linkkkk
题意:
给出长度为n的子序列,问长度为k的子序列有多少种
思路:
很容易想到d p [ i ] [ j ]表示从前i个字符里选j个字母的子序列种类;
根据第i个字符选或不选,分为d p [ i − 1 ] [ j ] + d p [ i − 1 ] [ j − 1 ]
最后要减去重复的部分:d p [ i ] [ j ] − d p [ p r e [ i ] − 1 ] [ j − 1 ]
类似:传送门
代码:
// Problem: 操作集锦 // Contest: NowCoder // URL: https://ac.nowcoder.com/acm/contest/4853/C // Memory Limit: 524288 MB // Time Limit: 2000 ms // // Powered by CP Editor (https://cpeditor.org) #include<bits/stdc++.h> using namespace std; typedef long long ll;typedef unsigned long long ull; typedef pair<ll,ll>PLL;typedef pair<int,int>PII;typedef pair<double,double>PDD; #define I_int ll inline ll read(){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;} #define read read() #define rep(i, a, b) for(int i=(a);i<=(b);++i) #define dep(i, a, b) for(int i=(a);i>=(b);--i) ll ksm(ll a,ll b,ll p){ll res=1;while(b){if(b&1)res=res*a%p;a=a*a%p;b>>=1;}return res;} const int maxn=1100,maxm=1e6+7; const ll mod=1e9+7; ll n,k,pre[maxn],pos[maxn],dp[maxn][maxn]; char s[maxn]; int main(){ n=read,k=read; cin>>s+1; for(int i=1;i<=n;i++){ pre[i]=pos[s[i]]; pos[s[i]]=i; } dp[0][0]=1; for(int i=1;i<=n;i++){ for(int j=0;j<=i;j++){ dp[i][j]=(dp[i-1][j]+dp[i][j])%mod; if(j>=1) dp[i][j]=(dp[i-1][j-1]+dp[i][j])%mod; if(pre[i]>=1) dp[i][j]=(dp[i][j]-dp[pre[i]-1][j-1]+mod)%mod; } } cout<<(dp[n][k]+mod)%mod; return 0; }