HDU 4632 Palindrome subsequence(区间DP求回文子序列数)

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<cctype>
 4 #include<cstring>
 5 #include<iostream>
 6 #include<algorithm>
 7 #include<vector>
 8 #include<queue>
 9 #include<set>
10 #include<map>
11 #include<stack>
12 #include<string>
13 #define lc(a) (a<<1)
14 #define rc(a) (a<<1|1)
15 #define MID(a,b) ((a+b)>>1)
16 #define fin(name)  freopen(name,"r",stdin)
17 #define fout(name) freopen(name,"w",stdout)
18 #define clr(arr,val) memset(arr,val,sizeof(arr))
19 #define _for(i,start,end) for(int i=start;i<=end;i++)
20 #define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
21 using namespace std;
22 typedef long long LL;
23 const int N=1e3+5;
24 const int MOD=10007;
25 const int INF=0x3f3f3f3f;
26 const double eps=1e-10;
27 
28 int dp[N][N];
29 char str[N];
30 
31 int main(){
32     int t,cas=0;
33     cin>>t;
34     while(t--){
35         cin>>str;
36         int n=strlen(str);
37         memset(dp,0,sizeof(dp));
38         for(int i=0;i<n;i++){
39             dp[i][i]=1;
40         }
41         for(int len=1;len<n;len++){
42             for(int i=0;i+len<n;i++){
43                 int j=i+len;
44                 dp[i][j]=(dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+MOD)%MOD;
45                 //如果str[i]==str[j],单独两个字符str[i],str[j]能组成一个回文串，同样也能跟[i-1,j+1]的所有回文串组成新的回文串
46                 if(str[i]==str[j])
47                     dp[i][j]+=dp[i+1][j-1]+1;
48             }
49         }
50         cout<<"Case "<<++cas<<": "<<dp[0][n-1]%MOD<<endl;
51     }
52     return 0;
53 }