POJ 题目3252 Round Numbers（组合数学）

Round Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10254		Accepted: 3741

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

转自：http://blog.youkuaiyun.com/acm_cxlove/article/details/7855433

转载请注明出处，谢谢 http://blog.youkuaiyun.com/ACM_cxlove?viewmode=contents           by---cxlove

题目：一个数转化成二进制之后，0的个数大于等于1的为round数，问一个区间内有多少round数。

http://poj.org/problem?id=3252

有点像是数位DP的题目。

转化成二进制后只有01两种情况，我们可以直接统计。

我们统计比N小的round数有多少。

对于长度比N小的数，比较简单，如果长度为L,那么高位肯定是1，然后枚举0的个数，利用组合数就能解决。

然后是长度和N一样，但是比N小的。

我们从高位开始枚举，如果出现1，则把这位看作0，那么枚举之后的低位，肯定是比原数小的。

依次下去，注意统计好高位已经出现的0，1个数。

[cpp] view plain copy print ?

1. #include<iostream>   
2. #include<cstdio>   
3. #include<ctime>   
4. #include<cstring>   
5. #include<cmath>   
6. #include<algorithm>   
7. #include<cstdlib>   
8. #include<vector>   
9. #define C    240   
10. #define TIME 10   
11. #define inf 1<<25   
12. #define LL long long   
13. using namespace std;  
14. int c[35][35];  
15. void Init(){  
16.     for(int i=0;i<33;i++){  
17.         c[i][0]=c[i][i]=1;  
18.         for(int j=1;j<i;j++)  
19.             c[i][j]=c[i-1][j]+c[i-1][j-1];  
20.     }  
21. }  
22. int slove(int n){  
23.     int len=0,bit[35];  
24.     while(n){  
25.         bit[++len]=n%2;  
26.         n/=2;  
27.     }  
28.     int sum=0;  
29.     for(int i=1;i<len;i++)  
30.         for(int j=(i+1)/2;j<i;j++)  
31.             sum+=c[i-1][j];  
32.     int one=1,zero=0;  
33.     for(int i=len-1;i;i--){  
34.         if(bit[i]){  
35.             //如果这位是1，则如果是0，枚举低位，肯定比原数小   
36.             zero++;  
37.             for(int j=max(0,(len+1)/2-zero);j<i;j++)  
38.                 sum+=c[i-1][j];  
39.             //统计结束后，要恢复   
40.             zero--;  
41.             one++;  
42.         }  
43.         else  
44.             zero++;  
45.     }  
46.     return sum;  
47. }  
48. int main(){  
49.     int l,r;  
50.     Init();  
51.     while(cin>>l>>r)  
52.         cout<<slove(r+1)-slove(l)<<endl;  
53.     return 0;  
54. }