做法：动态规划

思路

  • 用pre[i]存上一次'a'+i这个字母出现的位置，设dp[i][j]:前i个字符中长度为j的数量
  • 如果一个字母上次出现过，需要减去dp[pre[s[i]-'a']-1][j-1]重复的子序列
  • 每次删最少的字符即长度越长的子序列优先

代码

// Problem: Subsequences (hard version)
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/problem/113788
// Memory Limit: 524288 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp(aa,bb) make_pair(aa,bb)
#define _for(i,b) for(int i=(0);i<(b);i++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,b,a) for(int i=(b);i>=(a);i--)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define X first
#define Y second
#define lowbit(a) (a&(-a))
#define debug(a) cout<<#a<<":"<<a<<"\n"
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef long double ld;
const int N=105;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1.0);

int n;
int pre[N];
ll k,ans,num;
ll dp[N][N];//dp[i][j]:前i个字符中长度为j的数量
string s;

void solve(){
    cin>>n>>k;
    cin>>s;s=" "+s;
    rep(i,0,n) dp[i][0]=1;
    rep(i,1,n){
        rep(j,1,i){
            dp[i][j]=dp[i-1][j-1]+dp[i-1][j];
            if(pre[s[i]-'a']) dp[i][j]-=dp[pre[s[i]-'a']-1][j-1]; 
        }
        pre[s[i]-'a']=i;
    }
    rep(i,0,n) {
        num+=dp[n][i];
    }
    if(num<k){
        cout<<"-1\n";
    }
    else{
        per(i,n,0){
            if(dp[n][i]<k){
                k-=dp[n][i];
                ans+=dp[n][i]*(n-i);
            }
            else{
                ans+=k*(n-i);
                cout<<ans<<"\n";
                return;
            }
        }

    }
}


int main(){
    ios::sync_with_stdio(0);cin.tie(0);
//    int t;cin>>t;while(t--)
    solve();
    return 0;
}