题目连接 : 传送门 题意: 给定一个长度为的二进制串和一个长度为n的序列a[],我们能够依据这个二进制串得到它的Gray code. Gray code中假设第i项为1的话那么我们就能够得到a[i]的值,在原来的二进制串中有一些位置为? 表示能够为0, 也能够为1求最后所能得到的最大的值. 已知二进制码怎样得到Gray code请看:传送门 分析: 我们能够通过动态规划来解决问题,状态转移也很好找 ,dp[i][j]表示到第i个位置.第i个位置 取j所能得到的最大值.非常明显这个题的j仅仅能有…

Gray code Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 129    Accepted Submission(s): 68 Problem Description The reflected binary code, also known as Gray code after Frank Gray, is a binary n…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5375 题面: Gray code Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 626    Accepted Submission(s): 369 Problem Description The reflected binary cod…

Gray code Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 569    Accepted Submission(s): 337 Problem Description The reflected binary code, also known as Gray code after Frank Gray, is a binary…

Gray code Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 684    Accepted Submission(s): 402 Problem Description The reflected binary code, also known as Gray code after Frank Gray, is a binary…

题意:给一个二进制数(包含3种符号:'0'  '1'  '?'  ,问号可随意 ),要求将其转成格雷码,给一个序列a,若转成的格雷码第i位为1,则得分+a[i].求填充问号使得得分最多. 思路:如果了解格雷码的转换,相信能很快看出一些端倪.偷别人的图: 分析一下:所给的二进制数要转成格雷码,只与所给二进制有关.即不需要利用已经某些转换好的格雷码字. 接下来分析5个位的串 : (1)00?00 仅有1个问号,只会与后面那些连续且非问号的串转成格雷码有关 (2)00??0 有连续的1个问号,这才需要…

题意:给出一个二进制数,其中有些位的数字不确定,对于所有对应的格雷码,与一个序列a对应,第i位数字为1时得分a[i],求最大的得分. 解法:一个二进制数x对应的格雷码为x ^ (x >> 1),题解说是个dp……但其实就四种情况……判一下就好了 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h>…

dp[i][j]表示第i位取j的时候取得的最大的分数 然后分s[i]是不是问号,s[i-1]是不是问号这大的四种情况讨论 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; ; ]; int a[maxn]; char s[maxn]; int n; int main(){ int T; scanf("%d&qu…

题意:给你一串不确定的二进制码,其对应的格雷码的每一位有对应的权值,问转换成的格雷码的能取到的最大权值是多少. 思路:没有思路,乱搞也AC #pragma comment(linker, "/STACK:1000000000") #include <iostream> #include <cstdio> #include <fstream> #include <algorithm> #include <cmath> #incl…

题意:给一串字符串,里面可能出现0,1,?,当中问号可能为0或1,将这个二进制转换为格雷码后,格雷码的每位有一个权值,当格雷码位取1时.加上该位权值,求最大权值和为多少. 分析:比赛的时候愚了.竟然以为格雷码是由3个二进制字符转换的,于是妥妥wa了,事实上格雷码的求法是通过异或求的的,即Gi=Bi⊕Bi-1(G为格雷码,B为二进制数),那么仅仅要水水的DP一下就好了.详细DP是通过保存 DP[i][j],当中i为当前位.j表示当前为为0还是1,递推式为: 一.s[i]为0时 1.s[i-1]为0…

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int N=200000+5; const int inf=1<<24; int dp[N][2],a[N]; char s[2*N]; int main() { int n,m,i,_; scanf("%d",&_); for(int k=1;k<=_;k+…

题目传送门 /* 题意:给一个串,只能是0,1,?(0/1).计算格雷码方法:当前值与前一个值异或,若为1,可以累加a[i],问最大累加值 DP:dp[i][0/1]表示当前第i位选择0/1时的最大分数,那么分类讨论,情况太多,看代码,注意不可能的状态不要转移 */ /************************************************ * Author :Running_Time * Created Time :2015-8-11 15:49:03 * File N…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5375 编码规则:tmp = XOR(gr[i],gr[i-1]); 算是找规律的题目吧,考虑?前后字符和?数目的奇偶性就可以了,一个小trick就是当碰到需要减的时候是减问号区间内最小的那个, 然后就是调试的问题: 字符串从数组第二位开始存放,scanf("%s",s+1); 当用到min和max的时候记得每次使用前要清空. #include<stdio.h> ; int XO…

1.题目描写叙述:点击打开链接 2.解题思路:本题要求模拟俄罗斯方块游戏.然而比赛时候写了好久还是没过. 后来补题发现原来是第四步的逻辑实现写错了... 题目中要求假设一整行能够消除,那么仍然运行该步.否则才回到第一步.可是我的代码却是不论能否够消除,都回到第一步.. .补题时候还发现一个地方我的理解出错了.. (可能是我脑洞真的有点大).题目中说假设一整行能够消除,那么它上面的方格要下落.我的理解是下落的方格要一直降落到第一行或者某个支撑物上,最后发现依照这样写,例子都算不正确==.调试了一下…

1.题目描写叙述:点击打开链接 2.解题思路:本题利用字典树解决.本题要求查找全部的B[j]在A[i]中出现的总次数.那么我们能够建立一颗字典树,将全部的B[j]插入字典树,因为一个串的全部字串相当于它全部后缀的前缀. 因此在查找时候,仅仅须要查找A[i]的每个后缀就可以,然后累加这个后缀的前缀个数,就可以得到该后缀中子串的个数.全部后缀的值相加,就是终于的答案. 3.代码: #pragma comment(linker, "/STACK:1024000000,1024000000")…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5410 Problem Description Today is CRB's birthday. His mom decided to buy many presents for her lovely son.She went to the nearest shop with M Won(currency unit).At the shop, there are N kinds of pr…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5396 Problem Description Teacher Mai has n numbers a1,a2,⋯,anand n−1 operators("+", "-" or "*")op1,op2,⋯,opn−1, which are arranged in the form a1 op1 a2 op2 a3 ⋯ an. He wan…

题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5379 Problem Description Little sun is an artist. Today he is playing mahjong alone. He suddenly feels that the tree in the yard doesn't look good. So he wants to decorate the tree.(The tree has n…

题目链接:HDU 5355 http://acm.split.hdu.edu.cn/showproblem.php?pid=5355 Problem Description There are m soda and today is their birthday. The 1-st soda has prepared n cakes with size 1,2,…,n. Now 1-st soda wants to divide the cakes into m parts so that th…

题目链接 http://acm.hust.edu.cn/vjudge/contest/130883#problem/C Problem Description Zero Escape, is a visual novel adventure video game directed by Kotaro Uchikoshi (you may hear about ever17?) and developed by Chunsoft. Stilwell is enjoying the first ch…

原题链接 Problem Description There are n people and m pairs of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline friends (mostly using face-to-face communication). However, e…

原题链接 Problem Description Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special task to some staffs who were in the same group. In…

原题链接 Problem Description You are a "Problem Killer", you want to solve many problems. Now you have n problems, the i-th problem's difficulty is represented by an integer ai (1≤ai≤109).For some strange reason, you must choose some integer l and r…

HDOJ--4869--Turn the pokers[组合数学+快速幂] 题意:有m张扑克,开始时全部正面朝下,你可以翻n次牌,每次可以翻xi张,翻拍规则就是正面朝下变背面朝下,反之亦然,问经过n次翻牌后牌的朝向有多少种情况.我们可以把正面朝上理解为1,反面朝上理解为0,那么可以理解为求01串的不同的组合方式有几种. 解题思路:我们可以知道,每张牌假设起始状态都为0,如果翻奇数次,该牌最后的情况是1,如果翻偶数次,该牌的最后情况为0.根据n次翻牌的个数找出1的个数的下限和上限,然后再在这个范围…

题意:已知昨天天气与今天天气状况的概率关系(wePro),和今天天气状态和叶子湿度的概率关系(lePro)第一天为sunny 概率为 0.63,cloudy 概率 0.17,rainny 概率 0.2.给定n天的叶子湿度状态,求这n天最可能的天气情况 分析:概率dp设 dp[i][j] 表示第i天天气为j的最大概率,pre[i][j]表示第i天天气最可能为j的前一天天气,dp[i][j]=max(dp[i-1][k]+log(wePro[k][j])+log(lePro[j][lePos[i]]…

题意:给定一个长度为n的序列,规定f(l,r)是对于l,r范围内的某个数字a[i],都不能找到一个相应的j使得a[i]%a[j]=0.那么l,r内有多少个i,f(l,r)就是几. 问全部f(l,r)的总和是多少. 公式中给出的区间,也就是全部存在的区间. 思路:直接枚举每个数字,对于这个数字,假设这个数字是合法的i,那么向左能扩展的最大长度是多少,向右能扩展的最大长度是多少.那么i为合法的情况就是左长度*右长度(包括i且i是合法的区间总数). 统计左长度能够推断a[i]的约数是否在前面出现过-由…

Turn the pokers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 108    Accepted Submission(s): 21 Problem Description During summer vacation,Alice stay at home for a long time, with nothing to…

http://acm.hdu.edu.cn/showproblem.php?pid=5375 Gray code Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 988    Accepted Submission(s): 551 Problem Description The reflected binary code, also kn…

pid=5288">http://acm.hdu.edu.cn/showproblem.php?pid=5288 Problem Description OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now…

pid=5379">http://acm.hdu.edu.cn/showproblem.php? pid=5379 Problem Description Little sun is an artist. Today he is playing mahjong alone. He suddenly feels that the tree in the yard doesn't look good. So he wants to decorate the tree.(The tree has…