题解【POJ3252】Round Numbers

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

6

Source

USACO 2006 November Silver

Solution

简化版题意：求出一个区间[a,b]中有多少个“Round Number”，一个数是“Round Number”当且仅当它的二进制表示法中0的个数>=1的个数，其中\(1 \leqslant A, B \leqslant 2,000,000,000\)。

我们可以根据题意先写一个简单的暴力：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

inline int gi()

{

	int f = 1, x = 0;

	char c = getchar();

	while (c < '0' || c > '9')

	{

		if (c == '-')

			f = -1;

		c = getchar();

	}

	while (c >= '0' && c <= '9')

	{

		x = x * 10 + c - '0';

		c = getchar();

	}

	return f * x;

}

//以上不解释

inline bool pd(int x)//判断一个数是不是"Round Number"

{

	int y = 0, z = 0;//y是存二进制数中1的个数，z是存0的个数

	while (x > 0)

	{

		if (x & 1)//如果这一位是0

		{

			++y;//y就加1

		}

		else

		{

			++z;//否则z就加1

		}

		x = x >> 1;//x除以2

	}

	return z >= y;//这个语句的意思是：如果z>=y，就返回true，否则返回false。

}

int a, b, sum;//a、b是题目中的意思，sum是答案

int main()

{

	a = gi(), b = gi();//输入a、b

	for (int i = a; i <= b; i++)//从a到b枚举

	{

		if (pd(i))//如果i是“Round Number"

		{

			++sum;//sum就加一

		}

	}

	printf("%d", sum);//最后输出sum

	return 0;//结束

}

因为\(1 \leqslant A, B \leqslant 2,000,000,000\)，很明显，以上代码小数据能AC，但是大数据会TLE。

因此，我们要使用一个更加高效的算法——数位DP。

什么是数位DP呢？可以参考这篇文章：http://www.cnblogs.com/real-l/p/8540124.html

回到这一题：

我们设Rn[n,m]表示区间[n,m]中Round Number的个数，我们利用前缀和，就有：

Rn[a,b] = Rn[0, b] - Rn[0, a - 1]

记忆化搜索思路：

设dp[a][n0][n1]表示从高往低到达第a位时含有n0个0和n1个1在后面任意填时该状态下的总个数。

注意加一个变量flag来判断是否含有前导0。

直接DP思路：

先预处理出dp[i][j]表示前i位有j个0的方案数，然后从高位数位到低位数位DP。

Code

记忆化搜索：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

inline int gi()

{

    int f = 1, x = 0;

    char c = getchar();

    while (c < '0' || c > '9')

    {

        if (c == '-')

            f = -1;

        c = getchar();

    }

    while (c >= '0' && c <= '9')

    {

        x = x * 10 + c - '0';

        c = getchar();

    }

    return f * x;

}

int dp[35][35][35], wei[35];

int dfs(int a, int n0, int n1, int ddd, int flag)

{

    if (a == 0)

    {

        if (flag == 0)//这里不判断flag==0也可以，判不判断的区别在于是不是把0算上，判断就不把0算上了

        {

            if (n0 >= n1)

            {

            	return 1;

			}

            else

            {

            	return 0;

			}

        }

        return 0;

    }

    if (ddd == 0 && dp[a][n0][n1] != -1)

    {

        return dp[a][n0][n1];

    }

    int ed = ddd ? wei[a] : 1, ans = 0, nu0, nu1;

    for (int i = 0; i <= ed; i++)

    {

        if (flag && i == 0)

        {

            nu0 = nu1 = 0;

        }

        else

        {

            if (i == 0)

            {

                nu0 = n0 + 1, nu1 = n1;

            }

            else

            {

                nu0 = n0, nu1 = n1 + 1;

            }

        }

        ans = ans + dfs(a - 1, nu0, nu1, ddd && i == ed, flag && i == 0);

    }

    if (ddd == 0)

    {

        dp[a][n0][n1] = ans;

    }

    return ans;

}

int solve(int x)

{

    int tot = 0;

    while (x)

    {

        wei[++tot] = x & 1;

        x = x >> 1;

    }

    return dfs(tot, 0, 0, 1, 1);

}

int a, b;

int main()

{

    memset(dp, -1, sizeof(dp));

    a = gi(), b = gi();

    printf("%d\n", solve(b) - solve(a - 1));

    return 0;

}

动态规划代码：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

inline int gi()

{

    int f = 1, x = 0;

    char c = getchar();

    while (c < '0' || c > '9')

    {

        if (c == '-')

            f = -1;

        c = getchar();

    }

    while (c >= '0' && c <= '9')

    {

        x = x * 10 + c - '0';

        c = getchar();

    }

    return f * x;

}

int dp[40][40];

inline int getans(int x)

{

    int t = x, wei[40], len = 0, sum = 0, n0 = 0, n1 = 0;;

    while (t)

    {

        wei[++len] = t & 1;

        t = t >> 1;

    }

    for (int i = len - 1; i >= 1; i--)//这里先把第len位变为0，然后一次枚举最高的位数在第i位

    {

        for (int j = 0; j <= i - 1; j++)

        {

            if (j >= i - j)

            {

                sum = sum + dp[i - 1][j];

			}

        }

    }

    n1 = 1;

    for (int i = len - 1; i >= 1; i--)//这里是在第len位为1的情况下进行dp

    {

        if (wei[i] == 1)

        {

            if (i == 1)

            {

                if (n0 + 1 >= n1)

                {

                	++sum;

				}

            }

            else

            {

                for (int j = 0; j <= i - 1; j++)

                {

                    if (j + n0 + 1 >= n1 + i - 1 - j)

                    {

                        sum = sum + dp[i - 1][j];

                    }

                }

            }

            ++n1;

        }

        else

        {

        	++n0;

		}

    }

    return sum;

}

int a, b;

int main()

{

	a = gi(), b = gi();

    memset(dp, 0, sizeof(dp));

    dp[1][1] = 1, dp[1][0] = 1;

    for (int i = 1; i <= 32; i++)

    {

        for (int j = 0; j <= i; j++)

        {

            dp[i + 1][j] = dp[i + 1][j] + dp[i][j], dp[i + 1][j + 1] = dp[i + 1][j + 1] + dp[i][j];

        }

    }

    printf("%d", getans(b + 1) - getans(a));

    return 0;

}

巴特西