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+// 1140. Stone Game II
+#include "leetcode.h"
+
+/*
+ * alice and bob continue their games with piles of stones. there are a number
+ * of piles arranged in a row, and each pile has a positive integer number of
+ * stones 'piles[i]'. the objective of the game is to end with the most stones.
+ * alice and bob take turns, with alice starting first. initiall, 'M = 1'. on
+ * each player's turn, that player can take all the stones on the 'X' remaining
+ * piles, where '1 <= X <= 2M'. then, we set 'M = max(M, X)'. the game continues
+ * until all the stones have been take. assuming alice and bob play optimally,
+ * return maximum stones alice can get.
+ */
+
+class Solution {
+public:
+ int stoneGameII(vector<int> &piles) {
+ int n = piles.size();
+ vvd(int) dp(n + 1, vector<int>(n + 1, 0));
+ vector<int> sum(n + 1, 0);
+ for (int i = n - 1; i >= 0; i--)
+ sum[i] = sum[i + 1] + piles[i];
+ for (int i = 0; i <= n; i++)
+ dp[i][n] = sum[i];
+ for (int i = n - 1; i >= 0; i--)
+ for (int j = n - 1; j >= 1; j--)
+ for (int k = 1; k <= 2 * j && i + j <= n; j++)
+ dp[i][j] = max(dp[i][j], sum[i] - dp[i + k][max(j, k)]);
+ return dp[0][1];
+ }
+};
+
+int main() {
+ Solution obj;
+ vector<int> p1 = {2, 7, 9, 4, 4};
+ vector<int> p2 = {1, 2, 3, 4, 5, 100};
+ printf("%d\n", obj.stoneGameII(p1)); // expect: 10
+ printf("%d\n", obj.stoneGameII(p2)); // expect: 104
+}