poj 3252 Round Numbers 【数位dp】

Round Numbers

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 11086 Accepted: 4113

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：问你n-m的数字 满足二进制中0的个数不少于1的个数 的有多少个。

思路：记忆化，yes表示前面填好的位是否是当前数的上界，是的话，当前位只能填到bit[pos]，反之可以填到1。zero表示前面填的位是否全是0，因为前导0是不考虑的，需要清空情况0和1的状态重新搜。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#include <string>#define INF 1000000#define eps 1e-8#define MAXN (200000+10)#define MAXM (100000+10)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while((a)--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1#define PI acos(-1.0)#pragma comment(linker, "/STACK:102400000,102400000")#define fi first#define se secondusing namespace std;typedef pair<int, int> pii;int dp[40][40][40];//dp[i][j][k]表示 满足二进制长度为i 其中0的个数为j且1的个数为k的 合法数字int bit[40];int DFS(int pos, int num0, int num1, bool yes, bool zero){    if(pos == -1) return num0 >= num1;    if(!yes && dp[pos][num0][num1] != -1) return dp[pos][num0][num1];    int End = yes ? bit[pos] : 1;    int ans = 0;    for(int i = 0; i <= End; i++)        ans += DFS(pos-1, zero&&i==0 ? 0 : num0+(i==0), zero&&i==0 ? 0 : num1+(i==1), yes&&i==End, zero&&i==0);    if(!yes) dp[pos][num0][num1] = ans;    return ans;}int Count(int n){    int len = 0;    while(n)    {        bit[len++] = n & 1;        n >>= 1;    }    return DFS(len-1, 0, 0, 1, 1);}int main(){    int n, m; CLR(dp, -1);    while(scanf("%d%d", &n, &m) != EOF) Pi(Count(m) - Count(n-1));    return 0;}