Cerinta completa
A palindrome is a string that reads the same from left to right as it does from right to left.
Given a string, , of lowercase English letters, we define a -length rotation as cutting the first characters from the beginning of and appending them to the end of . For each , there are possible -length rotations (where ). See the Explanation section for examples.
Given and , find all -length rotations of ; for each rotated string, , print the maximum possible length of any palindromic substring of on a new line.
Input Format
The first line contains an integer, (the length of ).
The second line contains a single string, .
Constraints
Output Format
There should be lines of output, where each line contains an integer denoting the maximum length of any palindromic substring of rotation .
Sample Input 0
13
aaaaabbbbaaaa
Sample Output 0
12
12
10
8
8
9
11
13
11
9
8
8
10
Sample Input 1
7
cacbbba
Sample Output 1
3
3
3
3
3
3
3
Sample Input 2
12
eededdeedede
Sample Output 2
5
7
7
7
7
9
9
9
9
7
5
4
Explanation
Consider Sample Case 1, where .
The possible rotations, , for string are:
.
The longest palindromic substrings for each are:
and , so we print their length () on a new line.
, so we print its length () on a new line.
and , so we print their length () on a new line.
and , so we print their length () on a new line.
and , so we print their length () on a new line.
and , so we print their length () on a new line.
and , so we print their length () on a new line.
Limbajul de programare folosit: cpp14
Cod:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void solve(char *str,int *a);
void init( int n );
void range_increment( int i, int j, int val );
int query( int i );
int max(int x,int y);
void update(int x,int y,int z);
void sort_a2(int*a,int*b,int size);
void merge2(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int left_size, int right_size);
char str[1000001]={0};
int N,NN,a[2000004],tree[2000000],ans[500000],b[500000],c[500000];
int main(){
int i,j;
scanf("%d%s",&NN,str);
strncpy(str+NN,str,NN);
solve(str,a);
init(NN);
for(i=0;i<4*NN;i++)
if(a[i])
if(i%2)
update(i/2-a[i]/2,i/2+a[i]/2,a[i]);
else
update(i/2-a[i]/2,i/2+a[i]/2-1,a[i]);
for(i=0;i<NN;i++){
ans[i]=query(i);
b[i]=ans[i];
c[i]=i;
}
sort_a2(b,c,NN);
for(i=NN;i>=0;i--){
for(j=c[i];1;j=(j-1+NN)%NN)
if(ans[j]-ans[(j-1+NN)%NN]>2)
ans[(j-1+NN)%NN]=ans[j]-2;
else
break;
for(j=c[i];1;j=(j+1)%NN)
if(ans[j]-ans[(j+1)%NN]>2)
ans[(j+1)%NN]=ans[j]-2;
else
break;
}
for(i=0;i<NN;i++)
printf("%d\n",ans[i]);
return 0;
}
void solve(char *str,int *a){
char *p;
int len,R,Ri,i,j,mi;
len=strlen(str);
p=(char*)malloc(2*(len+1)*sizeof(char));
for(i=0;i<len;i++){
p[2*i]='#';
p[2*i+1]=str[i];
}
p[2*i]='#';
p[2*i+1]=0;
a[0]=R=Ri=0;
for(i=1;i<=len*2;i++)
if(i>=R){
if(p[i]!='#')
a[i]=1;
else
a[i]=0;
for(j=i+1;1;j++)
if(j<=2*len && 2*i-j>=0 && p[j]==p[2*i-j]){
if(p[j]!='#')
a[i]+=2;
}
else{
Ri=i;
R=j-1;
break;
}
}
else{
mi=2*Ri-i;
if(i+a[mi]>=R || mi==a[mi]){
a[i]=R-i;
for(j=R+1;1;j++)
if(j<=2*len && 2*i-j>=0 && p[j]==p[2*i-j]){
if(p[j]!='#')
a[i]+=2;
}
else{
Ri=i;
R=j-1;
break;
}
}
else
a[i]=a[mi];
}
free(p);
return;
}
void init( int n ){
N = 1;
while( N < n ) N *= 2;
int i;
for( i = 1; i < N + n; i++ ) tree[i] = 0;
}
void range_increment( int i, int j, int val ){
for( i += N, j += N; i <= j; i = ( i + 1 ) / 2, j = ( j - 1 ) / 2 )
{
if( i % 2 == 1 ) tree[i] = max(tree[i],val);
if( j % 2 == 0 ) tree[j] = max(tree[j],val);
}
}
int query( int i ){
int ans = 0,j;
for( j = i + N; j; j /= 2 ) ans = max(ans,tree[j]);
return ans;
}
int max(int x,int y){
return (x>y)?x:y;
}
void update(int x,int y,int z){
if(z>NN){
int m=x+z/2;
if(z%2)
if(NN%2)
update(m-NN/2,m+NN/2,NN);
else
update(m-NN/2+1,m+NN/2-1,NN-1);
else
if(NN%2)
update(m-NN/2,m+NN/2-1,NN-1);
else
update(m-NN/2,m+NN/2-1,NN);
}
if(y<NN){
range_increment(0,x,z);
range_increment(y+1,NN-1,z);
}
else
range_increment(y-NN+1,x,z);
return;
}
void sort_a2(int*a,int*b,int size){
if (size < 2)
return;
int m = (size+1)/2,i;
int*left_a,*left_b,*right_a,*right_b;
left_a=(int*)malloc(m*sizeof(int));
right_a=(int*)malloc((size-m)*sizeof(int));
left_b=(int*)malloc(m*sizeof(int));
right_b=(int*)malloc((size-m)*sizeof(int));
for(i=0;i<m;i++){
left_a[i]=a[i];
left_b[i]=b[i];
}
for(i=0;i<size-m;i++){
right_a[i]=a[i+m];
right_b[i]=b[i+m];
}
sort_a2(left_a,left_b,m);
sort_a2(right_a,right_b,size-m);
merge2(a,left_a,right_a,b,left_b,right_b,m,size-m);
free(left_a);
free(right_a);
free(left_b);
free(right_b);
return;
}
void merge2(int*a,int*left_a,int*right_a,int*b,int*left_b,int*right_b,int left_size, int right_size){
int i = 0, j = 0;
while (i < left_size|| j < right_size) {
if (i == left_size) {
a[i+j] = right_a[j];
b[i+j] = right_b[j];
j++;
} else if (j == right_size) {
a[i+j] = left_a[i];
b[i+j] = left_b[i];
i++;
} else if (left_a[i] <= right_a[j]) {
a[i+j] = left_a[i];
b[i+j] = left_b[i];
i++;
} else {
a[i+j] = right_a[j];
b[i+j] = right_b[j];
j++;
}
}
return;
}
Scor obtinut: 1.0
Submission ID: 464648495
Link challenge: https://www.hackerrank.com/challenges/circular-palindromes/problem