 5, 7, 11, 13]// hint use: Array.filterfunction returnPrimeNumbers(){ const numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]; }
Write a loop that sets each array element to the sum of itself and the next element, except for the last element which stays the same. Be careful not to index beyond the last element. Ex:Initial scores: 10, 20, 30, 40 Scores after the loop: 30, 50, 70, 40 The first element is 30 or 10 + 20, the second element is 50 or 20 + 30, and the third element is 70 or 30 + 40. The last element remains the same. #include <iostream>using namespace std; int main() { const int SCORES_SIZE = 4; int bonusScores[SCORES_SIZE]; int i; for (i = 0; i < SCORES_SIZE; ++i) { cin >> bonusScores[i]; } /* Your solution goes here */ for (i = 0; i < SCORES_SIZE; ++i) { cout << bonusScores[i] << " "; } cout << endl; return 0;}
Write a loop that sets each array element to the sum of itself and the next element, except for the last element which stays the same. Be careful not to index beyond the last element. Ex:Initial scores: 10, 20, 30, 40 Scores after the loop: 30, 50, 70, 40 The first element is 30 or 10 + 20, the second element is 50 or 20 + 30, and the third element is 70 or 30 + 40. The last element remains the same. #include <iostream>using namespace std; int main() {const int SCORES_SIZE = 4;int bonusScores[SCORES_SIZE];int i; for (i = 0; i < SCORES_SIZE; ++i) {cin >> bonusScores[i];} /* Your solution goes here */ for (i = 0; i < SCORES_SIZE; ++i) {cout << bonusScores[i] << " ";}cout << endl; return 0;} Please help me with this issue using c++.
JAVA Write an application containing three parallel arrays that hold 10 elements each. The first array hold four-digit student ID numbers, the second array holds first names, and the third array holds the students’ grade point averages. Use dialog boxes to accept a student ID number and display the student’s first name and grade point average. If a match is not found, display an error message that includes the invalid ID number and allow the user to search for a new ID number.
in java when you use an enhanced for loop with an array, you don’t need to use a/an ________________ variable to iterate through the elements of the array.
For any element in keysList with a value greater than 100, print the corresponding value in itemsList, followed by a space. Ex: If keysList = {42, 105, 101, 100} and itemsList = {10, 20, 30, 40}, print:20 30 Since keysList.at(1) and keysList.at(2) have values greater than 100, the value of itemsList.at(1) and itemsList.at(2) are printed. #include <iostream>#include <vector>using namespace std; int main() { const int SIZE_LIST = 4; vector<int> keysList(SIZE_LIST); vector<int> itemsList(SIZE_LIST); unsigned int i; for (i = 0; i < keysList.size(); ++i) { cin >> keysList.at(i); } for (i = 0; i < itemsList.size(); ++i) { cin >> itemsList.at(i); } /* Your solution goes here */ cout << endl; return 0;}
For any element in keysList with a value greater than 100, print the corresponding value in itemsList, followed by a space. Ex: If keysList = {42, 105, 101, 100} and itemsList = {10, 20, 30, 40}, print: 20 30 Since keysList.at(1) and keysList.at(2) have values greater than 100, the value of itemsList.at(1) and itemsList.at(2) are printed. #include <iostream>#include <vector>using namespace std; int main() {const int SIZE_LIST = 4;vector<int> keysList(SIZE_LIST);vector<int> itemsList(SIZE_LIST);unsigned int i; for (i = 0; i < keysList.size(); ++i) {cin >> keysList.at(i);} for (i = 0; i < itemsList.size(); ++i) {cin >> itemsList.at(i);} /* Your solution goes here */ cout << endl; return 0;} Please help me with this problem using c++.