Codeforces Global Round 19 (A-D题解)


源代码:ACM/OpenjudgeNow/Codeforces at master · abmcar/ACM (github.com)A. Sorting Parts题目大意:思路:如果前半段之中存在一个数大于后半段的其中一个数,则不可能完成排序可以使用一个前缀最大值和后缀最小值,枚举前半段的长度即

Codeforces Round #770 (Div. 2) (A-D题解)


源代码:ACM/OpenjudgeNow/Codeforces at master · abmcar/ACM (github.com)A. Reverse and Concatenate题目大意:思路:我们发现对于任意的字符串s,s+rev(s)为一个回文串,而对于一个回文串t,t+rev(t) =

Educational Codeforces Round 122 (Rated for Div. 2) (A-D题解)


源代码:ACM/OpenjudgeNow/Codeforces at master · abmcar/ACM (github.com)A. Div. 7题目大意:思路:对于任意一个数,在只更改最后一位的情况下,必定一种方式使得%7 == 0代码:void work(){ cin >>

Codeforces Round #769 (Div. 2) (A-D题解)


源代码:https://github.com/abmcar/ACM/tree/master/OpenjudgeNow/Codeforces/Codeforces%20Round%20%23769%20(Div.%202)A. ABC题目大意:思路:有这么几种情况字符串大小为1,此时无法获得长度大于1

Codeforces Round #768 (Div. 2) (A-D题解)


源代码:https://github.com/abmcar/ACM/tree/master/OpenjudgeNow/Codeforces/Codeforces%20Round%20%23768%20(Div.%202)A. Min Max Swap题目大意:思路:我们知道在两数之和相等的情况下,二

Educational Codeforces Round 121 (Rated for Div. 2) (A-C)题解


源代码:https://github.com/abmcar/ACM/tree/master/OpenjudgeNow/Codeforces/Educational%20Codeforces%20Round%20121%20(Rated%20for%20Div.%202)A. Equidistant

Codeforces Round #766 (Div. 2) (A-C) 题解


完整代码:https://github.com/abmcar/ACM/tree/master/OpenjudgeNow/Codeforces/Codeforces%20Round%20%23766%20(Div.%202)A. Not Shading题目大意:给你一个n*m的板子,每个格子分为黑色和

Codeforces Round #764 (Div. 3) (A-D)题解


完整代码:https://github.com/abmcar/ACM/tree/master/OpenjudgeNow/Codeforces/Codeforces%20Round%20%23764%20(Div.%203)A. Plus One on the Subset题目大意:给你一个数组,你可

Codeforces Round #765 (Div. 2) (A-C)题解


完整代码:https://github.com/abmcar/ACM/tree/master/OpenjudgeNow/Codeforces/Codeforces%20Round%20%23765%20(Div.%202)A. Ancient Civilization题目大意:给你n个二进制下长度小

Codeforces Round #761 (Div. 2) (A-D1) 题解


完整代码:https://github.com/abmcar/ACM/tree/master/OpenjudgeNow/Codeforces/Codeforces%20Round%20%23761%20(Div.%202)A. Forbidden Subsequence题目大意:给你一个字符串s和t