A. Today's Word
time limit per test: 1 second
memory limit per test: 256 megabytes
You have accepted an offer at VortaroEnMano Inc., a company committed to creating the most comprehensive Esperanto dictionary. Esperanto estas tre mojosa lingvo, so you work really hard to do your best -- mostly to keep your job in the depression of employment.
Today, you're assigned to refactor the function called "Hodiaŭa Vorto", i.e. "Today's Word" in English. This word is generated from a string, denoted as .
Here's how is generated:
1.The process begins with an initial string, , which is given. This string contains only lowercase English letters and has an even length.
2.For n≥1, is generated in the following way:
, where
is the length of
and +
is used to concatenate the strings. Note that the index of the string starts from 0.
The function next(S) increments each character in the string S to the next letter in the alphabet, i.e., a changes to b , b to c, and so on, with z changing to a. For instance, next(abz)=bca.
Your task is to determine the suffix of with a length of m.
Input
The first line contains two integers n
and m (1≤n,m≤), representing the length of
and the length of the desired suffix, respectively. It is guaranteed that n is an even number.
The second line contains a string, , composed of n lowercase English letters.
Output
Output a string of length m, which represents the suffix you are tasked to determine.
Example
input
6 10
bocchi
output
wrwxrwxsxy
Note
In the provided example, S1=bocbocchidij.
【思路分析】
递推+构造+模拟。直接模拟，从的后缀暴力构造，由于每次长度倍增，时间复杂度为
.
cpp
#include <iostream>
#include <vector>
#include <unordered_map>
#include <map>
#include <cmath>
#include <algorithm>
#include <climits>
#include <stack>
#include <cstring>
#define i64 long long
using namespace std;
void getString(string &s) {
for (auto &item: s) {
if (item + 16 <= 'z') item = item + 16;
else item = 16 - ('z' - item) + 'a' - 1;
}
}
void moveForwardString(string &s) {
for (auto &item: s) {
if (item - 1 < 'a') item = 'z';
else item -= 1;
}
}
void moveBackString(string &s) {
for (auto &item: s) {
if (item + 1 > 'z') item = 'a';
else item += 1;
}
}
void solve() {
i64 n, m;
cin >> n >> m;
string s, tmp;
cin >> s;
string sub = s.substr(n / 2, n / 2);
getString(sub);
int cnt = 4;
for (int i = 4 * n; i >= n / 2; i -= n / 2) {
tmp.insert(0, sub);
if (cnt == 4 || cnt == 2) moveForwardString(sub);
cnt--;
if (cnt == 0) {
cnt = 4;
moveBackString(sub);
}
}
for (int i = 0;i < 25;i++) {
cnt = 4;
sub = tmp;
if (tmp.length() >= m) break;
tmp.clear();
for (int i = 4 * n; i >= n / 2; i -= n / 2) {
tmp.insert(0, sub);
if (cnt == 4 || cnt == 2) moveForwardString(sub);
cnt--;
if (cnt == 0) {
cnt = 4;
moveBackString(sub);
}
}
}
cout<<tmp.substr(tmp.length()-m,m);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int t = 1;
// cin >> t;
while (t--) {
solve();
}
return 0;
}