hdu2899 ;<em>水</em>提,直接<em>三</em>分,其实求导后二分也可以。
https://www.u72.net/daima/8hh5.html - 2024-07-26 02:26:47 - 代码库<em>水</em>。。
https://www.u72.net/daima/405m.html - 2024-07-22 10:29:04 - 代码库(甲)自<em>三</em>峡七百里中,两岸连山,略无阙处;重岩叠嶂,隐天蔽日,自非亭午夜分不见曦月。至于夏<em>水</em>襄陵,沿溯阻绝。
https://www.u72.net/daima/2ckb.html - 2024-09-01 03:08:41 - 代码库分析:举个例子xxx(<em>三</em>位数)为魔力数,则xxx|(xxx&#43;1000*y),那么xxx|1000,这个就是结论同理:四位数xxxx|10000,五位数
https://www.u72.net/daima/nhkfe.html - 2024-08-02 12:40:10 - 代码库传说,Tango有一大“<em>水</em>王”,他不
https://www.u72.net/daima/40w4.html - 2024-07-22 10:10:24 - 代码库【分析】:还是刷<em>水</em>题心情好点~不过,不知为什么我的代码,G&#43;&#43;过了,而C&#43;&#43;却CE。// 748K
https://www.u72.net/daima/3a2r.html - 2024-07-20 19:31:25 - 代码库Description Yesterday your dear cousin Coach Pang gave you a new 100MB hard disk drive (HDD) as a gift because you will get married next ye
https://www.u72.net/daima/6kr.html - 2024-07-03 02:09:05 - 代码库http://acm.xidian.edu.cn/problem.php?id=1081 判断最大值和最小值。如果用二分的那种判断,会超long long。 #include<iostream>#include<cstri
https://www.u72.net/daima/cn76.html - 2024-08-17 10:23:44 - 代码库http://acm.xidian.edu.cn/problem.php?id=1094 细心过了样例就差不多了。 #include<iostream>#include<cstdio>#include<cstring>#include<al
https://www.u72.net/daima/ferb.html - 2024-08-17 07:43:23 - 代码库http://acm.xidian.edu.cn/problem.php?id=1041 简单规律。 #include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using
https://www.u72.net/daima/f0wn.html - 2024-08-17 00:13:24 - 代码库http://acm.xidian.edu.cn/problem.php?id=1040 简直智障,WA了好几次。 #include<iostream>#include<cstdio>#include<cstring>#include<algori
https://www.u72.net/daima/f0wk.html - 2024-08-17 00:13:39 - 代码库这场貌似是gcd专场?第一题很有意思,模拟gcd的过程即可。 1 //0924 candy 2 //by Cydiater 3 //2016.9.24 4 #include <iostream> 5 #include <cstdio> 6
https://www.u72.net/daima/k9hh.html - 2024-08-14 16:59:01 - 代码库1230多重hash练习一下,不用也可以//// main.cpp// codeves1230//// Created by Candy on 9/29/16.// Copyright &copy; 2016 Candy. All rights res
https://www.u72.net/daima/d0am.html - 2024-08-15 06:37:06 - 代码库http://acm.xidian.edu.cn/problem.php?id=1031 简单找规律,a[i][j] = (i+j)/gcd(i,j)。 #include<iostream>#include<algorithm>#include<cstring>#inc
https://www.u72.net/daima/frvn.html - 2024-08-16 19:35:01 - 代码库http://acm.xidian.edu.cn/problem.php?id=1021 这数据真刁钻,PI不能自己输入,要用atan定义。 #include<iostream>#include<cstring>#include<cstdio>#in
https://www.u72.net/daima/b8fw.html - 2024-08-16 09:27:41 - 代码库http://acm.xidian.edu.cn/problem.php?id=1035 本来想用goto优化一下的,不知道什么情况,加了goto就wa了。 #include<iostream>#include<cstring>#includ
https://www.u72.net/daima/b838.html - 2024-08-16 09:46:43 - 代码库http://acm.xidian.edu.cn/problem.php?id=1099 每次加上以当前为尾的区间数量。 #include<iostream>#include<cstdio>#include<cstring>#incl
https://www.u72.net/daima/f84s.html - 2024-08-17 06:28:40 - 代码库http://acm.xidian.edu.cn/problem.php?id=1062 遍历一遍就可以了。 #include<iostream>#include<cstring>#include<cstdio>#include<algorith
https://www.u72.net/daima/f7n6.html - 2024-08-17 05:14:26 - 代码库http://acm.xidian.edu.cn/problem.php?id=1129 保存累计的下标,直接输出。 #include<iostream>#include<cstring>#include<cstdio>#include<al
https://www.u72.net/daima/r43c.html - 2024-08-19 06:02:33 - 代码库开灯问题时间限制:3000 ms | 内存限制:65535 KB难度:1描述 有n盏灯,编号为1~n,第1个人把所有灯打开,第2个人按下所有编号为2 的倍数的开关(这些灯将
https://www.u72.net/daima/r5na.html - 2024-07-12 09:15:38 - 代码库