2015 Programming beauty Qualifying series number of palindrome subsequence

Time limit: 2000ms
Single point time limit: 1000ms
Memory Limit: 256MB
Describe
The number of palindrome sequences for a given string. The palindrome sequence reverses the character sequence and remains the same as the original sequence. For example, in the string aba, the palindrome sequence is "a", "a", "AA", "B", "ABA", a total of 5. Sub-sequences with different contents in the same location are counted as different subsequence.

Input
The first line is an integer t that represents the number of data groups. After that is the T-group data, each set of data is a row of strings.

Output
For each set of data output row, the format is "case #X: Y", X for the data number (starting from 1), and Y for the answer. Answer to 100007 modulo.

Data range
1≤t≤30
Small Data
String length ≤25
Big Data
String length ≤1000

Sample input
5
Aba
Abcbaddabcba
12111112351121
Ccccccc
Fdadfa
Sample output
Case #1:5
Case #2:277
Case #3:1333
Case #4:127
Case #5:17

Analysis:
1. Interval type DP, first of all, the state equation: if I to j this interval total number of palindrome subsequence, two cases ① when s[i] = = S[j], such as "a....a",d[i][j] = d[i][j-1] + d[i+1][j] + 1 (from d[i][j] = 2*d[i+1][j-1] + d[i][j-1] + d[i+1][j] + 1-2*d[i+1][j-1] simplification. Explanation: Known i+1 to J-1 have d[i+1][j-1] palindrome subsequence, and s[i] = = S[j], then can be with the middle (string i+1 to J-1) known palindrome sub-sequence and then constitute D[i+1][j-1] palindrome sub-sequence, plus the original middle string contains the palindrome sub-sequence altogether 2*d [I+1] [J-1], two a combination of "AA" is also a palindrome, so add 1, and then calculate the left and right side of a and the middle string of the palindrome sub-sequence, but this time is not over, note: In the calculation of the two sides of a and the middle string, and then two times the middle string contains the palindrome subsequence, so in minus 2*d[i+ 1][j-1] ② when s[i]! = s[j],d[i][j] = d[i][j-1]+d[i+1][j]-d[i+1][j-1]. This self should also be able to analyze, and similar to the above.
2. when the interval I to J is used, D[i+1][j-1], d[i][j-1], d[i+1][j]. This is why the interval DP generally starts from a smaller interval, and then does not widen the interval. Now the interval (i, j) used by the sub-range, before all have been finished, so or that sentence, direct use is good.
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#include <iostream>#include<cstdio>#include<string.h>#include<cstring>using namespacestd;Const intMOD =100007;intT, Len, d[1010][1010];Chars[1010];voiddp () { for(inti =1; i < Len; i++)    {         for(intj =0; J < (Len-i); J + +)        {            if(S[j] = = s[j+i]) d[j][i+J] = (d[j][i+j-1] + d[j+1][I+J] +1) %MOD; ElseD[j][i+J] = (d[j][i+j-1] + d[j+1][I+J]-d[j+1][i+j-1] + MOD)%MOD; }    }}intMain () {CIN>>T;  for(intQ =1; Q <= t; q++) {memset (s),0,sizeof(s)); memset (d,0,sizeof(d)); CIN>>s; Len=strlen (s);  for(inti =0; i < Len; i++) D[i][i] =1;        DP (); printf ("Case #%d:%d\n", Q, d[0][len-1] %MOD); }    return 0;}

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2015 Programming beauty Qualifying series number of palindrome subsequence