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- #include <iostream>
- #include <string>
- #include <vector>
- #include <algorithm>
- using namespace std;
- string s;
- long long n = 4010, p, m = 5000007;
- vector <unsigned long long> paw(n), h(n), rh(n), evenCount(n), oddCount(n), r({ 10000, 0 });
- vector <vector <unsigned long long>> dp(n, r);
- auto getHash(int L, int R) {
- auto result = h[R];
- if (L > 0) result -= h[L - 1];
- return result;
- }
- auto getRevHash(int L, int R) {
- auto result = rh[L];
- if (R < n - 1) result -= rh[R + 1];
- return result;
- }
- void rek(int prev, int tek) {
- if (tek != 0) {
- rek(tek, dp[tek][1]);
- }
- for (int i = tek; i < prev; i++) {
- cout << s[i];
- }
- cout << " ";
- }
- bool isPalindrome(int L, int R) {
- return (getHash(L, R) * paw[n - R - 1]) % m == (getRevHash(L, R) * paw[L]) % m;
- }
- int main()
- {
- iostream::sync_with_stdio(false);
- cin.tie(0), cout.tie(0);
- cin >> s;
- n = s.size(), p = 239017;
- paw[0] = 1;
- for (int i = 1; i < n; i++) {
- paw[i] = (paw[i - 1] * p) % m;
- }
- h[0] = s[0];
- for (int i = 1; i < n; i++) {
- h[i] = h[i - 1] + (paw[i] * s[i]) % m;
- }
- rh[n - 1] = s[n - 1];
- for (int i = n - 2, j = 1; i >= 0; i--, j++) {
- rh[i] = rh[i + 1] + (paw[j] * s[i]) % m;
- }
- for (long long i = 0; i < (n - 1); i++) {
- int left = 0, right = min(i + 1, n - i - 1);
- while (left < right) {
- int middle = (left + right) / 2;
- if (isPalindrome(i - middle, i + middle + 1)) {
- evenCount[i] = middle + 1;
- left = middle + 1;
- }
- else {
- right = middle;
- }
- }
- }
- for (long long i = 0; i < n; i++) {
- int left = 0, right = min(i + 1, n - i);
- while (left < right) {
- int middle = (left + right) / 2;
- if (isPalindrome(i - middle, i + middle)) {
- oddCount[i] = middle;
- left = middle + 1;
- }
- else {
- right = middle;
- }
- }
- }
- dp[0][0] = 0;
- dp[0][1] = 0;
- dp[1][0] = 1;
- dp[1][1] = 0;
- for (int i = 1; i <= n; i++) {
- for (int j = 1; j <= evenCount[i - 1]; j++) {
- if (dp[i - j][0] + 1 < dp[i + j][0]) {
- dp[i + j][0] = dp[i - j][0] + 1;
- dp[i + j][1] = i - j;
- }
- }
- for (int j = 0; j <= oddCount[i - 1]; j++) {
- if (dp[i - j - 1][0] + 1 < dp[i + j][0]) {
- dp[i + j][0] = dp[i - j - 1][0] + 1;
- dp[i + j][1] = i - j - 1;
- }
- }
- }
- cout << dp[n][0] << "\n";
- rek(n, n);
- }
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