牛客多校Round 1

Solved:1

rank:249

E. Removal

dp i,j表示前i个数删除了j个且选择了第i个的答案

类似字符串的dp 预处理一下nex i_k即i后面k第一次出现的位置  就好转移了

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + ;

int s[];
int nex[][];
int now[];
ll dp[][];

int main()
{
    int n, m, kk;
    while(~scanf("%d%d%d", &n, &m, &kk))
    {
        memset(dp, , sizeof(dp));
        for(int i = ; i <= n; i++) scanf("%lld", &s[i]);
        for(int i = ; i <= kk; i++) now[i] = n + ;

        for(int i = n; i >= ; i--)
        {
            for(int j = ; j <= kk; j++) nex[i][j] = now[j];
            now[s[i]] = i;
        }

        dp[][] = ; //dp[0][0] = 1;
        for(int i = ; i <= n; i++)
        {
            for(int j = ; j <= m; j++)
            {
                if(!dp[i][j]) continue;
                for(int k = ; k <= kk; k++)
                {
                    int t = nex[i][k];
                    if(t == n + ) continue;
                    if(t - i + j -  > m) continue;
                    (dp[t][t - i + j - ] += dp[i][j]) %= mod;
                }
            }
        }

        ll ans = ;
        for(int i = ; i <= n; i++)
            if(i - n + m >= )
                (ans += dp[i][i - n + m]) %= mod;
        printf("%lld\n",  ans);
    }
    return ;
}