
ACM - 分治
文章平均质量分 82
Freenm
这个作者很懒,什么都没留下…
展开
-
POJ 1854 - Evil Straw Warts Live
DescriptionA palindrome is a string of symbols that is equal to itself when reversed. Given an input string, not necessarily a palindrome, compute the number of swaps necessary to transform the st原创 2017-02-27 22:53:16 · 745 阅读 · 0 评论 -
POJ 2388 - Who's in the Middle
排序的水题,主要拿来测试了一下快速排序和归并排序哪个速度快一点。#include#includeusing namespace std;void merge(int num[],int st,int md,int ed){ int l_len=md-st+1,r_len=ed-md; int l[l_len+2],r[r_len+2]; for(int i=1;原创 2017-04-04 10:26:26 · 387 阅读 · 0 评论