$Result$

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$Hyperlink$

https://www.luogu.com.cn/problem/P2602

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$Description$

求$[L,R]$中各个数字出现的次数

数据范围：$L\le R\le 10^{12}$

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$Solution$

数位$dp$
首先枚举我们要求的数字，设$f_{rest,sum}$表示剩余$rest$位，这个数字出现了$sum$次的个数
考虑前导0和最高位的限制即可

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$Code$

#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;LL L,R;
int a[20],m;
LL f[20][20];
inline LL read()
{
	char c;LL d=1,f=0;
	while(c=getchar(),!isdigit(c)) if(c=='-') d=-1;f=(f<<3)+(f<<1)+c-48;
	while(c=getchar(),isdigit(c)) f=(f<<3)+(f<<1)+c-48;
	return d*f;
}
inline LL dfs(int rest,int sum,int x,bool lead,bool limit)//没有前导0时lead为True
{
	if(rest==0) return sum;
	if(lead&&limit==0&&f[rest][sum]!=-1) return f[rest][sum];
	int cancse=limit?a[rest]:9;
	LL res=0;
	for(register int i=0;i<=cancse;i++)
	 res+=dfs(rest-1,sum+((lead||i)&&(i==x)),x,lead||i,limit&&i==cancse);
 	if(lead&&limit==0) f[rest][sum]=res;
	return res;
}
inline LL solve(LL x,int y)
{
	m=0;
	while(x) a[++m]=x%10,x/=10;
	memset(f,-1,sizeof(f));
	return dfs(m,0,y,0,1);
}
signed main()
{
	L=read();R=read();
	for(register int i=0;i<=9;i++) printf("%lld ",solve(R,i)-solve(L-1,i));