poj 3252 Round Numbers

来源：互联网发布：geo数据库使用方法编辑：程序博客网时间：2024/05/16 05:10

Round Numbers

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 8194 Accepted: 2865

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 ofN 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 rangeStart..Finish

Sample Input

2 12

Sample Output

解题思路：

1、一个数n,长度为len的二进制数，先求出长度为len-1的round number；用排列组合求解，0的个数要大于等于长度的一半。切记：最高位不为0。

2、从次高位开始枚举，（若该位置为1，则把它改为0，则改变后的数必定小于n. 如：11***** 改为10*****）0的个数大于等于（len+1）/2,用排列组合计算。

#include<stdio.h>#define N 35#define Max(a,b) ((a)>(b)?(a):(b))int c[N][N];void inti(){int i,j;for(i=0;i<33;i++) //c[0][0]=1;当枚举到某个二进制数末位为0正好为round number，sum+=c[0][0]应该加一   {c[i][0]=c[i][i]=1; for(j=1;j<i;j++)c[i][j]=c[i-1][j]+c[i-1][j-1];}}int solve(int n){int i,j,len=0,sum=0;int bit[N];while(n){bit[++len]=n%2;n/=2;}for(i=1;i<len;i++)         //1、长度小于len时for(j=(i+1)/2;j<i;j++) //0的个数大于等于总数的一半，且不能全为0sum+=c[i-1][j];    //除去最高位1，i-1个位置，j个0 int one=1,zero=0;for(i=len-1;i>0;i--)       //2、枚举长度为len时{if(bit[i])    //这一位是1，改为0, {zero++;for(j=Max(0,(len+1)/2-zero);j<i;j++)   //0的个数不能为负sum+=c[i-1][j];       //第i位至len位已经固定，zero--;one++;}elsezero++;}return sum;}int main(){int a,b;inti();while(scanf("%d%d",&a,&b)!=-1){int t=solve(b+1)-solve(a);printf("%d\n",t);}return 0;}
                                                     0        0           	
					
					   poj 3252--Round Numbers
	  	   poj 3252 Round Numbers
	  	   poj 3252 Round Numbers
	  	   POJ 3252 Round Numbers
	  	   POJ 3252 Round Numbers
	  	   poj 3252 Round Numbers
	  	   poj-3252-Round Numbers
	  	   POJ 3252 Round Numbers
	  	   POJ 3252 Round Numbers
	  	   POJ 3252 Round Numbers
	  	   POJ 3252 Round Numbers
	  	   POJ 3252 Round Numbers
	  	   POJ 3252   Round Numbers
	  	   POJ 3252 Round Numbers
	  	   POJ 3252  Round Numbers
	  	   poj 3252 Round Numbers
	  	   poj 3252 Round Numbers
	  	   POJ 3252 Round Numbers
	     		  
	  	   jQuery Validate
	  	   20140409 Complex again（for lab）
	  	   KM算法
	  	   YARN/MRv2 Node Manager深入剖析—整体架构
	  	   URAL 1450 无环图最短路
	  	   poj 3252 Round Numbers
	  	   BNU 29379斩
	  	   ubuntu12.04 安装2013QQ
	  	   windows7,windows8,ubuntu三系统64位安装
	  	   单例模式的实现技巧
	  	   How advertising cookies let observers follow you across the web
	  	   jdbc操作mysql
	  	   阿里云CentOS 6.3 64位 配置PHP环境
	  	   2014-04-09工作日志:error：出现了一个问题，将导致程序停止工作。