1.给数组做反序 public class Ak01 { public static void main(String[] args) { int[] a = new int[]{22,48,41,2,7,9}; int start=0; int end=a.length-1; int size = a.length; for(int i = 0;i<size/2;i++) { int temp; temp = a[start]; a[start]=a[end]; a[end]=temp; st…
鉴于时间紧张...虽然知道博弈是个大课题但是花一个上午时间已经极限了... 希望省选过后再回过头来好好总结一遍吧. 接下来为了看着顺眼一点...还是按照难度顺序吧 POJ1082 一道最简单的博弈题,只需要DP就可以过. 在这道题里我尽情展示了多函数多过程的代码风格.. program poj1082; ..=[,,,,,,]; var n,i,x,y,z:longint; f:array[..,..,..]of boolean; function leap(x:longint):boole…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4772 题面: Zhuge Liang's Password Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1404 Accepted Submission(s): 926 Problem Description In the anc…