数位DP基础板子

一般的数位DP的题目都是在问我们从【l，r】满足条件的数字的个数 ——————我们按位来加快枚举

dfs模板

#include
using namespace std;
const int N=20;
int dp[N][N];
int a[N],len;
int n,m;

//pos 代表当前位置 limit是当前位置是否有这个枚举范围限制

int dfs(int pos,int pre,int limit){
	if(!pos)return 1;
	if(!limit&&~dp[pos][pre])return dp[pos][pre];
	
	int res = 0,up = limit?a[pos]:9;
	for(int i=0;i<=up;i++){
		if(i==4||(i==2&&pre==6))continue;
		res += dfs(pos-1,i,limit&&i==up);
	}
	
	return limit?res:dp[pos][pre] = res;
}

int cal(int x){
	memset(dp,-1,sizeof dp);
	len = 0;
	while(x)a[++len] = x%10,x/=10;
	return dfs(len,0,1);
}

int main()
{
	
	while(cin>>n>>m,n){
		cout<

带前导0

#include
using namespace std;
const int N=20;
int dp[N][N];
int a[N],len;


int dfs(int pos,int pre,int lead,int limit){
	if(!pos)return 1;
	if(!lead&&!limit&&~dp[pos][pre])return dp[pos][pre];
	
	int res = 0,up = limit?a[pos]:9;
	for(int i=0;i<=up;i++){
		if(abs(i-pre)<2)continue;
		if(lead&&!i)res+=dfs(pos-1,-2,lead&&!i,limit&&i==up);
		else res+=dfs(pos-1,i,lead&&!i,limit&&i==up);
	}
	
	return lead?res:(limit?res:dp[pos][pre] = res);
}

int cal(int x){
	memset(dp,-1,sizeof dp);
	len = 0;
	while(x)a[++len] = x%10,x/=10;
	return dfs(len,-2,1,1);
}

int n,m;
int main()
{
	cin>>n>>m;
	cout<

b进制下枚举

#include
using namespace std;
const int N=35;
int dp[N][N];
int a[N],len;
int n,m,k,b;

int dfs(int pos,int cnt,int limit){
	if(!pos)return cnt==k;
	if(!limit&&~dp[pos][cnt])return dp[pos][cnt];
	
	int res = 0,up = limit?a[pos]:b-1;
	for(int i=0;i<=up;i++){
		if(i>1||(i==1&&cnt==k))continue;
		res+=dfs(pos-1,cnt+(i==1),limit&&i==up);
	}
	
	return limit?res:dp[pos][cnt]=res;
	
}



int cal(int x){
	memset(dp,-1,sizeof dp);
	len = 0;
	while(x)a[++len] = x%b,x/=b;
	return dfs(len,0,1);
}

int main()
{
	cin>>n>>m>>k>>b;
	cout<

维护和

#include
using namespace std;
const int N=40;
int dp[N][10100];
int a[N],len;
int n,m,k;

int dfs(int pos,int sum,int limit){
	if(!pos)return sum%k==0;
	if(!limit&&~dp[pos][sum])return dp[pos][sum];
	
	int res = 0, up = limit?a[pos]:9;
	for(int i=0;i<=up;i++)
	 res+=dfs(pos-1,sum+i,limit&&i==up);
	
	
	return limit?res:dp[pos][sum] = res;
}
int cal(int x){
	memset(dp,-1,sizeof dp);
	len = 0;
	while(x)a[++len] = x%10,x/=10;
	return dfs(len,0,1);
}

int main()
{
	while(cin>>n>>m>>k){
		cout<