CF1183E/H Subsequences

思路：

dp好题，dp[i][j]表示到前i个字符为止并且以s[i]为结尾，共有多少个长度为j的不同的子序列。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 ll dp[][], sum[];
 int last[];
 int main()
 {
     int n; ll m; string s;
     while (cin >> n >> m >> s)
     {
         memset(dp, , sizeof dp);
         memset(sum, , sizeof sum);
         memset(last, -, sizeof last);
         for (int i = ; i <= n; i++)
         {
             dp[i][] = ;
             for (int j = ; j <= i; j++)
             {
                 for (int k = ; k < ; k++)
                 {
                     if (last[k] != -)
                     {
                         int p = last[k];
                         dp[i][j] += dp[p][j - ];
                     }
                 }
             }
             last[s[i - ] - 'a'] = i;
         }
         for (int i = ; i <= n; i++)
         {
             for (int j = ; j < ; j++)
             {
                 if (last[j] != -) sum[i] += dp[last[j]][i];
             }
         }
         sum[] = ;
         ll ans = ;
         bool flg = false;
         for (int i = n; i >= ; i--)
         {
             if (m - sum[i] <= ) { ans += m * ((ll)n - i); flg = true; break; }
             ans += sum[i] * ((ll)n - i); m -= sum[i];
         }
         cout << (flg ? ans : -) << endl;
     }
     return ;
 }