2019年11月5日
Codeforces Round #587 (Div. 3)
摘要： __ "Codeforces Round 587 (Div. 3)" __ __ "A. Prefixes" __ __思路：水题__ AC代码 阅读全文
posted @ 2019-11-05 20:50 Misuchii 阅读(82) 评论(0) 推荐(0) 编辑
Codeforces Round #589 (Div. 2)
摘要： __ "Codeforces Round 589 (Div. 2)" __ __ "A. Distinct Digits" __ __思路：水题__ AC代码 阅读全文
posted @ 2019-11-05 20:49 Misuchii 阅读(139) 评论(0) 推荐(0) 编辑
Codeforces Round #588 (Div. 2)
摘要： __ "Codeforces Round 588 (Div. 2)" __ __ "A. Dawid and Bags of Candies" __ __思路：水题__ AC代码 阅读全文
posted @ 2019-11-05 20:49 Misuchii 阅读(101) 评论(0) 推荐(0) 编辑
Codeforces Round #590 (Div. 3)
摘要： __ "Codeforces Round 590 (Div. 3)" __ __ "A. Equalize Prices Again" __ __思路：水题__ AC代码 阅读全文
posted @ 2019-11-05 20:48 Misuchii 阅读(129) 评论(0) 推荐(0) 编辑
2019年10月30日
Codeforces Round #591 (Div. 2, based on Technocup 2020 Elimination Round 1)
摘要： __ "Codeforces Round 591 (Div. 2, based on Technocup 2020 Elimination Round 1)" __ __ "A. CME" __ __思路：n为偶数答案为0 n为奇数答案为1 特判下n为2时答案为2__ AC代码 cpp includ 阅读全文
posted @ 2019-10-30 00:45 Misuchii 阅读(157) 评论(0) 推荐(0) 编辑
Educational Codeforces Round 74 (Rated for Div. 2)
摘要： __ "Educational Codeforces Round 74 (Rated for Div. 2)" __ __ "A. Prime Subtraction" __ __思路：任何大于等于二的数都可以由二和三构成__ AC代码 cpp include include include inc 阅读全文
posted @ 2019-10-30 00:45 Misuchii 阅读(110) 评论(0) 推荐(0) 编辑
AtCoder Beginner Contest 144
摘要： __ "AtCoder Beginner Contest 144" __ __ "A 9x9" __ __思路：水题__ AC代码 __ "E Gluttony" __ __思路：二分 上界为最大的最小$a$$ $最大$f$__ AC代码 cpp include include include in 阅读全文
posted @ 2019-10-30 00:44 Misuchii 阅读(145) 评论(0) 推荐(0) 编辑
Codeforces Round #592 (Div. 2)
摘要： __ "Codeforces Round 592 (Div. 2)" __ __ "A. Pens and Pencils" __ __思路：水题__ AC代码 cpp include include include include include include include include i 阅读全文
posted @ 2019-10-30 00:42 Misuchii 阅读(130) 评论(0) 推荐(0) 编辑
2019年10月27日
Educational Codeforces Round 75 (Rated for Div. 2)
摘要： __ "Educational Codeforces Round 75 (Rated for Div. 2)" __ __ "A. Broken Keyboard" __ __思路：最开始没搞清楚题意瞎搞 只需要判断前后两连续字符是否相同 不同则将前一个存入vector中 排序去个重输出即可__ A 阅读全文
posted @ 2019-10-27 15:49 Misuchii 阅读(190) 评论(0) 推荐(0) 编辑
Codeforces Global Round 5
摘要： __ "Codeforces Global Round 5" __ __ "A. Balanced Rating Changes" __ __思路：$n$为偶数 则这n个数里面奇数的个数一定为偶数 那么就每次令一个答案为$\frac{a_i}{2} 1$ 另一个答案为$\frac{a_i}{2} + 阅读全文
posted @ 2019-10-27 15:04 Misuchii 阅读(180) 评论(0) 推荐(0) 编辑