POJ 3252 Round Numbers（数位dp）
Round Numbers
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 17628 | Accepted: 7356 |
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
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 12Sample Output
6题意：
给你l,r(l<=r<=2e9)，求区间[l,r]中所有数的二进制表示中1和0的个数相等的数的个数。
思路：
考虑进行二进制数位dp。
为了方便记忆化，我们将dp第二维的初始值设置为50，碰到0就-1，1就+1，避免了记忆化的过程中负数下标的出现。
最后值还为50即符合条件。
注意前导0对答案的影响。
代码：
#include
#include
#include
#include
#include
#include
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
const int maxn=50;
ll a[maxn],c[maxn];
ll dp[maxn][maxn<<1][2];
ll dfs(int pos,int sum,bool limit,bool fg)
{
if(pos==-1) return (fg&&sum>=50);
if(!limit&&dp[pos][sum][fg]!=-1) return dp[pos][sum][fg];
int up=limit?a[pos]:1;
ll tmp=0;
for(int i=0;i<=up;i++)
{
if(fg)
{
if(i) tmp+=dfs(pos-1,sum-1,limit&&i==a[pos],fg);
else tmp+=dfs(pos-1,sum+1,limit&&i==a[pos],fg);
}
else if(i) tmp+=dfs(pos-1,sum-i,limit&&i==a[pos],1);
else tmp+=dfs(pos-1,sum,limit&&i==a[pos],fg);
}
if(!limit) dp[pos][sum][fg]=tmp;
return tmp;
}
ll solve(ll x)
{
if(!x) return 0;
int pos=0;
while(x)
{
a[pos++]=(x&1);
x>>=1;
}
ll ans=0;
ans=dfs(pos-1,50,1,0);
//cout<