题目链接：
codeforces 159D
题目大意：
给出一个字符出，求取这个字符串中互相不覆盖的两个回文子串的对数。
题目分析：统计枚举的结果就是最终的答案。 AC代码：;LL;int temp[MAX<<1];int Len[MAX<<1];int init ( char *st , int n ){int i;temp[0] = -1;for ( int i = 1 ; i <= 2*n ; i+=2 ){temp[i] = -2;temp[i+1] = st[i/2];}temp[2*n+1] = -2;temp[2*n+2] = -3;temp[2*n+3] = 0;return 2*n+1;}void manacher ( int *st , int len ){int mx = 0 , ans = 0 , po = 0;for ( int i = 1 ; i <= len ; i++ ){if ( mx > i )Len[i] = min ( mx – i , Len[2*po-i] );elseLen[i] = 1;while ( st[i-Len[i]] == st[i+Len[i]] )Len[i]++;if ( Len[i]+i > mx )mx = Len[i]+i , po = i;}}char s[MAX];LL a[MAX<<1],ans;int main ( ){while ( ~scanf ( “%s” , s ) ){ans = 0;memset ( a , 0 , sizeof ( a ) );int n = strlen ( s );manacher ( temp , init ( s, n ) );for ( int i = 1 ; i <= 2*n ; i++ ){int x = Len[i];if ( i&1 )for ( int j = 1 ; j < x ; j += 2 )a[i+j]++;elsefor ( int j = 0 ; j < x ; j += 2 )a[i+j]++;}a[0] = 0;for ( int i = 1 ; i <= 2*n ; i++ )a[i] += a[i-1];for ( int i = 1 ; i <= 2*n ; i++ ){int x = Len[i];if ( i&1 )for ( int j = 1 ; j < x ; j += 2 )ans += a[i-j-1];elsefor ( int j = 0 ; j < x ; j += 2 )ans += a[i-j-1];}printf ( “%lld\n” , ans );}}
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