POJ 3252：Round Numbers

POJ 3252：Round Numbers

Time Limit: 2000MS
Memory Limit: 65536K
Total Submissions: 10099
Accepted: 3669
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

题目给了一个区间，要求的是这个区间之内有多少个数，这种数满足转为二进制时0的数量大于等于1的数量。
也是头一次接触数位DP，捡起了很久没有用的C[j][i] = C[j-1][i-1] + C[j-1][i]。这个高中时做排列组合题目时会用到的公式。
求[start,finnish]这个区间之内有多少数符合条件，这样做比较麻烦，因为要考虑到start与finnish的各种情况，不如求0到start有多少个符合条件的数，求0到finnish+1有多少符合条件的数，之后两者相减即是结果。
所以这样就归结为求从0到n，符合条件的数的数量。
首先求转成二进制之后，位数本身就小于n的数量。这个求的时候很简单，满足0大于等于一般即可。
之后是位数相等的，由于n第一位一定是1，所以从左至右看0的数量，如果面对的是1，那么假设其是0。(因为此时，假设是0之后后面所有求出符合条件的数都是小于n的，所以符合题意。)求后面满足0的数量占整个数量一半以上的有多少个，不断相加。如果是0，则0的数量++即可。

代码：

#include <iostream>
using namespace std;

int c[33][33];
int bin[35];

void init()
{
    int i,j;
    memset(c,0,sizeof(c));

    for(i=0;i<=32;i++)
    {
        for(j=0;j<=i;j++)
        {
            if(j==0||j==i)
                c[i][j]=1;
            else
                c[i][j]=c[i-1][j-1]+c[i-1][j];
        }
    }
}

void dec_to_bin(int x)
{
    memset(bin,0,sizeof(bin));
    while(x!=0)
    {
        bin[++bin[0]]=x&1;
        x=x>>1;
    }
}

int result(int x)
{
    dec_to_bin(x);

    int i,j;
    int sum=0;

    for(i=0;i<bin[0]-1;i++)
    {
        for(j=i/2+1;j<=i;j++)
        {
            sum +=c[i][j];
        }
    }

    int zero=0;

    for(i=bin[0]-1;i>=1;i--)
    {
        if(bin[i])
        {
            for(j=(bin[0]+1)/2-(zero+1);j<=i-1;j++)
                sum += c[i-1][j];
        }
        else
            zero++;
    }
    return sum;
}
int main()
{
    init();
    int a,b;
    cin>>a>>b;

    cout<<result(b+1)-result(a)<<endl;

    return 0;
}

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