题意：

给一个161616进制的串NNN，让你求1−N1-N1−N中有多少个数有kkk个不同的数且没有前导零。

思路：

NNN很大，有2e52e52e5了，那么就比较明显是个数位dpdpdp了。首先没有前导零需要控制，还需要控制枚举是否能达到上界。根据这个题还需要存一个有多少个不同的数，我们这个可以状压一下。最后记一个位置即可。
设dp[pos][state][flag][lead]dp[pos][state][flag][lead]dp[pos][state][flag][lead]为到了pospospos位置，状态为statestatestate，是否能枚举到上界flagflagflag，以及是否存在前导零leadleadlead。让后跑裸的数位dpdpdp就好啦。
复杂度约为O(2e5∗k∗16)O(2e5*k*16)O(2e5∗k∗16)，实际跑起来很快。
当然也可以将[flag][lead][flag][lead][flag][lead]两维去掉，让后返回和赋值的时候特判一下就好啦。

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=200010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,k;
int a[N];
char s[N];
int f[N][20][2][2];int dfs(int pos,int state,int flag,int lead)
{int cnt=__builtin_popcount(state);if(cnt>k) return 0;if(pos==n+1) return cnt==k&&lead;if(f[pos][cnt][flag][lead]!=-1) return f[pos][cnt][flag][lead];int x=flag? 15:a[pos];int ans=0;for(int i=0;i<=x;i++)(ans+=dfs(pos+1,(!lead&&i==0)? state:state|(1<<i),flag||i<x,lead||i!=0))%=mod;f[pos][cnt][flag][lead]=ans;return ans;
}int main()
{//  ios::sync_with_stdio(false);
//  cin.tie(0);scanf("%s%d",s+1,&k);n=strlen(s+1);for(int i=1;i<=n;i++)if(s[i]>='A'&&s[i]<='Z') a[i]=s[i]-'A'+10;else a[i]=s[i]-'0';memset(f,-1,sizeof(f));printf("%d",dfs(1,0,0,0));return 0;
}
/**/

f[pos][pre]f[pos][pre]f[pos][pre]的做法。

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=200010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,k;
int a[N];
char s[N];
int f[N][20];int dfs(int pos,int state,int flag,int lead)
{int cnt=__builtin_popcount(state);if(cnt>k) return 0;if(pos==n+1) return cnt==k&&lead;if(flag&&lead&&f[pos][cnt]!=-1) return f[pos][cnt];int x=flag? 15:a[pos];int ans=0;for(int i=0;i<=x;i++)(ans+=dfs(pos+1,(!lead&&i==0)? state:state|(1<<i),flag||i<x,lead||i!=0))%=mod;if(flag&&lead) f[pos][cnt]=ans;return ans;
}int main()
{//  ios::sync_with_stdio(false);
//  cin.tie(0);scanf("%s%d",s+1,&k);n=strlen(s+1);for(int i=1;i<=n;i++)if(s[i]>='A'&&s[i]<='Z') a[i]=s[i]-'A'+10;else a[i]=s[i]-'0';memset(f,-1,sizeof(f));printf("%d",dfs(1,0,0,0));return 0;
}
/**/

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