题解：

#include <bits/stdc++.h>
using namespace std;

int a[11];
bool vis[11][2];
long long f[11][2], cnt[11][2];

long long dfs(int p, bool limit) {
	if (vis[p][limit])
		return f[p][limit];
	if (p == 0) {
		vis[p][limit] = true;
		int up = limit ? a[p] : 9;
		f[p][limit] = (1 + up) * up / 2;
		cnt[p][limit] = up + 1;
		return f[p][limit];
	}
	vis[p][limit] = true;
	int up = limit ? a[p] : 9;
	for (int i = 0; i <= up; i++) {
		// 第 p 位选为了 i
		int li = limit && i == up;
		dfs(p-1, li);
		f[p][limit] += f[p-1][li] + cnt[p-1][li] * i;
		cnt[p][limit] += cnt[p-1][li];
	}
	return f[p][limit];
}

long cal(int num) {
	memset(vis, 0, sizeof vis);
	memset(f, 0, sizeof f);
	memset(cnt, 0, sizeof cnt);
	int p = 0;
	for (; num; a[p++] = num % 10, num /= 10);
	return dfs(p-1, true);
}

int main() {
	int l, r;
	cin >> l >> r;
	cout << cal(r) - cal(l-1) << endl;
	return 0;
}

数位搜索（演示通过 dfs 的形式输出 0 ~ a）：

#include <bits/stdc++.h>
using namespace std;

int a[11], b[11];
void dfs(int p, bool limit) {
	if (p < 0) {
		cout << "number: ";
		for (int i = 10, f = 0; i >= 0; i--) {
			if (b[i] || i == 0) f = 1;
			if (f) cout << b[i];
		}
		cout << endl;
		return;
	}
	int up = limit ? a[p] : 9;
	for (int i = 0; i <= up; i++) {
		b[p] = i;
		dfs(p-1, limit && i == up);
	}
}

void cal(int num) {
	int p = 0;
	for (; num; a[p++] = num % 10, num /= 10);
	// a[p-1..0]
	dfs(p-1, true);
}

int main() {
	int a;
	cin >> a;
	cal(a);
	return 0;
}