Add each element in origList with the corresponding value in offsetAmount. Print each sum followed by a space. Ex: If origList = {40, 50, 60, 70} and offsetAmount = {5, 7, 3, 0}, print: 45 57 63 70 #include <iostream>#include <vector>using namespace std; int main() {const int NUM_VALS = 4;vector<int> origList(NUM_VALS);vector<int> offsetAmount(NUM_VALS);unsigned int i; for (i = 0; i < origList.size(); ++i) {cin >> origList.at(i);} for (i = 0; i < offsetAmount.size(); ++i) {cin >> offsetAmount.at(i);} cout << endl; return 0;} Please help me with this problem using c++.
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Add each element in origList with the corresponding value in offsetAmount. Print each sum followed by a space. Ex: If origList = {40, 50, 60, 70} and offsetAmount = {5, 7, 3, 0}, print: 45 57 63 70
#include <iostream>
#include <
using namespace std;
int main() {
const int NUM_VALS = 4;
vector<int> origList(NUM_VALS);
vector<int> offsetAmount(NUM_VALS);
unsigned int i;
for (i = 0; i < origList.size(); ++i) {
cin >> origList.at(i);
}
for (i = 0; i < offsetAmount.size(); ++i) {
cin >> offsetAmount.at(i);
}
cout << endl;
return 0;
}
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- #ifndef lab5ExF_h #define lab5ExF_h typedef struct point { char label[10]; double x ; // x coordinate for point in a Cartesian coordinate system double y; // y coordinate for point in a Cartesian coordinate system double z; // z coordinate for point in a Cartesian coordinate system }Point; void reverse (Point *a, int n); /* REQUIRES: Elements a[0] ... a[n-2], a[n-1] exists. * PROMISES: places the existing Point objects in array a, in reverse order. * The new a[0] value is the old a[n-1] value, the new a[1] is the * old a[n-2], etc. */ int search(const Point* struct_array, const char* target, int n); /* REQUIRES: Elements struct-array[0] ... struct_array[n-2], struct_array[n-1] * exists. target points to string to be searched for. * PROMISES: returns the index of the element in the array that contains an * instance of point with a matching label. Otherwise, if there is * no point in the array that its label matches the target-label, * it should return -1. * If there are more than…Add each element in origList with the corresponding value in offsetAmount. Print each sum followed by a comma (no spaces). Ex: If origList = {4, 5, 10, 12} and offsetAmount = {2, 4, 7, 3}, print: 6,9,17,15, #include <iostream>#include <string.h>using namespace std; int main() { const int NUM_VALS = 4; int origList[NUM_VALS]; int offsetAmount[NUM_VALS]; int i; cin >> origList[0]; cin >> origList[1]; cin >> origList[2]; cin >> origList[3]; cin >> offsetAmount[0]; cin >> offsetAmount[1]; cin >> offsetAmount[2]; cin >> offsetAmount[3]; /* Your code goes here */ cout << endl; return 0;}Write a for loop to print all NUM_VALS elements of vector courseGrades, following each with a space (including the last). Print forwards, then backwards. End with 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 = courseGrades.size() - 1 (Notes)Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element vector (vector<int> courseGrades(4)). See "How to Use zyBooks".Also note: If the submitted code has errors, 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>#include <vector>using namespace std; int main() { const int NUM_VALS = 4; vector<int> courseGrades(NUM_VALS); int i; for (i = 0; i < courseGrades.size(); ++i) { cin >> courseGrades.at(i); }…
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