POJ 3252 Round Numbers

Round Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12824		Accepted: 4946

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

大致题意：
输入两个十进制正整数a和b，求闭区间 [a ,b] 内有多少个Round number
所谓的Round Number就是把一个十进制数转换为一个无符号二进制数，若该二进制数中0的个数大于等于1的个数，则它就是一个Round Number

//dp[pos][num],到当前数位pos,0的数量减去1的数量为num的方案数
//最终的num>=0才能判合法,中间num可能<0,so 32当成'0' 
//注意本题有前导零的影响 
#pragma comment(linker,"/STACK:10240000,10240000")
#include<cstdio>
#include<cstring>
using namespace std;
const int N=64;
int a[N],dp[N][N];
int dfs(int pos,int num,bool lead,bool limit){
    if(!pos) return num>=32;
    if(!limit && !lead && dp[pos][num]!=-1) return dp[pos][num];
    int up=limit?a[pos]:1;
    int ans=0;
    for(int i=0;i<=up;i++){
        if(lead && !i) ans+=dfs(pos-1,num,lead && !i,limit && i==a[pos]);////有前导零就不统计在内 
        else ans+=dfs(pos-1,num+(!i?1:-1),lead && !i,limit && i==a[pos]);
    }
    if(!limit && !lead) dp[pos][num]=ans;
    return ans;
}
int solve(int x){
    int pos=0;
    for(;x;x>>=1) a[++pos]=x&1;
    return dfs(pos,32,true,true);
}
int main(){
    memset(dp,-1,sizeof dp);
    for(int a,b;~scanf("%d%d",&a,&b);){
        printf("%d
",solve(b)-solve(a-1));
    }
    return 0;
}