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 + 7;

int s[100005];
int nex[100005][12];
int now[12];
ll dp[100005][12];

int main()
{
    int n, m, kk;
    while(~scanf("%d%d%d", &n, &m, &kk))
    {
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++) scanf("%lld", &s[i]);
        for(int i = 1; i <= kk; i++) now[i] = n + 1;
        
        for(int i = n; i >= 0; i--)
        {
            for(int j = 1; j <= kk; j++) nex[i][j] = now[j];
            now[s[i]] = i;
        }
        
        dp[0][0] = 1; //dp[0][0] = 1;
        for(int i = 0; i <= n; i++)
        {
            for(int j = 0; j <= m; j++)
            {
                if(!dp[i][j]) continue;
                for(int k = 1; k <= kk; k++)
                {
                    int t = nex[i][k];
                    if(t == n + 1) continue;
                    if(t - i + j - 1 > m) continue;
                    (dp[t][t - i + j - 1] += dp[i][j]) %= mod;
                } 
            }
        }
        
        ll ans = 0;
        for(int i = 1; i <= n; i++)
            if(i - n + m >= 0)
                (ans += dp[i][i - n + m]) %= mod;
        printf("%lld
",  ans);
    }
    return 0;
}