Zoj 3725 painting storages DP + sort count

Reprinted on http://blog.csdn.net/zu_xu/article/details/9401497

Recurrence: WHEN n <m, there is no method that meets the conditions. Therefore, the result is 0; when n = m, the number of solutions is exactly 1. N> M, if 1 to the N-1 to meet the conditions, N can be any dyeing, the number of C (N-1) * 2; otherwise, there must be 1 to the N-M-1 does not meet the conditions, n-M is blue, N-M + 1 to N-1 is red, then dye n red, the number of the solution is 2 ^ (N-M-1)-C (N-M-1 ), that is, the number of dyeing schemes from 1 to N-M-1 minus the number of solutions that meet the conditions.

 # include 
  
    # include 
   
     # define mod 1000000007 typedef long ll; ll pow2 [100000]; ll ITER [100001]; int main () {int n, m; * pow2 = 1; for (INT I = 1; I <100000; ++ I) pow2 [I] = (pow2 [I-1] * 2) % MOD; while (~ Scanf ("% d", & N, & M) {If (n