poj 3252 RoundNumber

  題意是求把十進位數轉化成位元，0的個數大於等於1 的數，給定一個閉區間求出區間的這樣的數有多少個。

 

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

這是一個找規律的類比數學題，雖然WA了三次不過經過不懈的努力還是AC了。(*^__^*) 嘻嘻……

一定要注意細心，耐心。

代碼：

#include<iostream>using namespace std;int num[35],sum[35],numa[35],numb[35];int dp[35][35];int main(){    int a,b,i,j,la,lb,len,suma,sumb,cnt,k;    dp[0][0]=1;    for(i=1;i<=34;i++){dp[i][0]=1;dp[0][i]=0;}    for (i=1;i<34;++i)        for (j=1;j<=i;++j)            dp[i][j]=dp[i-1][j-1]+dp[i-1][j];    memset(sum,0,sizeof(sum));    for(i=2;i<=34;i++)    {      for(j=0;j<=i/2-1;j++)         sum[i]+=dp[i-1][j];    }    while(cin>>a>>b)    {      la=0;      int p=a;      while(p)      {        numa[la++]=p%2;        p=p/2;      }          suma=0;      k=0;      for(i=0;i<=la-1;i++) suma+=sum[i];      for(i=la-2;i>=0;i--)      {         if(numa[i]==1)         {           for(j=(la+1)/2-k-1;j<=i;j++)            suma+=dp[i][j];         }         else k++;      }      p=b+1;lb=0;      while(p)      {        numb[lb++]=p%2;        p=p/2;      }      sumb=0;k=0;      for(i=0;i<=lb-1;i++) sumb+=sum[i];      for(i=lb-2;i>=0;i--)      {          if(numb[i]==1)          {             for(j=(lb+1)/2-k-1;j<=i;j++)             {              sumb+=dp[i][j];             }          }          else k++;      }      cout<<sumb-suma<<endl;    }    return 0;}