Coloring Array Elements with Equal Parity Sums: C++ Dynamic Programming Solution

{/'title/':/'Coloring Array Elements with Equal Parity Sums: C++ Dynamic Programming Solution/',/'description/':/'This article explores a C++ solution for determining if you can color array elements in two colors such that the sums of each color have the same parity. The code utilizes dynamic programming to efficiently solve this problem./',/'keywords/':/'array, coloring, dynamic programming, c++, parity, equal sums, algorithm, programming/',/'content/':/'You are given an array consisting of n integers. Your task is to determine whether it is possible to color all its elements in two colors in such a way that the sums of the elements of both colors have the same parity and each color has at least one element colored.//n//nFor example, if the array is [1,2,4,3,2,3,5,4], we can color it as follows: [1,2,4,3,2,3,5,4], where the sum of the blue elements is 6 and the sum of the red elements is 18.//n//nc++代码内容：Here is a C++ code that solves the problem://n//ncpp//n#include <iostream>//n#include <vector>//n//nbool canColorArray(std::vector<int>& nums) {//n int sum = 0;//n for (int num : nums) {//n sum += num;//n }//n //n // If the sum is odd, it is not possible to color the array//n if (sum % 2 != 0) {//n return false;//n }//n //n int n = nums.size();//n std::vector<std::vector<bool>> dp(n+1, std::vector<bool>(sum/2+1, false));//n //n // Base cases//n dp[0][0] = true;//n for (int i = 1; i <= n; i++) {//n dp[i][0] = true;//n }//n //n for (int i = 1; i <= n; i++) {//n for (int j = 1; j <= sum/2; j++) {//n dp[i][j] = dp[i-1][j];//n if (j >= nums[i-1]) {//n dp[i][j] = dp[i][j] || dp[i-1][j-nums[i-1]];//n }//n }//n }//n //n return dp[n][sum/2];//n}//n//nint main() {//n std::vector<int> nums = {1, 2, 4, 3, 2, 3, 5, 4};//n //n if (canColorArray(nums)) {//n std::cout << /'It is possible to color the array/' << std::endl;//n } else {//n std::cout << /'It is not possible to color the array/' << std::endl;//n }//n //n return 0;//n}//n//n//nThe code uses dynamic programming to solve the problem. It creates a 2D boolean array dp where dp[i][j] represents whether it is possible to get a sum of j using the first i elements of the array. The base cases are dp[0][0] = true and dp[i][0] = true for all i. //n//nThen, it iterates through the array and for each element, it checks if it is possible to get the current sum j by including or excluding the current element. The final result is dp[n][sum/2], where n is the size of the array and sum is the total sum of the array. //n//nIf dp[n][sum/2] is true, it means it is possible to color the array in the desired way. Otherwise, it is not possible./