E - Round Numbers POJ - 3252

最新推荐文章于 2021-11-03 18:03:25 发布

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在没有手指的情况下，牛群采用了一种独特的决策方式：通过选择一个二进制表示中0的数量多于或等于1的数量的圆数来决定胜负。本篇详细介绍了如何通过数位动态规划的方法，计算在给定范围内满足条件的圆数数量，帮助一头名为Bessie的奶牛在比赛中作弊。

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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

题解：找区间中有多少个二进制中0的个数大于1的个数的数。

dp[i][j][k]代表区间长度为i，二进制中0和1的个数分别为j，k，然后对于给出来的数，将其转换为二进制然后进行数位dp即可，主要注意第一个1放的位置，这样就是避免前导零的多算。

#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
int a[50];
ll dp[50][50][50];
ll DFS(int pos,int zerosum,int onesum,int limit,int first)//first为第一个1放或者不放，为1表示已放，为0表示未放
{
    if(pos==0)return zerosum>=onesum;//返回逻辑值，满足则为1,否则为0
    if(!limit&&dp[pos][zerosum][onesum]!=-1)return dp[pos][zerosum][onesum];
    ll ans=0;
    int up=limit?a[pos]:1;//找到上限
    ans+=DFS(pos-1,first?zerosum+1:0,first?onesum:0,limit&&0==up,first?1:0);//这一位取0
    if(!limit||a[pos]==1)ans+=DFS(pos-1,zerosum,onesum+1,limit&&1==up,1);//这一位取1
    if(!limit)dp[pos][zerosum][onesum]=ans;
    return ans;
}
ll solve(ll n)
{
    ll top=0;
    while(n){
        a[++top]=n&1;
        n/=2;
    }
    return DFS(top,0,0,1,0);
}
int main()
{
    memset(dp,-1,sizeof(dp));
    ll l,r;
    cin>>l>>r;
    cout<<solve(r)-solve(l-1)<<endl;
    return 0;
}