POJ - 3252 Round Numbers【数位dp】

最新推荐文章于 2021-05-28 22:57:56 发布

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最新推荐文章于 2021-05-28 22:57:56 发布

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在一场独特的博弈中，两头奶牛通过选择小于二十亿的整数进行比赛，如果两个数字都是“圆数”，则先手胜利。一个正整数被称为“圆数”，当其二进制表示中零的数量不少于一的数量。为了作弊，Bessie想计算给定范围内“圆数”的总数，这需要将数字转换为二进制并比较零和一的数量。

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Round Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16896		Accepted: 7010

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

Source

USACO 2006 November Silver

由于此题要求的是二进制，故分解位数时分解为2进制。

记录前导0，0的个数和1的个数即可。

#include "cstdio"
#include "cstring"
#include "queue"
#include "iostream"
#include "vector"
#include "algorithm"
#include "map"
using namespace std;
const int inf = 0x3f3f3f3f;
int dig[70];
int dp[70][70][70][2];
int dfs(int pos,int zero,int one,bool lead,bool lim)//zero0的个数，one1的个数，lead前导0
{
    if(pos==-1)return zero>=one&&lead;
    if(!(lim)&&dp[pos][zero][one][lead]!=-1)return dp[pos][zero][one][lead];
    int maxi=lim?dig[pos]:1;
    int ans=0;
    for (int i = 0; i <= maxi; ++i) {
        ans+=dfs(pos-1,zero+(!(i&1)&(lead)),one+(i&1),lead||i,lim&&i==maxi);
    }if(!lim)dp[pos][zero][one][lead]=ans;
    return ans;
}
int getn(int n)
{
    int cnt=0;
    while(n)
    {
        dig[cnt++]=n&1;
        n>>=1;
    }
    return dfs(cnt-1,0,0,0,1);
}
int main()
{
    int n,m;
    cin>>n>>m;
    memset(dp,-1, sizeof(dp));
    printf("%d\n",getn(m)-getn(n-1));
}