CodeChef 3-Palindromes(Manacher+dp)

3-Palindromes   Problem code: PALIN3

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Mike likes strings. He is also interested in algorithms. A few days ago he discovered for himself a very nice problem:

You are given a digit string S. You need to count the number of substrings of S, which are palindromes.

Do you know how to solve it? Good. Mike will make the problem a little bit more difficult for you.

You are given a digit string S. You need to count the number of substrings of S, which are palindromes without leading zeros and can be divided by 3 without a remainder.

A string is a palindrome if it reads the same backward as forward. A
string is a palindrome without leading zeros if it reads the same
backward as forward and doesn't start with symbol '0'. A string is a
digit string, if it doesn't contain any symbols except '0', '1', '2',
..., '9'.

Please, note that you should consider string "0" as a palindrome without leading zeros.

Input

The first line of the input contains a digit string S.

Output

Your output should contain the only integer, denoting the number of substrings of S, which are palindromes without leading zeros and can be divided by 3 without a remainder.

Constraints

1 ≤ |S| ≤ 1 000 000

Example

Input:
1045003

Output:
4

Explanation

In the example you should count S[2..2] = "0", S[5..5] = "0", S[6..6] = "0" and S[7..7] = "3".

给出一个数字串。

问有多少个子串既是回文串也能被3整除~

先用Manacher处理好串的回文串长度。

然后用一个数组 cnt[i][j] 表示sigma( 1 ~  i-1 到i 的数 ) %3 == j 串的个数。

#include <bits/stdc++.h>

using namespace std;
typedef long long LL ;
typedef pair<LL,LL> pii;
#define X first
#define Y second
const int N = ;
char Ma[N] , s[N];
int Mp[N] , len ;
void Manacher( char s[] , int len ) {
    int l =  ;
    Ma[l++] = '$' ; Ma[l++] = '#' ;
    for( int i =  ; i < len ; ++i ) {
        Ma[l++] = s[i];
        Ma[l++] = '#' ;
    }
    Ma[l] =  ; int mx =  , id =  ;
    for( int i =  ; i < l ; ++i ) {
        Mp[i] = mx>i?min(Mp[*id-i],mx-i):;
        while( Ma[i+Mp[i]] == Ma[i-Mp[i]] ) {
            Mp[i]++;
        }
        if( i + Mp[i] > mx ) {
             mx = i + Mp[i];
            id = i ;
        }
    }
}

bool is_dig( char op ) {
    if( op >= '' && op <= '' ) return true ;
    return false ;
}
LL cnt[N][] , sum[N] ;

void Run() {
    int n = strlen(s) ;
    Manacher(s,n);
    len =  * n +  ;
    memset( sum ,  , sizeof sum );
    memset( cnt ,  , sizeof cnt );
    for( int i =  ; i < len ; ++i ){
           sum[i] = sum[i-];
        if( is_dig(Ma[i]) ) sum[i] += ( Ma[i] - '' );
    }
    for( int i =  ; i < len ; ++i ) {
        if( !is_dig(Ma[i]) || Ma[i] == '' ) {
            for( int j =  ; j <  ; ++j )
                cnt[i][j] = cnt[i-][j];
        }
        else {
            int x = Ma[i] - '' ;
            for( int j =  ; j <  ; ++j ) {
                int _j = (j+x)%;
                cnt[i][_j] += cnt[i-][j] ;
            }
            cnt[i][x%]++;
        }
    }
    LL ans =  ;
    for( int i =  ; i < len ; ++i ) {
        int x =  , tmp = sum[i-] - sum[i-Mp[i]] ;
        if( is_dig(Ma[i]) ) {
            x = Ma[i] - '' ;
            if( x %  ==  ) ans++ ;
        }
        for( int j =  ; j <  ; ++j ) {
            if( ( *j + x )% ==  ) {
                ans += cnt[i-][j];
                for( int z =  ; z <  ; ++z ) if( (z+tmp)% == j ){
                    ans -= cnt[i-Mp[i]][z];
                }
            }
        }
    }
    printf("%lld\n",ans);
}

int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    int _ , cas =  ;
    while( scanf("%s",s) != EOF )Run();
}