Zhuge Liang's Mines Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 44    Accepted Submission(s): 9 Problem Description In the ancient three kingdom period, Zhuge Liang was the most famous and…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4747 思路: 比赛打得太菜了,不想写....线段树莽一下 实现代码: #include<iostream> #include<cstdio> #include<map> #include<cmath> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|…

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这个题目虽然在比赛的时候苦思无果,但是赛后再做就真的是个水题,赤果果的水题. 题目的意思是给n个数构成的环,两只兔子从任一点开始分别顺逆时针跳,每次可以调到任意一个数(最多不会跳过一圈). 求最多能跳多少步. 这个题目是个dp,两个不同的方向开始跳跃,而且跳过的数字相同,那么就是要你求回文子串啦, 可以这样考虑,用dp的方法求出从一个位置到另一个位置的最长回文子串的长度,这样等于把[1-n]个数分成两个区间[1-x]和[x+1,n], 而答案就是两个区间最长回文串之和.(仔细理解这里,这是关键)…

啦啦啦. 这是杭州网赛的一个题目,当时没做出来,当然这个想法确实比较难想到. 题目质量很高,这个题目也很特别,以前都没做过类似的题目.让我又一次体验了线段树的强大力量. 题目的意思是给你n个数a1-an.对于任何一个区间[l,r],它所对应的值为这个区间内没有出现的最小非负整数,求所有1<=L<=R<=n的区间所对应的值的总和. 比赛的时候苦思无果,以为是什么高端的数据结构或者很奇怪的算法,后来才发现自己坑了. 题目其实是这样的,在最开始预处理所有1开头的(即L=1的所有区间的值,然后从…

Description Long long ago, there lived two rabbits Tom and Jerry in the forest. On a sunny afternoon, they planned to play a game with some stones. There were n stones on the ground and they were arranged as a clockwise ring. That is to say, the firs…

Description At the end of the 200013 th year of the Galaxy era, the war between Carbon-based lives and Silicon civilization finally comes to its end with the Civil Union born from the ruins. The shadow fades away, and the new-born Union is opening a…

Problem Description Mex is a function on a set of integers, which is universally used for impartial game theorem. For a non-negative integer set S, mex(S) is defined as the least non-negative integer which is not appeared in S. Now our problem is abo…

http://acm.hdu.edu.cn/showproblem.php?pid=4710 Balls Rearrangement Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 735    Accepted Submission(s): 305 Problem Description Bob has N balls and A b…

http://acm.hdu.edu.cn/showproblem.php?pid=4708 Rotation Lock Puzzle Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Alice was felling into a cave. She found a strange door with a number square m…

http://acm.hdu.edu.cn/showproblem.php?pid=4715 Difference Between Primes Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description All you know Goldbach conjecture.That is to say, Every even integer great…

http://acm.hdu.edu.cn/showproblem.php?pid=4712 Hamming Distance Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 1610    Accepted Submission(s): 630 Problem Description (From wikipedia) For bina…

hduoj 4706 Children's Day 2013 ACM/ICPC Asia Regional Online —— Warmup Herding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2005    Accepted Submission(s): 563 Problem Description Little Joh…

http://acm.hdu.edu.cn/showproblem.php?pid=4707 Pet Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description One day, Lin Ji wake up in the morning and found that his pethamster escaped. He searched in th…

http://acm.hdu.edu.cn/showproblem.php?pid=4706 Children's Day Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others) Problem Description Today is Children's Day. Some children ask you to output a big letter 'N'. 'N' is c…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4749 题目大意:给一个原序列N,再给出一个序列M,问从N中一共可以找出多少个长度为m的序列,序列中的数的相对大小关系与序列M中相对大小关系相同.(序列之间不能重叠) 解题思路:从头开始以i为起点暴搜,不断找长度为m的序列,判断是否满足条件.若满足,跳到i+m之后继续搜,若不满足,向后移一位继续搜. 判断方法压缩数据比较相对大小. #include<cstdio> #include<cstri…

http://acm.hdu.edu.cn/showproblem.php?pid=4750 题意: 定义f(u,v)为u到v每条路径上的最大边的最小值..现在有一些询问..问f(u,v)>=t的点对有所少对,注意(1,2)和(2,1)是不同的点对 分析: 原来最小生成树有一个很鬼畜的结论,那就是一个图的最小生成树中任意两个点的路径中的最大边一定最小.(妈蛋,完全不知道这个) 然后此题就变得很明朗了,用kruskal算法,加边的时候此边连接的两个集合的路径中的最大边就是这个边,存储下来,询问的时…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4751 题目大意:判断一堆人能否分成两组,组内人都互相认识. 解题思路:如果两个人不是相互认识,该两人之间连边.最终构成一张图,二分匹配. #include<cstdio> #include<cstring> #include<iostream> using namespace std; #define maxn 105 #define maxm 20010 int n,e;…

Minimum palindrome Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 586    Accepted Submission(s): 110 Problem Description Setting password is very important, especially when you have so many "in…

这个纯粹是一个细节题啊!!! 由于某个地方的浮点数比较写错了,WA了无数次啊…… 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> using namespace std; ],str2[]; double h,s,l,r,g,b,v; bool is(double h,double a,double b)…

比赛的时候,被题目误导了,题目最后说结果可能很大,要取模,那时就想直接求会TLE的!!! 赛后才知道,坑啊………… 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> #define ll long long #define M 80001 #define mod 1000000007 using names…

1003 Rotation Lock Puzzle 找出每一圈中的最大值即可 代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector> #define ll __int64 #define pi acos(-1.0) #…

HDU 4716 A Computer Graphics Problem 水题.略 HDU 4717 The Moving Points 题目:给出n个点的起始位置以及速度矢量,问任意一个时刻使得最远的两点的距离最小. 分析:显然只有两点的话,答案满足三分性质.对于多个点,画个图分析一下,其实也满足三分性质.因此,先构造n*(n-1)/2个二次函数,于是三分枚举时间即可. #include <iostream> #include <cstdio> #include <cstr…

第一眼就想到DP,然后想了N久就想不到可以不重算的DP  最后没办法了 先算出来 再去重.. 因为最多只有三个 对于三个来说有三种组合情况 x+y+z, x*y*z, x*y+z 那要么 x,y,z都不同 要么 有两个相同 要么有三个相同 对都不同情况我的DP结果会重复两次 对于有两个相同的会重复一次 统计出都相同的 两个相同的 最后减掉..有点乱 不过A了 先预处理 时间差不多4S多 再O(1)询问 #include <iostream> #include<cstdio> #in…

第一个被板刷的题 取余 依次算在周几 #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<cmath> #include<map> using namespace std; double ll,ss; ]; ][] = {"Monday", "…

找了很久的规律,只看十进制数字,各种乱七八糟的规律=没规律!看了别人的解题报告,虽然看懂了,可是怎么发现的这个规律呢T.T~想了很久很久~ 以下是转载的别人的图,自己再画太麻烦了~全部看出0~2n-1-1,下标从0开始!!! 每次洗牌的时候,奇数在后偶数在前时,只需循环右移一下,如下:0(000) -->0(000) -->0(000) -->0(000)1(001) -->4(100) -->2(010) -->1(001)2(010) -->1(001) --…

AC了,但是不知道为什么,但是恶心的不得了~最近写代码,思路都非常清晰,但是代码各种bug~T.T~说说思路吧:二分~330ms~ 小队友fribbi的思路是离线250msAC~ 预处理solve函数(让能求出来的尽量都求出来)-->a[2]和a[n-3]已知 ①已知a[i]可知a[i+3]②已知a[i] a[i-1] 或者a[i] a[i+1]或者a[i+1] a[i-1]可知全部 ③已知a[0]或a[1]或a[n-2]或a[n-1]可知全部 提问: ①已知则直接输出,不存在a[i+1] a[…

题目描述 略... 题解 注意控制精度即可....变量全部定义成double,结果round就行....妈蛋....被这题目恶心死了.... 代码: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> using namespace std; #define ES…

哈哈哈 #include <iostream> #include <cstring> #include <string> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <queue> #include <map> #define inf 0x3f3f3f3f using names…

Minimum palindrome Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 309    Accepted Submission(s): 150 Problem Description Setting password is very important, especially when you have so many "i…