Round Numbers(数位DP)
Round Numbers
http://poj.org/problem?id=3252
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 17293 | Accepted: 7188 |
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 to be milked first. They can't even flip a coin because it's so hard to toss using hooves.
They have thus resorted to "round number" matching. The first cow picks an integer less than two billion. The second cow does the same. If the numbers are both "round numbers", the first cow wins,
otherwise the second cow wins.
A positive integer N is said to be a "round number" if the binary representation of N has as many or more zeroes than it has ones. For example, the integer 9, when written in binary form, is 1001. 1001 has two zeroes and two ones; thus, 9 is a round number. The integer 26 is 11010 in binary; since it has two zeroes and three ones, it is not a round number.
Obviously, it takes cows a while to convert numbers to binary, so the winner takes a while to determine. Bessie wants to cheat and thinks she can do that if she knows how many "round numbers" are in a given range.
Help her by writing a program that tells how many round numbers appear in the inclusive range given by the input (1 ≤ Start < Finish ≤ 2,000,000,000).
Input
Output
Sample Input
2 12
Sample Output
6
#include<iostream>
#include<cstring>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pb push_back
#define eb emplace_back
#define maxn 13000005
#define eps 1e-8
#define pi acos(-1.0)
#define rep(k,i,j) for(int k=i;k<j;k++)
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<long long,int>pli;
typedef pair<char,int> pci;
typedef pair<pair<int,string>,pii> ppp;
typedef unsigned long long ull;
const long long MOD=1e9+;
/*#ifndef ONLINE_JUDGE
freopen("1.txt","r",stdin);
#endif */ int dp[][];
int a[];
int num; int dfs(int pos,int sta,bool lead,bool limit){
if(pos==-) return sta>=;
if(!limit&&!lead&&dp[pos][sta]!=-) return dp[pos][sta];
int up=limit?a[pos]:;
int ans=;
for(int i=;i<=up;i++){
if(lead&&i==) ans+=dfs(pos-,sta,lead,limit&&i==a[pos]);
else ans+=dfs(pos-,sta+(i==?:-),lead&&i==,limit&&i==a[pos]);
}
if(!limit&&!lead) dp[pos][sta]=ans;
return ans;
}
int solve(int x){
int pos=;
while(x){
a[pos++]=x%;
x>>=;
}
return dfs(pos-,,true,true);
} int main(){
#ifndef ONLINE_JUDGE
// freopen("1.txt","r",stdin);
#endif
std::ios::sync_with_stdio(false);
int n,m;
memset(dp,-,sizeof(dp));
while(cin>>n>>m){
cout<<solve(m)-solve(n-)<<endl;
}
}
Round Numbers(数位DP)的更多相关文章
- poj 3252 Round Numbers(数位dp 处理前导零)
Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, P ...
- 4-圆数Round Numbers(数位dp)
Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 14947 Accepted: 6023 De ...
- POJ3252 Round Numbers —— 数位DP
题目链接:http://poj.org/problem?id=3252 Round Numbers Time Limit: 2000MS Memory Limit: 65536K Total Su ...
- POJ 3252 Round Numbers(数位dp&记忆化搜索)
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 ...
- poj3252 Round Numbers (数位dp)
Description The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, P ...
- POJ - 3252 - Round Numbers(数位DP)
链接: https://vjudge.net/problem/POJ-3252 题意: The cows, as you know, have no fingers or thumbs and thu ...
- poj 3252 Round Numbers 数位dp
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 ...
- 【poj3252】 Round Numbers (数位DP+记忆化DFS)
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 ...
- $POJ$3252 $Round\ Numbers$ 数位$dp$
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 ...
随机推荐
- day04-完整性约束
完整性约束 关键字: not null 与 default unique primary auto_increment foreign key 1.介绍 约束条件与数据类型的宽度一样,都是可选参数作用 ...
- python 中的比较==和is
Python 中的比较:is 与 == 在 Python 中会用到对象之间比较,可以用 ==,也可以用 is .但是它们的区别是什么呢? is 比较的是两个实例对象是不是完全相同,它们是不是同一个对象 ...
- 一个简单的SignalR例子
本文介绍如何使用SignalR的Hub制作一个简单的点赞页面.不同浏览器(或者不同窗口)打开同一个页面,在任何一个页面点赞,所有页面同时更新点赞数. 1.使用Visual Studio Communi ...
- Spring AOP demo 和获取被CGLIB代理的对象
本文分为两部分:1)给出Spring AOP的一个例子(会使用CGLIB代理):2)给出获取被CGLIB代理的原始对象. 1.Spring AOP Demo 这部分参考了博文(http://www.v ...
- webpack 自动发现 entry 的配置和引用方式
假定我们的项目目录为如下的样子: - root/ - assets/ - app/ - global.js - index/ - index.js - auth/ - login.js - regis ...
- KVM虚拟化技术(二)KVM介绍
KVM:Kernel Virtual Machine KVM是基于虚拟化扩展的x86硬件,是Linux完全原生的全虚拟化解决方案.部分半虚拟化支持,主要是通过半虚拟网络驱动程序的形式用于Linux和W ...
- 循环取到json中的字段数据,加到html中
$.ajax({ type:'post', data:{specialName:specialName,count:count}, url:"admin/pcAdminGetArticleL ...
- eval 用法
计算 eval('1+1') # 2 在字典中提取键 的值 eval('a',{'a':1}) # 1 计算 Boolean 值 eval( 'True',{'a':1}) # True eval(' ...
- tensorflow serving 中 No module named tensorflow_serving.apis,找不到predict_pb2问题
最近在学习tensorflow serving,但是运行官网例子,不使用bazel时,发现运行mnist_client.py的时候出错, 在api文件中也没找到predict_pb2,因此,后面在网上 ...
- js 乘法 4.39*100 出现值不对问题解决
https://www.jianshu.com/p/a026245661bb //除法函数,用来得到精确的除法结果 //说明:javascript的除法结果会有误差,在两个浮点数相除的时候会比较明显. ...