并且,只要有一种情况下Alex能赢,那么Alex就必赢,因为两人以最优策略取石堆且Alex先手。 boolstoneGame(vector<int>&piles) {intdp[505][505][2];inthead=0,tail=piles.size()-1; memset(dp,0,sizeof(dp));boolflag=0;for(inti=head; i<=tail; i++)for(intj=tail; j>=i; j--) {intlen=j...
Alex and Lee play a game with piles of stones. There are an even number of piles arranged in a row, and each pile has a positive integer number of stonespiles[i]. The objective of the game is to end with the most stones. The total number of stones is odd, so there are no ties....
其中rival = dp[i + k][max(k, j)] 时间复杂度为 O(n ^ 3), 空间复杂度为 O(n ^ 2) 代码: C++: class Solution{public:intstoneGameII(vector<int>&piles){intn=piles.size(),sum=0;vector<vector<int>>dp(n,vector<int>(n+1));for(inti=n-1;i>-1;i--){sum+=piles[i];for(int...
Leetcode 292 取石子博弈游戏(数学题) 4个石头,必输5个石头,拿一个,对手四个,必赢6个石头,拿两个,对手四个,必赢7个石头,拿三个,对手四个,必赢8个石头,不论拿多少个,对手总能拿到4,必输。 由此总结出规律,只要是4的倍数,那么必输,否则,就必赢。
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Can you solve this real interview question? Stone Game II - 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
We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the stones have weights x and y with x <= y. The result of this smash is: If x == y, both stones are destroyed, and ...
877. Stone Game class Solution { public: bool stoneGame(vector<int>& piles) { int l = piles...[ i ][ i ] = piles[i]; for( int d = 1 ; d < l ; d ++ ){ for( int i = 0 ; i < l - d ; i ++ ) dp leetcode 96. 不同的二叉搜索树 Solution(object): def numTre...
URAL 1180. Stone Game (博弈 + 规律)[通俗易懂] 大家好,又见面了,我是全栈君。 1180. Stone Game Time limit: 1.0 second Memory limit: 64 MB Two Nikifors play a funny game. There is a heap ofNstones in front of them. Both Nikifors in turns take some stones from the heap. One ...
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