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文章目录Fibonacci数列Ackerman函数排列问题整数划分问题总结Fibonacci数列public static int fibonacci(int n){ if(n <= 1) { return 1; } return fibonacci(n-1)+fibonacci(n-2);}Ackerman函数排列问题public static void perm(Object [] list,int k,int m){ if(k == m) { for.