题目链接:HDU4632 Palindrome subsequence
题意:找出字符串里回文子序列的个数;
分析:由差分可知dp[l][r]=(dp[l][r-1]+dp[l+1][r]-dp[l+1][r-1]+mod)%mod
当s[l]=s[r]时,dp[l][r]=(dp[l][r]+dp[l+1][r-1]+1)%mod,[l,r]可由[l+1,r-1]转移来,并且构成一个新的回文序列,再+1;
#include<bits/stdc++.h>
using namespace std;
const int maxn=1007;
const int mod=10007;
typedef long long ll;
char s[maxn];
int dp[maxn][maxn],n;
int rua()
{
scanf("%s",s+1);
n=strlen(s+1);
for(int i=1;i<=n;i++) dp[i][i]=1;
for(int i=1;i<n;i++)
if(s[i]==s[i+1]) dp[i][i+1]=3;
else dp[i][i+1]=2;
for(int i=3;i<=n;i++)
for(int l=1;l+i-1<=n;l++)
{
int r=l+i-1;
dp[l][r]=(dp[l][r-1]+dp[l+1][r]-dp[l+1][r-1]+mod)%mod;
if(s[l]==s[r]) dp[l][r]=(dp[l][r]+dp[l+1][r-1]+1)%mod;
}
return dp[1][n];
}
int main()
{
int t;scanf("%d",&t);
for(int id=1;id<=t;id++) printf("Case %d: %d\n",id,rua());
return 0;
}