UVA 517 - Word (周期规律+位运算)

Word

Dr. R. E. Wright's class was studying modified L-Systems. Let us explain necessary details. As a model let us have words of length n over a two letter alphabet 

.
The words are cyclic, this means we can write one word in any of n forms we receive by cyclic shift, whereby the first and the last letters in the word are considered to be neighbours.
Rewriting rules rewrite a letter at a position i, depending on letters at the positions i - 2, i, i+1. We rewrite all letters of the word in one step. When we have a given starting
word and a set of rewriting rules a natural question is: how does the word look after s rewriting steps?
Help Dr. R. E. Wright and write a program which solves this task.

Input 

There are several blocks in the input file, each describing one system. There is an integer number n, 2
< n< 16 the length of the input word in the first line. There is a word in the next line. The word contains only lowercase letters a and b.
There are four characters c1 c2 c3 c4 in
the next eight lines. Each quadruple represents one rewriting rule with the following meaning: when the letter at the position i - 2 is c1 and
the letter at the position i is c2 and
the letter at the position i + 1 is c3 then
the letter at the position i after rewriting will be c4.
Rewriting rules are correct and complete. There is an integer numbers, 

,
in the last line of the block.

Output 

There is one line corresponding to each block of the input file. The line contains a word which we receive after s rewriting
steps from the corresponding starting word using given rewriting rules. As we mentioned above, the word can be written in any of n cyclic shifted
forms. The output file contains the lexicographically smallest word, assuming that a < b.

Sample Input 

5
aaaaa
aaab
aabb
abab
abbb
baab
babb
bbab
bbbb
1

Sample Output 

bbbbb

题意：给定一个字符串，和8种变化方式，变化方式代表如果i - 2, i , i + 1位置字符对应的变化方式，把i位置变化为字符。所有字符只有a或b。给出s，问变化s次后的字符串是什么。

思路：s很大，但是n只有16，由于只有a和b，可以把a当成0，b当成1，这样就是一个二进制数，最多2^15种字符串，那么周期长度肯定不超过2^15，所以利用周期去查找。过程利用位运算。然后注意一个坑点，按字典序输出。因为题目给的是环形的字符串，所以比如bbbaaa是要输出aaabbb才对。

代码：位运算写得有点乱。。。不太熟练

#include <stdio.h>
#include <string.h>
#include <set>
#include <vector>
using namespace std;

int n, s, state, change, way[8], vis[(1<<16) + 10];
vector<int> zq;
void init() {
zq.clear();
memset(vis, -1, sizeof(vis));
state = 0;
char str[20];
scanf("%s", str);
for (int i = 0; i < n; i++)
state += (str[i] - 'a') * (1<<i);
for (int j = 0; j < 8; j++) {
change = 0;
scanf("%s", str);
for (int k = 0; k < 3; k++)
change = (change<<1) + str[k] - 'a';
way[change] = str[3] - 'a';
}
}

void tra() {
int save[20], i;
for (i = 0; i < n; i++) {
int change = 0;
change = (change<<1) + ((state&(1<<((i - 2 + n)%n)))?1:0);
change = (change<<1) + ((state&(1<<i))?1:0);
change = (change<<1) + ((state&(1<<((i + 1)%n)))?1:0);
save[i] = way[change];
}
state = 0;
for (i = 0; i < n; i++)
state += save[i] * (1<<i);
}

void solve() {
int num = 0;
int len;
scanf("%d", &s);
while (1) {
if (vis[state] != -1) {
len = num - vis[state];
break;
}
zq.push_back(state);
vis[state] = num++;
tra();
}
int st;
if (s <= vis[state]) st = zq[s];
else st = zq[(s - vis[state]) % len + vis[state]];
char ans[20] = {"bbbbbbbbbbbbbbb"};
char save[20];
memset(save, 0, sizeof(save));
for (int i = 0; i < n; i++) {
for (int j = i; j < i + n; j++) {
if (st&(1<<(j%n))) save[j - i] = 'b';
else save[j - i] = 'a';
}
if (strcmp(save, ans) < 0) strcpy(ans, save);
}
printf("%s\n", ans);
}

int main() {
while (~scanf("%d", &n)) {
init();
solve();
}
return 0;
}