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 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…

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…

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)…

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 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…

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: 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…