数位DP集合

hdu2089不要62

#include<bits/stdc++.h>
using namespace std;
#define long long ll
int a[20];//存数字 
int dp[20][2];
 
int dfs(int count/*位数*/,int pre/*用于排除62的情况*/,int sta/*状态*/,bool limit/*判断是否有上限*/){
	if(count == -1)
		return 1;
	if(!limit && dp[count][sta] != -1)
		return dp[count][sta];//退出递归
	int up = limit ? a[count] : 9;
	int tmp = 0;
	for(int i = 0;i <= up;i++){
		if(pre == 6 && i == 2)/*排除62的情况*/
			continue;
		if(i == 4)/*排除4的情况*/
			continue;
		tmp += dfs(count-1,i,i == 6,limit && i == a[count]);
	}
	if(!limit)
		dp[count][sta] = tmp;
	return tmp;
}
 
int solve(int x){//将数拆开存在数组里 
	int count = 0;
	while(x){
		a[count++] = x % 10;
		x /= 10;
	}
	return dfs(count-1,-1,0,true);
}
 
int main(){
	int a,b;
	while(~scanf("%d%d",&a,&b) && a + b){
		memset(dp,-1,sizeof(dp));
		printf("%d\n",solve(b) - solve(a-1));
	}
	return 0;
}

hdu4389X mod f(x)

#include<bits/stdc++.h>
using namespace std;
#define long long ll
int a[20];//存数字 
int dp[20][2];
 
int dfs(int count/*位数*/,int pre/*用于排除62的情况*/,int sta/*状态*/,bool limit/*判断是否有上限*/){
	if(count == -1)
		return 1;
	if(!limit && dp[count][sta] != -1)
		return dp[count][sta];//退出递归
	int up = limit ? a[count] : 9;
	int tmp = 0;
	for(int i = 0;i <= up;i++){
		if(pre == 6 && i == 2)/*排除62的情况*/
			continue;
		if(i == 4)/*排除4的情况*/
			continue;
		tmp += dfs(count-1,i,i == 6,limit && i == a[count]);
	}
	if(!limit)
		dp[count][sta] = tmp;
	return tmp;
}
 
int solve(int x){//将数拆开存在数组里 
	int count = 0;
	while(x){
		a[count++] = x % 10;
		x /= 10;
	}
	return dfs(count-1,-1,0,true);
}
 
int main(){
	int a,b;
	while(~scanf("%d%d",&a,&b) && a + b){
		memset(dp,-1,sizeof(dp));
		printf("%d\n",solve(b) - solve(a-1));
	}
	return 0;
}

来源：https://blog.csdn.net/weixin_43252274/article/details/99083621