Here are four complete C++ programs, each taking input from the user.

1. Coin Changing (Dynamic Programming)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cout << "Enter number of coin denominations: ";
    cin >> n;

    vector<int> coins(n);
    cout << "Enter the coin denominations: ";
    for (int i = 0; i < n; i++) cin >> coins[i];

    int amount;
    cout << "Enter the target amount: ";
    cin >> amount;

    vector<int> dp(amount + 1, INT_MAX);
    vector<int> lastCoin(amount + 1, -1);
    dp[0] = 0;

    for (int i = 1; i <= amount; i++) {
        for (int j = 0; j < n; j++) {
            if (coins[j] <= i && dp[i - coins[j]] != INT_MAX) {
                if (dp[i - coins[j]] + 1 < dp[i]) {
                    dp[i] = dp[i - coins[j]] + 1;
                    lastCoin[i] = coins[j];
                }
            }
        }
    }

    if (dp[amount] == INT_MAX) {
        cout << "It is not possible to form the amount with given coins.\n";
    } else {
        cout << "Minimum number of coins required: " << dp[amount] << "\n";
        // Reconstruct the coins used
        vector<int> used;
        int cur = amount;
        while (cur > 0) {
            used.push_back(lastCoin[cur]);
            cur -= lastCoin[cur];
        }
        cout << "Coins used: ";
        for (int c : used) cout << c << " ";
        cout << "\n";
    }

    return 0;
}

2. Coin Changing (Greedy Approach)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cout << "Enter number of coin denominations: ";
    cin >> n;

    vector<int> coins(n);
    cout << "Enter the coin denominations: ";
    for (int i = 0; i < n; i++) cin >> coins[i];

    int amount;
    cout << "Enter the target amount: ";
    cin >> amount;

    // Sort coins in descending order for greedy selection
    sort(coins.rbegin(), coins.rend());

    vector<int> result;
    int remaining = amount;

    for (int i = 0; i < n && remaining > 0; i++) {
        while (coins[i] <= remaining) {
            remaining -= coins[i];
            result.push_back(coins[i]);
        }
    }

    if (remaining != 0) {
        cout << "Greedy approach could not form the exact amount ";
        cout << "(this coin system may not be canonical).\n";
    } else {
        cout << "Number of coins used: " << result.size() << "\n";
        cout << "Coins used: ";
        for (int c : result) cout << c << " ";
        cout << "\n";
    }

    return 0;
}

3. Fractional Knapsack (Greedy Approach)

#include <bits/stdc++.h>
using namespace std;

struct Item {
    double weight;
    double value;
    double ratio;
};

int main() {
    int n;
    cout << "Enter number of items: ";
    cin >> n;

    vector<Item> items(n);
    for (int i = 0; i < n; i++) {
        cout << "Enter weight and value of item " << i + 1 << ": ";
        cin >> items[i].weight >> items[i].value;
        items[i].ratio = items[i].value / items[i].weight;
    }

    double capacity;
    cout << "Enter the knapsack capacity: ";
    cin >> capacity;

    // Sort items by value/weight ratio in descending order
    sort(items.begin(), items.end(), [](const Item &a, const Item &b) {
        return a.ratio > b.ratio;
    });

    double totalValue = 0.0;
    double remainingCapacity = capacity;

    cout << "\nItems selected (fraction taken):\n";
    for (int i = 0; i < n && remainingCapacity > 0; i++) {
        if (items[i].weight <= remainingCapacity) {
            // Take the whole item
            remainingCapacity -= items[i].weight;
            totalValue += items[i].value;
            cout << "Item (w=" << items[i].weight << ", v=" << items[i].value
                 << ") -> Taken fully (100%)\n";
        } else {
            // Take fraction of the item
            double fraction = remainingCapacity / items[i].weight;
            totalValue += items[i].value * fraction;
            cout << "Item (w=" << items[i].weight << ", v=" << items[i].value
                 << ") -> Taken " << fraction * 100 << "%\n";
            remainingCapacity = 0;
        }
    }

    cout << fixed << setprecision(2);
    cout << "\nMaximum value obtainable: " << totalValue << "\n";

    return 0;
}

4. Bin Packing (First Fit Approach)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cout << "Enter number of items: ";
    cin >> n;

    vector<int> items(n);
    cout << "Enter sizes of the items: ";
    for (int i = 0; i < n; i++) cin >> items[i];

    int binCapacity;
    cout << "Enter bin capacity: ";
    cin >> binCapacity;

    vector<int> binRemaining;   // remaining space in each used bin
    vector<vector<int>> binContents;

    for (int i = 0; i < n; i++) {
        bool placed = false;

        // Try to place item in the first bin that has enough space
        for (int j = 0; j < (int)binRemaining.size(); j++) {
            if (binRemaining[j] >= items[i]) {
                binRemaining[j] -= items[i];
                binContents[j].push_back(items[i]);
                placed = true;
                break;
            }
        }

        // If it doesn't fit in any existing bin, open a new bin
        if (!placed) {
            binRemaining.push_back(binCapacity - items[i]);
            binContents.push_back({items[i]});
        }
    }

    cout << "\nTotal bins used: " << binContents.size() << "\n";
    for (int j = 0; j < (int)binContents.size(); j++) {
        cout << "Bin " << j + 1 << ": ";
        int sum = 0;
        for (int item : binContents[j]) {
            cout << item << " ";
            sum += item;
        }
        cout << "(used " << sum << "/" << binCapacity << ")\n";
    }

    return 0;
}

Notes

• Coin Changing (DP) always finds the optimal (minimum) number of coins, even for non-canonical coin systems.
• Coin Changing (Greedy) is faster but can fail on non-standard denominations (e.g., coins {1, 3, 4} for amount 6).
• Fractional Knapsack sorts by value/weight ratio and allows partial items, so greedy is provably optimal here.
• Bin Packing uses the First Fit heuristic — simple and reasonably good, but not always optimal (bin packing is NP-hard in general).