POJ 3252 Round Numbers (组合数学+数位dp)

Round Numbers

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 13888 Accepted: 5483

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

Source

USACO 2006 November Silver

利用组合数

计算满足二进制中0的个数大于1的个数的数，分两类：

（1）位数小于len时：长度为1~len-1，首位为1，所以1<=i<=len-2，j表示从i个数里取j个0，所以(i/2+1)<=j<=i，ans+=c[i][j];

（2）位数等于len时：需考虑最大上限，若第i位为1，可计算其为0的时候；若第i位为0，zero++,其影响在后几位中需取0的个数

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>#define maxn 2520typedef long long ll;using namespace std;int s,f,cnt;int c[35][35];int b[35];void bin(int x){cnt=0;while(x){b[++cnt]=x%2;x/=2;}return ;}int work(int x){int i,j,ans=0;bin(x);for(i=1;i<=cnt-2;i++){for(j=i/2+1;j<=i;j++){ans+=c[i][j];}}int z=0;for(i=cnt-1;i>=1;i--){if(b[i]){for(j=(cnt+1)/2-(z+1);j<=i-1;j++){ans+=c[i-1][j];}}elsez++;}return ans;}int main (){int i,j;    c[0][0]=1;for(i=1;i<35;i++){c[i][1]=i;c[i][0]=1;}for(i=2;i<35;i++){for(j=2;j<35;j++){c[i][j]=c[i-1][j-1]+c[i-1][j];}}scanf("%d %d",&s,&f);printf("%d\n",work(f+1)-work(s));return 0;}