Round Numbers POJ - 3252（数位dp）

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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
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 range Start… Finish
Sample Input
2 12
Sample Output
6

题意： 求二进制0比1的的数字
思路： 数位dp，dp(i,j)代表到了第i位，0的个数和1的个数差为j-32的数目。

#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;int a[105];
int dp[70][70];int dfs(int pos,int num,bool lead,bool limit)
{if(!pos)return num >= 32;if(!lead && !limit && dp[pos][num] != -1)return dp[pos][num];//前面有1且没有达到上线的状态才能记录下来int up = limit ? a[pos] : 1;int ans = 0;for(int i = 0;i <= up;i++){if(lead && !i)ans += dfs(pos - 1,num,lead && !i,limit && i == a[pos]);else ans += dfs(pos - 1,num + (!i ? 1 : -1),lead && !i,limit && i == a[pos]);}if(!lead && !limit)dp[pos][num] = ans;return ans;
}int solve(int x)
{int pos = 0;for(;x;x>>=1)a[++pos] = x & 1;return dfs(pos,32,true,true);
}int main()
{int a,b;memset(dp,-1,sizeof(dp));while(~scanf("%d%d",&a,&b)){printf("%d\n",solve(b) - solve(a - 1));}return 0;
}