codeforces215 E. Periodical Numbers（数位dp）

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 ≤ 10181 ≤ l ≤ r ≤ 10^{18}). 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 %I64d specifier.
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

Input

25 38

Output

Note

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

In the second sample periodic numbers are 31 and 36.

思路：假如说现在处理的是1-x，那么枚举所有可能的长度，然后枚举k 的大小，对于小于极限长度的长度，那么直接可以用1<<(j-1)算出，因为假如现在枚举的k为j，相当于处理最高位一样，剩下的可以随便变化；对于极限值，用cal计算

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=70;
int dig[maxn];
LL x,y;
LL dp[maxn];
LL cal(int len,int k,LL x)
{LL a=0,b=0;for(int i=0;i<k;i++)a+=(dig[len-i]<<(k-1-i));b=a;for(int i=1;i<len/k;i++)b<<=k,b+=a;return a-(1<<(k-1))+1/*此处加1是因为全0的原因*/-(b>x);
}
LL solve(LL x)
{int len=0;LL n=x;while(x){dig[++len]=x%2;x>>=1;}LL ans=0;for(int i=2;i<=len;i++){memset(dp,0,sizeof(dp));for(int j=1;j<i;j++){if(i%j)continue;if(i<len)dp[j]=(1<<(j-1));else dp[j]=cal(len,j,n);for(int k=1;k<j;k++)if(j%k==0)dp[j]-=dp[k];ans+=dp[j];}}return ans;
}
int main()
{scanf("%I64d%I64d",&x,&y);printf("%I64d\n",solve(y)-solve(x-1));return 0;
}

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