请用动态规划法编程实现0-1背包问题。

好的，以下是0-1背包问题的动态规划代码：

#include <iostream>
#include <vector>

using namespace std;

int zeroOneKnapsack(vector<int>& weights, vector<int>& values, int maxWeight) {
    int n = weights.size();
    vector<vector<int>> dp(n + 1, vector<int>(maxWeight + 1, 0));
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= maxWeight; j++) {
            if (weights[i - 1] <= j) {
                dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weights[i - 1]] + values[i - 1]);
            } else {
                dp[i][j] = dp[i - 1][j];
            }
        }
    }
    return dp[n][maxWeight];
}

int main() {
    vector<int> weights = {1, 3, 4, 5};
    vector<int> values = {1, 4, 5, 7};
    int maxWeight = 7;
    int result = zeroOneKnapsack(weights, values, maxWeight);
    cout << result << endl;
    return 0;
}

这段代码实现了动态规划的解法，解决了0-1背包问题。