Round Numbers - POJ 3252 数位dp

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最新推荐文章于 2021-11-03 18:03:25 发布

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分类专栏： POJ dp动态规划和递推

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本文介绍了一种使用数位DP算法解决特定问题的方法，即计算给定区间内二进制表示中0的数量不少于1的数量的整数个数。

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Round Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8971		Accepted: 3179

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 12

Sample Output

题意：找到一个区间内的在二进制中0比1的个数多的数字的个数。

思路：数位dp，具体实现看代码。

#include<cstdio>
#include<cstring>
using namespace std;
int dp[100][200],num[100];
//pos代表当前要处理的位置，l代表0比1多的数目，因为会为负数，所有加上100
//f为是否有大小限制，p为是否到目前位置，前面处理的数都是0
int dfs(int pos,int l,int f,int p)
{ int i,j,k,m,ans=0;
  if(pos+l<100)
   return 0;
  if(pos==0)
  { if(l>=100)
     return 1;
    else
     return 0;
  }
  if(f==1)
   m=num[pos];
  else
   m=1;
  if(f==0 && p==0 && dp[pos][l]!=-1)
   ans=dp[pos][l];
  else
  for(i=0;i<=m;i++)
  { if(i==0)
     k=1;
    else
     k=-1;
    if(f==1 && i==m)
    { if(p==1 && i==0)
       ans+=dfs(pos-1,l,1,1);
      else
       ans+=dfs(pos-1,l+k,1,0);
    }
    else
    { if(p==1 && i==0)
       ans+=dfs(pos-1,l,0,1);
      else
       ans+=dfs(pos-1,l+k,0,0);
    }
  }
  if(f==0 && p==0)
   dp[pos][l]=ans;
  return ans;
}
int solve(int n)
{ if(n<0)
   return 0;
  int i,j,k,ret=n,len=0;
  while(ret)
  { num[++len]=ret%2;
    ret/=2;
  }
  return dfs(len,100,1,1);
}
int main()
{ int l,r;
  memset(dp,-1,sizeof(dp));
  scanf("%d%d",&l,&r);
  printf("%d\n",solve(r)-solve(l-1));
}