題目大意：求區間$[l,r]$中數字$0\sim9$出現個數

題解：數位$DP$

卡點：無

C++ Code：

#include <cstdio>
#include <iostream>
struct node {
	long long s[10], sum;
	inline node() {for (register int i = 0; i < 10; i++) s[i] = 0; sum = 0;}
	inline node(int x) {for (register int i = 0; i < 10; i++) s[i] = 0; sum = x;}
	inline friend node operator + (const node &lhs, const node &rhs) {
		node res;
		res.sum = lhs.sum + rhs.sum;
		for (register int i = 0; i < 10; i++) res.s[i] = lhs.s[i] + rhs.s[i];
		return res;
	}
	inline friend node operator - (const node &lhs, const node &rhs) {
		node res;
		res.sum = lhs.sum - rhs.sum;
		for (register int i = 0; i < 10; i++) res.s[i] = lhs.s[i] - rhs.s[i];
		return res;
	}
	inline friend std::ostream & operator << (std::ostream &Fout, const node __node) {
		for (register int i = 0; i < 10; i++) {
			Fout << __node.s[i];
			Fout << (i == 9 ? '\n' : ' ');
		}
		return Fout;
	}
};

int num[20], tot;
node f[20];
bool vis[20];
node calc(int x, int lim, int lead) {
	if (!x) return node(1);
	if (!lim && lead && vis[x]) return f[x];
	node F;
	for (int i = lim ? num[x] : 9, op = 1; ~i; i--, op = 0) {
		node tmp = calc(x - 1, lim && op, lead || i);
		F = F + tmp;
		if (i || lead) F.s[i] += tmp.sum;
	}
	if (!lim && lead) f[x] = F, vis[x] = true;
	return F;
}
node solve(long long x) {
	if (x < 0) return node();
	tot = 0;
	while (x) {
		num[++tot] = x % 10;
		x /= 10;
	}
	return calc(tot, 1, 0);
}
long long l, r;
int main() {
	std::ios::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	std::cin >> l >> r;
	std::cout << solve(r) - solve(l - 1) << std::endl;
	return 0;
}