To find the last element pair with a quantity not equal to 'quantityThreshold', one can modify the code as:
C++
#include <iostream>
#include <vector>
using namespace std;
int main() {
int numElements;
int quantityThreshold;
int lastPosition = -1;
int i;
cin >> numElements;
vector<string> colorList(numElements);
vector<int> quantityList(numElements);
for (i = 0; i < colorList.size(); ++i) {
cin >> colorList.at(i);
cin >> quantityList.at(i);
}
cin >> quantityThreshold;
// Update lastPosition to the index of the last element pair with a quantity not equal to quantityThreshold
for (i = numElements - 1; i >= 0; --i) {
if (quantityList.at(i) != quantityThreshold) {
lastPosition = i;
break;
}
}
// Output the color and quantity of the last element pair with a quantity not equal to quantityThreshold
if (lastPosition != -1) {
cout << colorList.at(lastPosition) << " " << quantityList.at(lastPosition) << endl;
cout << "last match found at # " << lastPosition << endl;
} else {
cout << "No result" << endl;
}
return 0;
}
So, based on the code in the question, the main change is that the loop starts from the end of the 'colorList' and 'quantityList' vectors and iterates backwards.
This ensures that the last element pair with a quantity not equal to 'quantityThreshold' is found and assigned to 'lastPosition'.