poj 3252 Round Numbers(数位dp)

Round Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11681		Accepted: 4392

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

题意是求一个范围中的Round数，这种数的二进制形式中的0多于1。把数字转换为二进制，然后简单的搞数位dp。发现：数位dp中判断条件和0有关的话，要用一个flag来防止数字倒过来以后开头的0，这种0显然是不合适的。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <bitset>
using namespace std;
#define ll long long

int dig[33];
ll dp[33][33][33];

ll dfs(int pos, int cnt0, int cnt1, int flag, int lim) {
    //if(pos == -1) printf("%d %d
", cnt0, cnt1);
    if(pos == -1) return cnt0 > cnt1;
    if(!lim && !flag && dp[pos][cnt0][cnt1] != -1) return dp[pos][cnt0][cnt1];
    int End = lim ? dig[pos] : 1;
    ll ret = 0;
    for(int i = 0; i <= End; i++) {
        if(flag) {
            if(i) ret += dfs(pos - 1, 0, 0, 0, lim && (i == End));
            else ret += dfs(pos - 1, 0, 0, 1, lim && (i == End));
        }
        else {
            if(i) ret += dfs(pos - 1, cnt0, cnt1 + 1, 0, lim && (i == End));
            else ret += dfs(pos - 1, cnt0 + 1, cnt1, 0, lim && (i == End));
        }
    }
    if(!lim && !flag) dp[pos][cnt0][cnt1] = ret;
    return ret;
}

ll func(ll num) {
    int n = 0;
    //bitset<32> btt(num);
    while(num) {
        dig[n++] = num & 1;
        num >>= 1;
    }
    return dfs(n - 1, 0, 0, 1, 1);
}

int main() {
    ll n, m;
    memset(dp, -1, sizeof(dp));
    while(~scanf("%I64d %I64d", &n, &m)) {
        printf("%I64d
", func(m) - func(n - 1));
    }
}