[poj3252]Round Numbers_数位dp】的更多相关文章

Round Numbers poj3252 题目大意:求一段区间内Round Numbers的个数. 注释:如果一个数的二进制表示中0的个数不少于1的个数,我们就说这个数是Round Number.给定区间l,r<=$2\cdot 10^9$. 想法:又是一道数位dp裸题.我们先来设状态:dp[i]表示二进制表示下有i为而且第一位是1的Round Number的个数. 这题的特殊之处在于我们并不需要转移?因为我们可以直接求出任意的dp[i].显然,我们的i位数的第一位是1,所以,后面0的个数一定…
题目链接:http://poj.org/problem?id=3252 Round Numbers Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14640   Accepted: 5881 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (al…
Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets…
地址 拆成2进制位做dp记搜就行了,带一下前导0,将0和1的个数带到状态里面,每种0和1的个数讨论一下,累加即可. WA记录:line29. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #define dbg(x) cerr<<#x<<" = "<<x<…
题目大意:给你一个区间$[l,r]$,求在该区间内有多少整数在二进制下$0$的数量$≥1$的数量.数据范围$1≤l,r≤2*10^{9}$. 第一次用记忆化dfs写数位dp,感觉神清气爽~(原谅我这个蒟蒻,原先写的四不像数位dp至少需2h,用真记忆化dfs不到半小时写出) 我们用$f[i][j]$表示在最后的$i+j$为中,用了$i$个$0$,$j$个$1$的方案数(第$i+j$位也可以是$0$).该方程转移显然为$f[i][j]=f[i-1][j]+f[i][j-1]$. 于是我们用记忆化df…
Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets…
题目链接:[kuangbin带你飞]专题十五 数位DP E - Round Numbers 题意 给定区间.求转化为二进制后当中0比1多或相等的数字的个数. 思路 将数字转化为二进制进行数位dp,由于一个二进制数的最高位必须为1.所以设置变量first记录前面位是否有1,若有1,则可随意放,否则,仅仅可放1. 同一时候.上面的推断决定了搜索时len的大小与二进制本身的长度不一定相等,所以需两个变量对1和0的个数进行记录. 用dp[a][b][c]保存长度a,b个0,c个1的数字个数.记忆化搜索.…
题目链接 找一个范围内二进制中0的个数大于等于1的个数的数的数量.基础的数位dp #include<bits/stdc++.h> using namespace std; #define mem1(a) memset(a, -1, sizeof(a)) ], dp[][][], len; int dfs(int len, int num0, int num1, int f, int first) { //first记录前面是否全部为0 if(!len) { return num0>=nu…
正解:数位$dp$ 解题报告: 传送门$w$ 沉迷写博客,,,不想做题,,,$QAQ$口胡一时爽一直口胡一直爽$QAQ$ 先港下题目大意嗷$QwQ$大概就说,给定区间$[l,r]$,求区间内满足二进制中0的个数小于等于1的个数的数个数 数位$dp$板子,,,? 首先区间转成$[1,r]-[1,l-1]$然后十进制转二进制这个就不港了,,, 然后就考虑$dfs$中要记录哪些东西$QwQ$? 首先依然是一个$pos$一个$lim$,因为01的个数对结果会有影响,所以显然考虑还要记一个$num$表示实…
Round Numbers Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14947   Accepted: 6023 Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors',…