poj 3252 Round Numbers(数位DP,4级)
来源:互联网 发布:ipv4 ipv6无网络权限 编辑:程序博客网 时间:2024/06/07 11:12
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
Source
#include<cstdio>#include<cstring>#include<iostream>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))#define LL __int64using namespace std;int dp[40][40][40];//位数,1,0int bit[40],pos;int DP(int pp,int one,int zero,bool nozero,bool big){ if(pp==0)return zero>=one; if(big&&dp[pp][one][zero]!=-1)return dp[pp][one][zero]; int kn=big?1:bit[pp]; int ret=0; FOR(i,0,kn) { ///已经有非0 才算法 ret+=DP(pp-1,one+(i==1),zero+(nozero&&i==0),nozero||i!=0,big||kn!=i); } if(big)dp[pp][one][zero]=ret; return ret;}int get(int x){ pos=0; while(x) { bit[++pos]=x&1;x/=2; } clr(dp,-1); return DP(pos,0,0,0,0);}int main(){ int n,m; while(cin>>n>>m) { if(n==0&&m==0)break; cout<<get(m)-get(n-1)<<endl; }}
- poj 3252 Round Numbers(数位DP,4级)
- POJ 3252 Round Numbers 数位DP
- poj 3252 Round Numbers (数位DP)
- poj 3252 Round Numbers(数位dp)
- POJ 3252 Round Numbers (数位DP)
- poj 3252 Round Numbers(数位dp)
- Round Numbers - POJ 3252 数位dp
- 【数位DP】【poj 3252】Round Numbers
- poj 3252 Round Numbers 数位dp
- POJ 3252 Round Numbers (数位DP)
- POJ 3252 round numbers(数位DP)
- 【数位DP】POJ 3252 Round Numbers
- POJ 3252 Round Numbers 数位dp(入门
- poj 3252 Round Numbers(数位dp)
- POJ 3252 Round Numbers(数位DP)
- POJ 3252 Round Numbers (简单数位DP)
- POJ 3252 Round Numbers(数位dp)
- POJ 3252 Round Numbers (数位dp)
- centos 下面安装的firefox 出现中文乱码
- [poj 2186]Popular Cows[Tarjan强连通分量]
- 【解题报告】HDU 4638 Group - 树状数组 + 求一段区间连续数字的段数
- linux write 命令 退格键无法生效
- 360阻止程序
- poj 3252 Round Numbers(数位DP,4级)
- 关于java参数传递
- JQuery之focus函数
- 在vs2012中添加控件
- nyoj-589-糖果
- maven 中配置运行指定的测试文件
- hdu 2102 A计划(优先队列+dfs)
- 骨牌铺方格 sdut 1018
- linux系统进程状态理解