Palindrome-subsequence-HDU-4632

好菜這麼簡單的題都想不出來

dp式代表i，j內迴文串字數最多的次數

dp[i][j] = dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1]及是有左端右端減去中間重合部分

如果i,j相等說明可以與[i+1，j-1]中的字符串組合成新的子串，並且str[i] 與str[j]也能組合成一個字符串

#include<cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define fst first
#define sec second
#define sci(num) scanf("%d",&num)
#define scl(num) scanf("%lld",&num)
#define scf(num) scanf("%lf",&num)
#define mem(a,b) memset(a,b,sizeof a)
#define cpy(a,b) memcopy(a,b,sizeof b)
typedef long long LL;
typedef pair<int,int> P;
const int MAX_E = 1e6 + 100;
const int MAX_N = 1e3 + 100;
const LL MOD = 10007;
const int  INF = 0x3f3f3f3f;
char str[MAX_N];
int dp[MAX_N][MAX_N];
int main(){
    int T;
    sci(T);
    for (int cs = 1; cs <= T;cs++) {
        scanf("%s",str + 1);
        int N = strlen(str + 1);
        mem(dp,0);
        for (int i = 1;i <= N;i++)
            dp[i][i] = 1;
        for (int len = 1;len <= N;len++) {
            for (int i = 1;i + len <= N;i++) {
                int j = i + len;
                dp[i][j] = (dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1] + MOD) % MOD;
                if (str[i] == str[j]) {
                    dp[i][j] += dp[i + 1][j - 1] + 1;
                    dp[i][j] %= MOD;
                }
            }
        }
        printf("Case %d: %d\n",cs,dp[1][N]);


    }

    return 0;
}