操作集锦 牛客练习赛60-C（dp）

题意：

给一个小写字母组成的串，问长度为$K$的不重复的子序列有多少个。

思路：

$dp[i][j]$表示长度为$i$的最后一位为$j$的子序列的个数。

转移方程为：$dp[i][j]=sum_{k=0}^{25}{dp[i-1][k]}$。

特判$K=0$时个数为$1$。

代码：

#include<bits/stdc++.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define ls root<<1
#define rs root<<1|1
#define Q(a) cout<<a<<endl

using namespace std;
const int inf = 2e9 + 7;
const ll Inf = 1e18 + 7;
const int maxn = 1e6 + 5;
const int mod = 1e9 +7;

ll gcd(ll a, ll b)
{
    return b ? gcd(b, a % b) : a;
}

ll lcm(ll a, ll b)
{
    return a / gcd(a, b) * b;
}

ll read()
{
    ll p = 0, sum = 0;
    char ch;
    ch = getchar();
    while (1)
    {
        if (ch == '-' || (ch >= '0' && ch <= '9'))
            break;
        ch = getchar();
    }

    if (ch == '-')
    {
        p = 1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        sum = sum * 10 + ch - '0';
        ch = getchar();
    }
    return p ? -sum : sum;
}

ll dp[maxn][50];
char s[maxn];

int main()
{
    int n,k;
    scanf("%d %d",&n,&k);
    scanf("%s",s+1);
    if(k==0)
    {
        printf("1
");
        return 0;
    }
    for(int i=1;i<=n;++i)
    {
        int x=s[i]-'a';
        for(int j=i;j>=1;--j)
        {
            ll res=0;
            for(int o=0;o<26;++o)
            {
                res+=dp[j-1][o];
                res%=mod;
            }
            dp[j][x]=res;
        //    cout<<j<<" "<<(char)(x+'a')<<" "<<dp[j][x]<<endl;
        }
        if(dp[1][x]==0) dp[1][x]=1;
    }
    ll ans=0;
    for(int i=0;i<26;++i)
    {
        ans+=dp[k][i];
        ans%=mod;
    }
    printf("%lld
",ans);
}