POJ3252 数位DP Round Numbers

Round Numbers

Time Limit: 2000MS

Memory Limit: 65536K

Total Submissions: 14014

Accepted: 5564

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

思路： 先将数分解成二进制，进行数位DP，但由于状态中sta会小于0，且最小不超过30几， 因此以40作为0开始。并且注意这题需要判断前导0.

#include <iostream>#include <cstdio>#include <string.h>#include <string>#define LL long longusing namespace std;LL dp[50][100];int a[50];LL dfs( int pos , int sta , bool lead , bool limit ){if( pos == -1 ) return sta  >= 40 ;if( !limit && !lead && dp[pos][sta] != -1 ) return dp[pos][sta];int up = limit ? a[pos] : 1;LL ans = 0 ;for( int i = 0 ; i <= up ; i ++ ){if( lead && i == 0 ) ans += dfs( pos - 1 , sta , lead , limit && i == up ) ;else ans += dfs( pos - 1 , sta + (i==0?1:-1) , lead && i == 0 , limit && i == up );}if( !limit && !lead ) dp[pos][sta] = ans ;return ans ;}LL solve( LL x ){int tot = 0 ;while( x ){a[tot++] = x & 1 ;x >>= 1 ;}return dfs( tot-1 , 40 , true , true );}int main(){ios_base::sync_with_stdio(false); cin.tie(0);LL l , r ;memset(dp,-1,sizeof(dp));while ( cin >> l >> r ) {cout << solve(r) - solve(l-1) << endl; }return 0;}