CodeForces - 600C Make Palindrome 贪心】的更多相关文章

A string is called palindrome if it reads the same from left to right and from right to left. For example "kazak", "oo", "r" and "mikhailrubinchikkihcniburliahkim" are palindroms, but strings "abb" and &qu…
要保证变化次数最少就是出现次数为奇数的相互转化,而且对应字母只改变一次.保证字典序小就是字典序大的字母变成字典序小的字母. 长度n为偶数时候,次数为奇数的有偶数个,按照上面说的搞就好了. n为奇数时,要考虑最后中间那个字母.交换法可以证明,其实是贪心最后没有转化掉的字母. #include<bits/stdc++.h> using namespace std; typedef long long ll; ; char s[LEN]; ]; //#define LOCAL int main()…
codeforces 704B - Ant Man 贪心 题意:n个点,每个点有5个值,每次从一个点跳到另一个点,向左跳:abs(b.x-a.x)+a.ll+b.rr 向右跳:abs(b.x-a.x)+a.lr+b.rl,遍历完所有的点,问你最后的花费是多少 思路:每次选一个点的时候,在当前确定的每个点比较一下,选最短的距离. 为什么可以贪心?应为答案唯一,那么路径必定是唯一的,每个点所在的位置也一定是最短的. #include <bits/stdc++.h> using namespace…
CodeForces - 50A Domino piling (贪心+递归) 题意分析 奇数*偶数=偶数,如果两个都为奇数,最小的奇数-1递归求解,知道两个数都为1,返回0. 代码 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <sstream> #include <set> #include <map…
C. Make Palindrome Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/600/problem/C Description A string is called palindrome if it reads the same from left to right and from right to left. For example "kazak", "oo&q…
题目链接:http://codeforces.com/contest/600/problem/C C. Make Palindrome time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A string is called palindrome if it reads the same from left to right an…
A string is called palindrome if it reads the same from left to right and from right to left. For example "kazak", "oo", "r" and "mikhailrubinchikkihcniburliahkim" are palindroms, but strings "abb" and &qu…
C. Palindrome Transformation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningl…
[链接] 我是链接,点我呀:) [题意] 题意 [题解] 计算出来每个字母出现的次数. 把字典序大的奇数出现次数的字母换成字典序小的奇数出现次数的字母贪心即可. 注意只有一个字母的情况 然后贪心地把字典序小的字母放在前面就好 [代码] #include <bits/stdc++.h> #define rep1(i,a,b) for (int i = a;i <= b;i++) #define rep2(i,a,b) for (int i = a;i >= b;i--) #defin…
题意:给定k个长度为n的字符串,每个字符串有一个魅力值ai,在k个字符串中选取字符串组成回文串,使得组成的回文串魅力值最大. 分析: 1.若某字符串不是回文串a,但有与之对称的串b,将串a和串b所有的魅力值分别从大到小排序后,若两者之和大于0,则可以放在回文串的两边. 2.若某字符串是回文串,将其魅力值从大到小排序后,两两依次分析:(mid---可能放在回文串中间的串的最大魅力值) (1)若两个数都是正的,那么就将其放在两边,并将结果计入ans.(ans---回文串两边的串的魅力值之和) (2)…