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chuanjiaoshi
- C/C++求解野人与传教士问题。自定义野人和传教士人数。-C / C for savage problems with the missionaries. Since the definition of savage and the number of missionaries.
chuanjiaoshi
- C/C++求解野人与传教士问题。自定义野人和传教士人数。-C/C for savage problems with the missionaries. Since the definition of savage and the number of missionaries.
chuanjiaoshi
- 一个传教士与野人的程序，希望大家多多指教，-A missionary with the wild man of the procedure, hope that the exhibitions, I would like to thank
chuanjiaoshi
- 有N个传教士和N个野人要过河，现在有一条船只能承载K个人（包括野人），K<N，在任何时刻，如果有野人和传教士在一起，必须要求传教士的人数多于或等于野人的人数。-N and N have a savage missionary to cross the river, there is now a boat can only hold K individuals (including Savage), K <N，在任何时刻，如果
chuanjiaoshi
- 1. 野人与传教士过河问题：有传教士和野人各三人需要过河，他们都会划船。现只有一条船，一次只能载两人。假设野人多于传教士时传教士就会被吃掉。编程求出所有能确保全部安全的过河的计划，并给出Visual Prolog程序-1. Savage and missionaries across the river problem: There are three missionaries and savages the need to cross
chuanjiaoshi
- 有3个传教士和3个野人要过河，只有一艘船，这艘船每次只能载2个人过河，且无论哪边野人的数量大于传教士的数量时，野人就会吃掉传教士。怎样让他们都安全过河？-Three missionaries and three wild man to cross the river, only one ship, the ship can only set two individuals across the river, and regardless
chuanjiaoshi
- 三个传教士与三个野人渡河，只有传教士会划船，野人在岸上多于传教士会吃了传教士，一条船，只能带两个人，，用c语言实现野人与传教士问题-Three missionaries and three savage cross the river, only missionaries will be boating, Savage ashore than missionaries will eat the missionaries, a boat,
chuanjiaoshi
- 图论中的搜索算法，传教士过河问题，找出最佳搜索路径-Graph theory search algorithm, missionaries across the river, to find the best search path
chuanjiaoshi
- 传教士 自动加血补血 用于魔力宝贝 私服 官服-Magic baby Hang up the detailed code