UVa 1339 题目大意:给定两个长度相同且不超过100个字符的字符串,判断能否把其中一个字符串重排后,然后对26个字母一一做一个映射,使得两个字符串相同 解题思路:字母可以重排,那么次序便不重要,可以分别统计两个字符串中的各个字母出现的次数,得到两个cnt[26]数组, 又由于可以进行映射,则可以直接对两个数组进行排序后判断是否相等(相当于原来相等的值的两个地方做映射) /* UVa 1339 Ancient Cipher --- 水题 */ #include <cstdio> #incl…

大意:读入两个字符串(都是大写字母),字符串中字母的顺序可以随便排列.现在希望有一种字母到字母的一一映射,从而使得一个字符串可以转换成另一个字符串(字母可以随便排列)有,输出YES:否,输出NO:exp: 输入HAHAHEHE 输出YES 可以将A→E,或E→A: 关键在于,问题简化:因为不用考虑字母位置,所以可以分别统计两个字符串中不同字母出现次数,计入数组counts1和counts2.那么,怎么知道谁和谁对应呢?这里面有个隐含条件:如果A→B,那么A在1数组出现的次数和B在2数组中出现的次…

题目链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4085 13855995 1339 Ancient Cipher Accepted C++ 0.012 2014-07-09 12:35:33 Ancient Cipher Ancient Roman empire had a strong government system wit…

1339 - Ancient Cipher Ancient Roman empire had a strong government system with various departments, including a secret service department. Important documents were sent between provinces and the capital in encrypted form to prevent eavesdropping. The…

1.链接地址: http://poj.org/problem?id=2159 http://bailian.openjudge.cn/practice/2159 2.题目: Ancient Cipher Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28064   Accepted: 9195 Description Ancient Roman empire had a strong government system…

这题就真的想刘汝佳说的那样,真的需要想象力,一开始还不明白一一映射是什么意思,到底是有顺序的映射?还是没顺序的映射? 答案是没顺序的映射,只要与26个字母一一映射就行 下面给出代码 //Uva1339 //Ancient Cipher //easy #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { freopen("Ancient.i…

Ancient Cipher Description Ancient Roman empire had a strong government system with various departments, including a secret service department. Important documents were sent between provinces and the capital in encrypted form to prevent eavesdropping…

POJ2159 Ancient Cipher Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38430   Accepted: 12515 Description Ancient Roman empire had a strong government system with various departments, including a secret service department. Important doc…

2017-08-31 20:11:39 writer:pprp 一开始说好这个是个水题,就按照水题的想法来看,唉~ 最后还是懵逼了,感觉太复杂了,一开始想要排序两串字符,然后移动之类的,但是看了看 好像没有什么规律... 然后就去膜大神code了 其实转换了一个思路,对两个字符串分别统计每个的个数, 然后分别排序,如果每个个数都可以对的上就说明可以通过两个操作得到 代码如下: /* @theme:poj 2159 ancient cipher @writer:pprp @declare: @be…

Ancient Cipher Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 36074   Accepted: 11765 Description Ancient Roman empire had a strong government system with various departments, including a secret service department. Important documents w…