代码+注释 1 //我感觉这道题最扯的就是我因为输入而TLE了半天,懵逼死了,想破脑袋也没想到因为输入TLE了半天 2 //题意:求区间内数字满足"奇数各数出现偶数次,偶数各数出现奇数次"的数字的个数. 3 //题解: 4 //dp[x][y]表示长度为x的时候0~9各出现的状态情况,因为可能有未出现的情况,如果这个y用二进制 5 //保存的话那么未出现的偶数可能会因为0而出现误判,所以应该多一个状态,就用三进制.0表示未出现, 6 //1表示出现了奇数次,2表示出现了偶数次. 7 /…

题目链接:http://www.spoj.com/problems/BALNUM/en/ Time limit: 0.123s Source limit: 50000B Memory limit: 1536MB Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if: 1)      Every even digit…

Balanced Numbers Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if: 1)      Every even digit appears an odd number of times in its decimal representation 2)      Every odd digit app…

链接: https://vjudge.net/problem/SPOJ-BALNUM 题意: Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if: 1) Every even digit appears an odd number of times in its decimal representation 2)…

BALNUM - Balanced Numbers Time limit:123 ms Memory limit:1572864 kB Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if: 1)      Every even digit appears an odd number of times in its…

一个数字是Balanced Numbers,当且仅当组成这个数字的数,奇数出现偶数次,偶数出现奇数次 一下子就相到了三进制状压,数组开小了,一直wa,都不报re, 使用记忆化搜索,dp[i][s] 表示长度为i,状态为s,时,满足条件的balance number的个数 #include <stdio.h> #include <string.h> #include <stdlib.h> #include <algorithm> #include <io…

Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if: 1)      Every even digit appears an odd number of times in its decimal representation 2)      Every odd digit appears an even numb…

SPOJ - BALNUM 题意: Balanced Numbers:数位上的偶数出现奇数次,数位上的奇数出现偶数次(比如2334, 2出现1次,4出现1次,3出现两次,所以2334是 Balanced Numbers) ,求一个区间内有多少Balanced Numbers. 解题思路:看题很容易想到数位dp,但是怎么记录某数字出现的次数呢?由于某数字出现的次数只与奇偶有关,与大小没有关系,所以我们可以分别用0,1,2来表示某数字没有出现,出现奇数次,出现偶数次,然后就可以用三进制来记录所有数字…

Balanced Numbers https://vjudge.net/contest/287810#problem/K Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if: 1) Every even digit appears an odd number of times in its decimal rep…

Description Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if: 1)      Every even digit appears an odd number of times in its decimal representation 2)      Every odd digit appears…

题目:http://www.spoj.com/problems/BALNUM/en/ 题意:找出区间[A, B]内所有奇数字出现次数为偶数,偶数字出现次数为计数的数的个数. 分析: 明显的数位dp题,首先,只有3种状态(0:没出现过, 1:数字出现奇数次, 2:数字出现偶数次),所以, 0~9 出现的次数就可以用3进制表示,最大的数就是 310 ,那么我们就可以把1019 哈希到310 内了.其中,我们可以假设: (0:30  ,1:31 , 2:32 , .... , 9: 39 ) 当第一次…

思路: 把0~9的状态用3进制表示,数据量3^10 代码: #include<cstdio> #include<map> #include<set> #include<queue> #include<cstring> #include<string> #include<cmath> #include<cstdlib> #include<iostream> #include<algorithm&…

题意: 求出所给范围内满足其数位上的奇数出现偶数次,数位上的偶数出现奇数次(或不出现)的数的个数. 思路: 对于0 ~ 9 每个数有3种情况. 1.没出现过 2.出现奇数次 3.出现偶数次 那么就可以用三进制来表示状态. #include <iostream> #include <cstring> #include <cstdio> #include <cmath> using namespace std; typedef long long ll; ; i…

题意: 平衡树定义为“一个整数的某个数位若是奇数,则该奇数必定出现偶数次:偶数位则必须出现奇数次”,比如 222,数位为偶数2,共出现3次,是奇数次,所以合法.给一个区间[L,R],问有多少个平衡数? 思路: 这题比较好解决,只有前导零问题需要解决.如果枚举到011,那么其前导零(偶数)出现了1次而已,而此数11却是平衡数,所以不允许前导零的出现! 由于dfs时必定会枚举到前导零,否则位数较少的那些统计不到.状态需要3维or2维也行,3维的比较容易处理,用一维表示数位出现次数,另一维表示数位是否…

题目链接:http://www.spoj.com/problems/BALNUM/en/ 题意:问你在[A,B]的闭区间内有几个满足要求的数,要求为每个出现的奇数个数为偶数个,每个出现的偶数个数为奇数个. 显然dfs很好想到dfs(int len , int even[] , int odd[] , int flag , int first) even表示前len位出现偶数的总状态 , odd表示前len位出现奇数的总状态,first表示是否为首位(特判全为0的结果) 但是这样写dfs dp不好…

题目链接 一个数称为平衡数, 满足他各个数位里面的数, 奇数出现偶数次, 偶数出现奇数次, 求一个范围内的平衡数个数. 用三进制压缩, 一个数没有出现用0表示, 出现奇数次用1表示, 出现偶数次用2表示, 这样只需要开一个20*60000的数组. #include<bits/stdc++.h> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y)…

题意 求[X,Y]区间内能被其各位数(除0)均整除的数的个数. CF 55D 有些时候因为问题的一些"整体性"而导致在按位统计的过程中不能顺便计算出某些量,所以只能在枚举到最后一位确定数字时才能计算相应的统计. 在本题中,我们无法在过程中确定到底有哪些数位,以及这个数本身,所以这些计算都要放在最后.所以首先我们需要参数num传递当前搜索确定的数字,以及判断该数字是否能被其数位整除.而判断各位整除只要数字整除各位数的最小公倍数lcm即可. 但我们随后发现了问题:各位数最小公倍数最大可能为…

一开始题看错了...dp[pos][sets][viss],其中sets表示出现次数,viss表示出现没有. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ][][],bit[],ret=; void get_bit(long long x) { ret=; ;x/=;} } long long check(long…

题意:求给定区间,一个数的数位上每个奇数出现偶数次,每个偶数出现奇数次,这样数的个数 分析:先考虑状态,但总是想不全,所以要把状态压缩一下,用三进制,0 该数不放  1 放了奇数次 2放了偶数次 dp[i][j] 长度为i 状态是j的数字个数,需要前导0判断,前导0不能计入偶数出现的次数. #include <map> #include <set> #include <list> #include <cmath> #include <queue>…

题意 求区间[A,B]上的平衡数个数.平衡数是这样的数:在数的各个位上,奇数数字出现偶数次,偶数数字出现奇数次. 思路 很明显我们需要记录每一位出现的次数.分别记录是不明智的,而我们又只需要记录奇数次或者偶数次即可.所以我们可以用一个<=1024的数state表示0~9这10个数字出现的次数奇偶性,当奇数出现偶数次则相应位为1,当偶数出现奇数次相应位为1,最后判断是不是1023.但是这样它的初试状态不好确定.很明显初始时我们需要把state赋值成1010101010,即682(0次算偶次).但是…

题意:求某一区间内的平衡数个数(指一个数,其中出现过的数,如果是偶数,那么必须出现奇数次,反之偶数次) 题解:用三进制来枚举(0到9)所有情况,0代表没有出现,1代表出现奇数次,2代表出现偶数次dp[i][j]i代表位数,j代表状态,在记忆化搜索的时候要记录0是否出现过 (因为之前很少写3进制的状态,写的大多数是2进制的,所以很不熟练,还是要多练,刚开始是把题意理解错了,以为是所有的奇数出现偶数次 这样,所以搞了半天也不对) #include<bits/stdc++.h> #define C…

题意:一个数,如果满足奇数的数字出现偶数次,偶数的数字出现奇数次, 就是符合的数,注比如:12313  就满足,因为1 3出现了偶数次.2出现了奇数次 思路,对于这道题,就是状态压缩加dp: 对于一个数字来说,0~9每一位,都只有3种状态量,要么从未出现,要么出现次数为奇 要么为偶,所以可以用3进制来表示,通过3进制表示出其状态,然后到pos-1的时候 判断0~9的各个状态是否符合条件即可 #include<cstdio> #include<string.h> using name…

数位DP... Balanced Number Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 1337    Accepted Submission(s): 583 Problem Description A balanced number is a non-negative integer that can be balanced…

[HDU3709]Balanced Number 试题描述 A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit…

Balanced Number Problem Description A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some…

Balanced Number Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 3798    Accepted Submission(s): 1772 Problem Description A balanced number is a non-negative integer that can be balanced if a pi…

 Balanced Number Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Description A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box wi…

Balanced Number Problem Description A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some…

Problem Description A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the nu…

Balanced Number Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 9036    Accepted Submission(s): 4294 Problem Description A balanced number is a non-negative integer that can be balanced if a pi…