Search - 4.2.3-Win

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[Data structs] 3_15 DL : 0: 问题的提出：赌博游戏世界无所不在，其中“掷双骰”游戏家喻户晓，其游戏规则如下：每次掷两个骰子，每个骰子的6面上分别标有1、2、 3、4、5、6，两个骰子停止滚动后，计算其向上的点数之和。假如首次掷出后点数之和为7和11，那么玩家赢（庄家输）。假如首次掷出后点数之和为2、3或12，那么玩家输（庄家赢）。假如首次掷出后点数之和为4、5、6、8、9和10，那么这些数字会立即成为玩家的“目标点” （即下次如遇到此点数为赢）。要想赢，必须不断地掷两个骰子，直到点数与目标点数相等为止，但在这之前，假如不幸地掷出7点，那么玩家马上就输，试编程模拟此游戏。-the problem : the world of gambling games omnipresent, "throwing two dice" game known its rules of the game are as follows : two each throwing dice, each of the six dice surface were labeled with 1, 2, 3, 4, 5, 6, two dice stopped rolling, the calculation of its progressive and Numbers. If the points after the first allocate the sum of 7 and 11, then the player wins (Makers T). If the points after the first allocate the sum of two, three or 12, then the players lose (Makers Win). If the points after the first allocate the sum of 4, 6, 8 and 9 and 10, So these figures will immediately become the player of the "target spots" (that is, the next case of this points to win). To win, to be constantly throwing two dice, until Decimal points with the same goal, but before this,
Date : 2026-01-07 Size : 1kb User : 萝卜
[Data structs] stone_game_dp DL : 0: 现有4堆石子,两个人轮流取石子,他们有n种可能的取法,取法表示从第1堆取A1个石子,从第2堆中取B1个,第3堆取C1个,第4堆取D1个.如果取的时候某一堆的石子数比所要取的石子数要少,则这种取法是不可行的.取到最后没有可行取法的人就算输了. 现给出4堆石子的石子数目以及n种取法,请问如果两个人都采用最优取法,先取的人是赢还是输.-Four existing heap of stones, two people take turns stones, they have emulated n possible, take the A1 emulated said a stone from a pile of rubble taken from the second B1, 3rd stacker C1, 4th heap Take a D1 and if it' s a pile of stones take a few less than the number of stones to take it, then this is not feasible emulated Take to the last man, even if there is no feasible emulated lost. heap of stones now gives 4 n kinds of stones as well as the number of emulated, what if two people are using the optimal emulated, first take the people win or lose.
Date : 2026-01-07 Size : 1kb User : lean