C++ please help I will give you a good rating!!!!! Implement the following function by using subsets() developed in step 1. /* Check if given value given be expressed by K or less coins chosen from the given set of coins @param coins: all coins we can choose from @param first, last: specify the range of coins to choose from, i.e., coins[first…last] @param value: the value to express @param K: the maximum # of coins to use @precondition: all coins have positive values @postcondition: return true of false depending on the checking result */ bool CoinChangeK (const vector & coins, int first, int last, int value, int K) Hint: call subsets( ) function to return all possible subsets, and then go through them to see if any subsets with size <=K as a sum equal to value or not. Code #include #include #include using namespace std; void PrintVector (const vector & v){ cout <<"["; for (auto e:v){ cout< & L, int first, int last, int value, vector & used) { if (value==0) { used.clear(); return true; } if (first>last) //no more coins to use { used.clear(); return false; } if (value<0) { used.clear(); return false; } //general case below vector used1; bool ret1= CoinChange (L, first, last-1, value-L[last], used1); if (ret1) // used1 include all values from L[first...last-1] that add up to valeu-L[last] used1.push_back (L[last]); //now: used1 include all coins used from L[first...last} that add up to value vector used2; // If not using L[last]... bool ret2 = CoinChange (L, first, last-1, value, used2); if (ret1 && ret2) { if (used1.size() > used2.size()) used = used2; else used = used1; return true; } else if (ret1) { used = used1; return true; } else if (ret2){ used = used2; return true; } else { used.clear(); return false; } } bool CoinChangeK (const vector & coins, int first, int last, int value, int K) { return true; } bool UnlimitedCoinChange (const vector & coins, int value, vector& bestSolution) { return true; } int main() { vector coins{2,5,3,10}; vector used; vector values{4, 6,15, 18, 30, 41}; //use this to test //This part demo the CoinChange function: optimization problem /* for (auto v: values) { //Todo: replace CoinChange with your CoinChangeUnlimited function... if (CoinChange (coins, 0, coins.size()-1, v, used)) { cout <<"value="<> command; if (command=="coinchangek"){ //we cannot make 20 using 2 or fewer coins if (CoinChangeK (coins, 0, coins.size()-1, 20, 2)!=false || CoinChangeK (coins, 0, coins.size()-1, 5, 1)!=true) { cout <<"fail coinchangek tests\n"; return 1; //faild coinchangeK test } else{ cout <<"pass coinchangek tests\n"; return 0; //pass coinchangeK test } } else if (command=="unlimited"){ //Test UnlimitedCoinChange vector bestSolution; if (UnlimitedCoinChange (coins, 1,bestSolution)!=false) { cout <<"Failed UnlimitedCoinChange case 1\n"; return 1; //failed unlimited test } if (UnlimitedCoinChange (coins, 15, bestSolution)!=true) { cout <<"Failed UnlimitedCoinChange case 2\n"; return 1; } vector expectedSol{5,10}; sort (bestSolution.begin(), bestSolution.end()); if (bestSolution!=expectedSol){ cout <<"Failed UnlimitedCoinChange case 2\n"; return 1; } if (UnlimitedCoinChange (coins, 30, bestSolution)!=true) { cout <<"Failed UnlimitedCoinChange case 3\n"; return 1; } vector expectedSol3{10,10,10}; sort (bestSolution.begin(), bestSolution.end()); if (bestSolution!=expectedSol3){ cout <<"Failed UnlimitedCoinChange case 3\n"; return 1; } cout <<"Pass unlimitedCoinChange cases\n"; return 0; } }
C++ please help I will give you a good rating!!!!!
Implement the following function by using subsets() developed in step 1.
/* Check if given value given be expressed by K or less coins chosen from the given set of coins
@param coins: all coins we can choose from
@param first, last: specify the range of coins to choose from, i.e., coins[first…last]
@param value: the value to express
@param K: the maximum # of coins to use
@precondition: all coins have positive values
@postcondition: return true of false depending on the checking result */
bool CoinChangeK (const
Hint: call subsets( ) function to return all possible subsets, and then go through them to see if any subsets with size <=K as a sum equal to value or not.
Code
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
void PrintVector (const vector<int> & v){
cout <<"[";
for (auto e:v){
cout<<e<<" ";
}
cout <<"]";
}
/* check if we can use values in L[left...right] to make a sum of value, and find
the best solution, i.e., smallest set of coins tht make this value
@param L, first, last: specify a sub-vector where coins/values are chosen from
@param value: the sum/value we want to make
@pre-condition: all parameters are initialized, L[] and value are non-negative
@post-condition: return true/false depending on the checking result, if return true,
used vector contains coins that make up the value, with the minimul # of elements from
L [first...last]
*/
bool CoinChange (vector<int> & L, int first, int last, int value, vector<int> & used)
{
if (value==0)
{
used.clear();
return true;
}
if (first>last) //no more coins to use
{
used.clear();
return false;
}
if (value<0)
{
used.clear();
return false;
}
//general case below
vector<int> used1;
bool ret1= CoinChange (L, first, last-1, value-L[last], used1);
if (ret1)
// used1 include all values from L[first...last-1] that add up to valeu-L[last]
used1.push_back (L[last]);
//now: used1 include all coins used from L[first...last} that add up to value
vector<int> used2;
// If not using L[last]...
bool ret2 = CoinChange (L, first, last-1, value, used2);
if (ret1 && ret2) {
if (used1.size() > used2.size())
used = used2;
else
used = used1;
return true;
} else if (ret1) {
used = used1;
return true;
} else if (ret2){
used = used2;
return true;
} else {
used.clear();
return false;
}
}
bool CoinChangeK (const vector<int> & coins, int first, int last, int value, int K)
{
return true;
}
bool UnlimitedCoinChange (const vector<int> & coins, int value, vector<int>& bestSolution)
{
return true;
}
int main()
{
vector<int> coins{2,5,3,10};
vector<int> used;
vector<int> values{4, 6,15, 18, 30, 41}; //use this to test
//This part demo the CoinChange function: optimization problem
/*
for (auto v: values) {
//Todo: replace CoinChange with your CoinChangeUnlimited function...
if (CoinChange (coins, 0, coins.size()-1, v, used))
{
cout <<"value="<<v <<" True\n";
//display used vector
for (int i=0;i<used.size();i++)
cout <<used[i]<<" ";
cout<<endl;
}
else
cout <<"Value=" << v<<" False"<<endl;
}
*/
//Test CoinChangeK
cout <<"Enter coinchangek or unlimited to test the corresponding function:";
string command;
cin >> command;
if (command=="coinchangek"){
//we cannot make 20 using 2 or fewer coins
if (CoinChangeK (coins, 0, coins.size()-1, 20, 2)!=false ||
CoinChangeK (coins, 0, coins.size()-1, 5, 1)!=true)
{
cout <<"fail coinchangek tests\n";
return 1; //faild coinchangeK test
}
else{
cout <<"pass coinchangek tests\n";
return 0; //pass coinchangeK test
}
} else if (command=="unlimited"){
//Test UnlimitedCoinChange
vector<int> bestSolution;
if (UnlimitedCoinChange (coins, 1,bestSolution)!=false) {
cout <<"Failed UnlimitedCoinChange case 1\n";
return 1; //failed unlimited test
}
if (UnlimitedCoinChange (coins, 15, bestSolution)!=true) {
cout <<"Failed UnlimitedCoinChange case 2\n";
return 1;
}
vector<int> expectedSol{5,10};
sort (bestSolution.begin(), bestSolution.end());
if (bestSolution!=expectedSol){
cout <<"Failed UnlimitedCoinChange case 2\n";
return 1;
}
if (UnlimitedCoinChange (coins, 30, bestSolution)!=true) {
cout <<"Failed UnlimitedCoinChange case 3\n";
return 1;
}
vector<int> expectedSol3{10,10,10};
sort (bestSolution.begin(), bestSolution.end());
if (bestSolution!=expectedSol3){
cout <<"Failed UnlimitedCoinChange case 3\n";
return 1;
}
cout <<"Pass unlimitedCoinChange cases\n";
return 0;
}
}
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