随笔分类 -  数学 容斥

hdu 6169 Senior PanⅡ Miller_Rabin素数测试+容斥
摘要：Senior PanⅡ Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others) Problem Description Senior Pan had just failed in his m 阅读全文
posted @ 2017-08-22 17:16 jhz033 阅读(271) 评论(2) 推荐(0)
hdu 5768 Lucky7 中国剩余定理+容斥+快速乘
摘要：Lucky7 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Description When ?? was born, seven crows flew in and 阅读全文
posted @ 2016-07-29 14:39 jhz033 阅读(266) 评论(0) 推荐(0)
Educational Codeforces Round 13 A、B、C、D
摘要：A. Johny Likes Numbers time limit per test 0.5 seconds memory limit per test 256 megabytes input standard input output standard output A. Johny Likes 阅读全文
posted @ 2016-06-21 15:40 jhz033 阅读(203) 评论(0) 推荐(0)
hdu 2204 Eddy's爱好 容斥原理
摘要：Eddy's爱好 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description Ignatius 喜欢收集蝴蝶标本和邮票，但是Eddy的爱好很特别，他对数字比较 阅读全文
posted @ 2016-05-24 16:56 jhz033 阅读(267) 评论(0) 推荐(0)
xtu 1242 Yada Number 容斥原理
摘要：Yada Number Problem Description: Every positive integer can be expressed by multiplication of prime integers. Duoxida says an integer is a yada number 阅读全文
posted @ 2016-05-24 16:05 jhz033 阅读(164) 评论(0) 推荐(0)
codeforces 300 div2 B.Pasha and Phone 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 @ 2016-05-24 14:04 jhz033 阅读(175) 评论(0) 推荐(0)
hdu 2841 Visible Trees 容斥原理
摘要：Visible Trees Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description There are many trees forming a m * 阅读全文
posted @ 2016-05-21 20:11 jhz033 阅读(131) 评论(0) 推荐(0)
hdu 4135 Co-prime 容斥原理
摘要：Co-prime Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description Given a number N, you are asked to count 阅读全文
posted @ 2016-05-11 11:01 jhz033 阅读(166) 评论(0) 推荐(0)
hdu 1796 How many integers can you find 容斥定理
摘要：How many integers can you find Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Problem Description Now you get a num 阅读全文
posted @ 2016-05-11 00:05 jhz033 阅读(171) 评论(0) 推荐(0)
广州工业大学2016校赛 F 我是好人4 dfs+容斥
摘要：Problem F: 我是好人4 Description 众所周知，我是好人！所以不会出太难的题，题意很简单 给你n个数，问你1000000000（含1e9）以内有多少个正整数不是这n个数任意一个的倍数 最后友情提供解题代码（我真是太好人了） void solve(int p[], int n) { 阅读全文
posted @ 2016-04-11 21:34 jhz033 阅读(191) 评论(0) 推荐(0)