POJ 3252 Round Numbers（组合数学）

Round Numbers

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 10223 Accepted: 3726

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

     题意：问在闭区间[n,m]中有多少个数是round numbers，所谓round numbers就是把闭区间中的某一个十进制的数字转换成二进制后0的个数大于等于1的个数，那么这个数就是round numbers

<pre name="code" class="cpp">#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>using namespace std;int c[33][33] = {0};int bin[35];int n,m;void updata()   ///计算n个里面取m个的方法数{    for(int i=0;i<=32;i++)    {        for(int j=0;j<=i;j++)        {            if(j == 0 || i == j)            {                c[i][j] = 1;            }            else            {                c[i][j] = c[i-1][j-1] + c[i-1][j];            }        }    }}void upbin(int x)   /// 将要求的数转化为二进制数并且逆序存储{    bin[0] = 0;    while(x)    {        bin[++bin[0]] = x%2;        x = x / 2;    }    return ;}int qurry(int x)   ///计算0-n之间的Round Number的个数{    int sum = 0;    upbin(x);    ///求二进制长度小于len的所有二进制数中Round Number的个数      for(int i=1;i<bin[0]-1;i++)    {        for(int j=i/2+1;j<=i;j++)        {            sum += c[i][j];        }    }    int zero = 0;    ///求二进制长度等于len的所有二进制数中Round Number的个数      for(int i=bin[0]-1;i>=1;i--)    {        if(bin[i])   ///当前位的值为1        {            for(int j=(bin[0]+1)/2-(zero+1);j<=i-1;j++)  ///看懂这里就行了            {                sum += c[i-1][j];            }        }        else        {            zero++;        }    }    return sum;}int main(){    updata();    scanf("%d%d",&n,&m);    printf("%d\n",qurry(m+1)-qurry(n));    return 0;}