codeforces215E（数位DP，规律水过）

地址：http://codeforces.com/contest/215/problem/E

E. Periodical Numbers

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

A non-empty string s is called binary, if it consists only of characters "0"
and "1". Let's number the characters of binary string s from
1 to the string's length and let's denote the i-th character in string s as si.

Binary string s with length n is periodical,
if there is an integer 1 ≤ k < n such that:

k is a divisor of number n

for all 1 ≤ i ≤ n - k, the following condition fulfills: si = si + k

For example, binary strings "101010" and "11" are periodical and "10" and "10010" are not.

A positive integer x is periodical, if its binary representation (without leading zeroes)
is a periodic string.

Your task is to calculate, how many periodic numbers are in the interval from l to r (both
ends are included).

Input

The single input line contains two integers l and r (1 ≤ l ≤ r ≤ 1018).
The numbers are separated by a space.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Output

Print a single integer, showing how many periodic numbers are in the interval from l to r (both
ends are included).

Sample test(s)

input

1 10

output

3

input

25 38

output

2

Note

In the first sample periodic numbers are 3, 7 and 10.

In the second sample periodic numbers are 31 and 36.

题意：寻找周期数，例如10的二进制是1010，是以10为循环节的周期数。

思路：通过确定第一节来找周期数，这题核心在于去重。

           例如以2个二进制数为一小节生成的周期数11 11 11 11、10 10 10 10与以4个二进制数为一小节生成的周期数1111 1111、1010 1010重复。

           自己试了多种方法，但每次都不能很好的将置顶情况与不置顶情况分开。

           看了下别人的思路，每次都重置数组来去重，略犀利。

代码：

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL __int64
LL num[100],dp[100];
LL getdp(int len,int k,LL m) //对于置顶情况的处理
{
LL x=0;
for(int i=0;i<k;i++) x<<=1,x+=num[len-i];
LL y=x;
for(int i=1;i<len/k;i++) y<<=k,y+=x; //这里y是把极值生成了
return x-(1<<(k-1))+1-(y>m); //将极限值与限定值相比较，从而判断极值是否在限定范围内
}
LL getans(LL m)
{
int len;
LL n=m,ans=0;
for(len=0;n;n>>=1) num[++len]=n&1;
for(int i=2;i<=len;i++)
{
memset(dp,0,sizeof(dp)); //每次重置数组，这样可以对付多种情况。我做这题时只想到了开二进制来去重，没想到滚动更方便
for(int j=1;j<i;j++)
{
if(i%j) continue;
if(i<len) dp[j]=1<<(j-1);
else dp[j]=getdp(len,j,m);
for(int k=1;k<j;k++) //这里去重
if(!(j%k)) dp[j]-=dp[k];
ans+=dp[j];
}
}
return ans;
}
int main()
{
LL l,r;
scanf("%I64d%I64d",&l,&r);
printf("%I64d\n",getans(r)-getans(l-1));
return 0;
}