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 #include using namespace std; int main() { const int SIZE_LIST = 4; vector keysList(SIZE_LIST); vector 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 <
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;
}
Trending now
This is a popular solution!
Step by step
Solved in 4 steps with 2 images
