[Digital DP] spoj 10606 balanced numbers

Question:

For each bit of a number.

For each odd number, an even number must appear.

For each even number, an odd number of times is required.

Ideas:

Store the status of each number in a three-in-one mode. values 0 and 1 are odd and 2 are even.

Then the others are the same as the common digit DP.

Code:

# Include "cstdlib" # include "cstdio" # include "cstring" # include "cmath" # include "queue" # include "algorithm" # include "iostream" using namespace STD; // February 10: 03: 08 long DP [22] [180000]; int num [22]; int OK (INT X) {int V [50], CNT = 0; while (x) {v [CNT ++] = x % 3; X/= 3 ;}for (INT I = 0; I <CNT; I ++) {If (! V [I]) continue; if (I % 2 = 0 & V [I]! = 1) return 0; if (I % 2! = 0 & V [I]! = 2) return 0;} return 1;} int JS (int n, int X) {int V [50], CNT = 0, TEP = 0; memset (V, 0, sizeof (v); While (n) {v [CNT ++] = n % 3; N/= 3;} If (V [x] <2) V [x] ++; else V [x] = 1; for (INT I = 9; I> = 0; I --) TEP = TEP * 3 + V [I]; return TEP;} long DFS (INT site, int N, int zero, int f) {If (Site = 0) {If (zero) return 0; Return OK (n);} If (! F &&! Zero &&~ DP [site] [N]) return DP [site] [N]; int Len = f? Num [site]: 9; long ans = 0; For (INT I = 0; I <= Len; I ++) {If (zero) {if (I = 0) ans + = DFS (site-1, n, zero & I = 0, F & I = Len ); else ans + = DFS (site-1, JS (n, I), zero & I = 0, F & I = Len );} else ans + = DFS (site-1, JS (n, I), zero & I = 0, F & I = Len);} If (! F &&! Zero) DP [site] [N] = ans; return ans;} long solve (long x) {int CNT = 0; while (X) {num [++ CNT] = x % 10; X/= 10;} return DFS (CNT, 0, 1);} int main () {int T; cin> T; memset (DP,-1, sizeof (DP); While (t --) {long X, Y; scanf ("% LLD ", & X, & Y); printf ("% LLD \ n", solve (y)-solve (x-1);} return 0;} // September 25, 2014 10:39:37

[Digital DP] spoj 10606 balanced numbers