POJ - 3252 Round Numbers

最新推荐文章于 2021-11-03 18:03:25 发布

brucehb

最新推荐文章于 2021-11-03 18:03:25 发布

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分类专栏：数位dp

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本文介绍了一种使用数位动态规划(DP)的方法来解决一个关于“圆数”匹配的游戏问题。在这个游戏中，两个参与者选择小于两亿的整数，如果两个数都是“圆数”，则第一个参与者获胜。一个正整数被称为“圆数”，如果它的二进制表示中零的数量不少于一的数量。文章详细解释了数位DP的实现，包括状态定义、递归函数和边界条件，并提供了一个C++代码示例。

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

思路：

数位dp。dp[i][j][k]，i表示第i位，j表示在i位前的高位中含有0的数量，k表示在i位前的高位中含有1的数量。

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;

int a[40], dp[40][40][40];

int dfs(int pos, int num0, int num1, bool limit)
{
    if (pos < 0) 
	{
		return num0 >= num1;
	}
    if (!limit && dp[pos][num0][num1] != -1) 
	{
		return dp[pos][num0][num1];
	}

    int up = limit ? a[pos] : 1;
	int result = 0;
    for (int i = 0; i <= up; i++)
	{
		int c = num1 > 0? num0 + (i == 0) : 0;
		int d = num1 + (i == 1);
        result += dfs(pos - 1, c, d, limit && i == up);
    }

    if (!limit)
	{
		dp[pos][num0][num1] = result;
	}

    return result;
}

int solve(int x)
{
    int pos = 0;
	while (x > 0)
	{
		a[pos] = x % 2;
		x >>= 1;
		pos++;
	}

    return dfs(pos - 1, 0, 0, true);
}

int main()
{
    int m, n;
    memset(dp, -1, sizeof(dp));
    while (~scanf("%d%d", &m, &n))
	{
        printf("%d\n", solve(n) - solve(m - 1));
    }

    return 0;
}