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随笔分类 -  容斥原理

HDU 1796 容斥原理
摘要:How many integers can you find Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7439 Accepted Sub 阅读全文
posted @ 2016-10-30 18:59 半根毛线 阅读(446) 评论(0) 推荐(0)
Codeforces Round #330 (Div. 2) B 容斥原理
摘要:B. Pasha and Phone time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output B. Pasha and Phone tim 阅读全文
posted @ 2016-07-20 20:52 半根毛线 阅读(227) 评论(0) 推荐(0)
Codeforces Round #345 (Div. 2) C (multiset+pair )
摘要:C. Watchmen time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Watchmen are in a danger and 阅读全文
posted @ 2016-03-08 18:58 半根毛线 阅读(243) 评论(0) 推荐(0)
HDU2841 (队列容斥)
摘要:Visible TreesTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2213Accepted Submission(s): 908 Prob... 阅读全文
posted @ 2015-12-01 00:12 半根毛线 阅读(166) 评论(0) 推荐(0)
15ecjtu校赛1006 (dfs容斥)
摘要:Problem Description在平面上有一个n*n的网格,即有n条平行于x轴的直线和n条平行于y轴的直线,形 成了n*n个交点(a,b)(1using namespace std;int que[100010][20];int jishu[100010];void Init(){ me... 阅读全文
posted @ 2015-12-01 00:09 半根毛线 阅读(303) 评论(0) 推荐(0)

 
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