Uva11127-triple-free binary strings (DFS + bit operation)

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Question:A string with the length of N is given. The string consists of '1', '0 ','*', Wherein '*'Can be replaced with '1', '0' at will to find the maximum number of strings without three consecutive substrings.

Ideas:We can use binary to represent strings for DFS. The Left shift of bitwise operations indicates that '1' is placed on the '*' position. Note that three consecutive identical substrings exist in the recursion process.

Code:

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> using namespace STD; const int maxn = 32; char STR [maxn]; int N, ans; bool judge (int s, int d) {int K = (1 <D)-1; int A = S & K; S = s> D; int B = S & K; S = s> D; int c = S & K; if (a = B & B = c) Return true; else return false;} // use the bitwise operation to shift the D bit to the right to determine whether three substrings are the same void DFS (int s, int CNT) {int num = CNT/3; int t = n-CNT-1; for (INT I = 1; I <= num; I ++) {If (Judge (S> (t + 1 ), i) return;} If (CNT = N) {ans ++; return;} else if (STR [CNT] = '0') {DFS (S, CNT + 1);} else if (STR [CNT] = '1') {DFS (S + (1 <t), CNT + 1 );} else if (STR [CNT] = '*') {DFS (S, CNT + 1); DFS (S + (1 <t), CNT + 1 );} return;} int main () {int CAS = 1; while (scanf ("% d", & N) {scanf ("% s", STR ); ans = 0; DFS (0, 0); printf ("case % d: % d \ n", CAS ++, ANS);} return 0 ;}