Malek Dance Club（递推）

Malek Dance Club

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

As a tradition, every year before IOI all the members of Natalia Fan Club are invited to Malek Dance Club to have a fun night together. Malek Dance Club has 2n members and coincidentally Natalia Fan Club also has 2nmembers. Each member of MDC is assigned a unique id i from 0 to 2n - 1. The same holds for each member of NFC.

One of the parts of this tradition is one by one dance, where each member of MDC dances with a member of NFC. A dance pair is a pair of numbers (a, b) such that member a from MDC dances with member b from NFC.

The complexity of a pairs' assignment is the number of pairs of dancing pairs (a, b) and (c, d) such that a < c andb > d.

You are given a binary number of length n named x. We know that member i from MDC dances with member  from NFC. Your task is to calculate the complexity of this assignment modulo 1000000007 (109 + 7).

Expression  denotes applying «XOR» to numbers x and y. This operation exists in all modern programming languages, for example, in C++ and Java it denotes as «^», in Pascal — «xor».

注意到n很小，如果能够求出递推公式，问题将很容易得到解决。

x长度为n，f(x)表示complexity，显然f（0）=0，f（1）=1。当n>1时，f（0x）和f(1x)可由f(x)推出。

（1）求f(0x): i 分别取 0,1,...,2^n-1，j分别取2^n,...,2^(n+1) - 1, 统计（i，i  xor 0x）与（j, j xor 0x）能组成多少对，注意j xor 0x的第一位是1，而i xor 0x的第一位是0，故而 j xor 0x > i  xor 0x，而 j > i，故（i，i  xor 0x）与（j, j xor 0x）不能配对。统计（j, j xor 0x）内部能组成多少对，所有j的第一位相同， 导致j xor 0x的第一位都相同，故而j的第一位是没有比较意义的，去掉没有影响，故（j, j xor 0x）的配对数为f(x)。所以f(0x)=2f(x)

(2)求f(1x):  i 分别取 0,1,...,2^n-1，j分别取2^n,...,2^(n+1) - 1, 统计（i，i  xor 1x）与（j, j xor 1x）能组成多少对，注意j xor 1x的第一位是0，而i xor 1x的第一位是1，故而 i  xor 1x > j xor 1x ，而 i < j，故（i，i  xor 1x）与（j, j xor 1x）之间能产生2^(2n)对。统计（j, j xor 0x）内部能组成多少对，所有j的第一位都相同， 导致j xor 1x的第一位都相同，故而j的第一位是没有比较意义的，去掉没有影响，故（j, j xor 0x）的配对数为f(x)。所以f(1x)=2f(x)+2^(2n)

综上：

• f(0x) = 2f(x)
• f(1x) = 2f(x) + 2^2n

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 #include <map>
 5 #include <vector>
 6 #include <cstdio>
 7 #include <cmath>
 8 #include <cstring>
 9 using namespace std;
10 
11 char x[101];
12 const int m = 1000000007;
13 int n;
14 
15 long long POW(long long a, long long b)
16 {
17     if(!b) return 1;
18     long long c = POW(a, b>>1);
19     c = (c * c) % m;
20     if(b & 1)
21     {
22         c = (c * a) % m;
23     }
24     return c;
25 }
26 
27 int f(int k)
28 {
29     if(k == n - 1)
30     {
31         if(x[k] == '0') return 0;
32         else return 1;
33     }
34     if(x[k] == '0') return (2 * f(k + 1)) % m;
35     else return ((2 * f(k + 1)) % m + POW(4, n - k - 1)) % m;
36 }
37 
38 int main()
39 {
40     while(scanf("%s", x) != EOF)
41     {
42         n = strlen(x);
43         printf("%d\n", f(0));
44     }
45     return 0;
46 }