codeforces215E（數位DP，規律水過）

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.

題意：尋找週期數，例如10的二進制是1010，是以10爲循環節的週期數。

思路：通過確定第一節來找週期數，這題核心在於去重。

           例如以2個二進制數爲一小節生成的週期數11 11 11 11、10 10 10 10與以4個二進制數爲一小節生成的週期數1111 1111、1010 1010重複。

           自己試了多種方法，但每次都不能很好的將置頂情況與不置頂情況分開。

           看了下別人的思路，每次都重置數組來去重，略犀利。

代碼：

#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)  //對於置頂情況的處理
{
    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;  //這裏y是把極值生成了
    return x-(1<<(k-1))+1-(y>m);  //將極限值與限定值相比較，從而判斷極值是否在限定範圍內
}
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));  //每次重置數組，這樣可以對付多種情況。我做這題時只想到了開二進制來去重，沒想到滾動更方便
        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++)  //這裏去重
                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;
}