poj3252Round Numbers非递归数位dp解

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9894 Accepted: 3569

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 ofN 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

Line 1: Two space-separated integers, respectively Start and Finish.

Output

Line 1: A single integer that is the count of round numbers in the inclusive rangeStart..Finish

Sample Input

2 12

Sample Output

6

Source

USACO 2006 November Silver

输入两个十进制正整数a和b，求闭区间 [a ,b]内有多少个Round number

所谓的Round Number就是把一个十进制数转换为一个无符号二进制数，若该二进制数中0的个数大于等于1的个数，则它就是一个Round

第一次独立完成数位dp，后来发现网上都是用递归做的，显然速度上没有我的优，很开森

dp[i][j][k][l]:对于长度为i的二进制数，最高位为j，有k个0，l个1时有多少个数字

#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<iostream>#include<algorithm>#include<bitset>#include<climits>#include<list>#include<iomanip>#include<stack>#include<set>using namespace std;int dp[40][2][40][40];int bit[40],tail;void cg(int x){tail=0;for(int i=0;x>>i;i++)bit[tail++]=(x>>i)&1;}int work(int x){if(x==0)return 0;cg(x);int ans=0;for(int i=1;i<tail;i++)for(int j=(i+1)/2;j<=i;j++)ans+=dp[i][1][j][i-j];int sum0=0,sum1=0;for(int i=tail-1;i>0;i--){if(bit[i])sum1++;elsesum0++;if(bit[i-1])for(int j=0;j<=i;j++)if(j+sum0>=i-j+sum1)ans+=dp[i][0][j][i-j];}if(bit[0])sum1++;elsesum0++;if(sum0>=sum1)ans++;return ans;}int main(){dp[1][0][1][0]=1;dp[1][1][0][1]=1;for(int i=1;i<32;i++)for(int j=0;j<=i;j++){dp[i+1][0][j+1][i-j]+=dp[i][0][j][i-j]+dp[i][1][j][i-j];dp[i+1][1][j][i-j+1]+=dp[i][0][j][i-j]+dp[i][1][j][i-j];}int a,b;while(cin>>a>>b)cout<<work(b)-work(a-1)<<endl;}