ps：枚举翻转区间的端点的数值，因为每个数都在0~9范围内，所以只需要枚举45种情况（C（10，2）组合数）。b串：0 1 2 3 4 5 6 7 8 9，枚举翻转区间的端点的数值3，6，得到新的b串：0 1 2 3 6 5 4 3 6 7 8 9，让这个串与原串跑一遍普通的最长公共子序列就能得到答案。注意保留一份端点值。怎么更新翻转区间的位置我还没搞明白~

int solve() {

    dp[0][0] = 0;

    for (int i = 1; i <= n; ++i) dp[i][0] = 0;

    for (int i = 1; i <= m; ++i) dp[0][i] = 0;

    for (int i = 1; i <= n; ++i) {

        for (int j = 1; j <= m; ++j) {

            //dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) + (a[i] == b[j]);

            dp[i][j] = dp[i - 1][j];

            l[i][j] = l[i - 1][j];

            r[i][j] = r[i - 1][j];

            if (a[i] == b[j]) {

                dp[i][j]++;

                if (j == L && l[i][j] == 0) l[i][j] = i;

                if (j == R) r[i][j] = i;

            }

            if (dp[i][j] < dp[i][j - 1]) {

                dp[i][j] = dp[i][j - 1];

                l[i][j] = l[i][j - 1];

                r[i][j] = r[i][j - 1];

            }

        }

    }

    return dp[n][m];

}