G.小苯的数位MEX
思路
比较模板的数位dp，虽然我不会
代码
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using piii = pair<int, pii>;
using pll = pair<ll, ll>;
using plll = pair<ll, pll>;
using plii = pair<ll, pii>;
#define fi first
#define se second
const int INF = 0x3f3f3f3f;
const ull mod1 = (1ull << 61) - 1, mod2 = 1e9 + 7;
const ull base1 = 131, base2 = 13331;
// mt19937 rnd(time(0));const int mod = 998244353;vector<int> nums;
int mex;
int f[11][1100]; //f[dep][state]到第dep位（从高位到低位），该位之前的状态为state(不包括该位)//分别为层数 状态 最高位限制 有无前导零
int dfs(int dep, int state, bool limit, bool lead0) {if (dep == nums.size()) {if (lead0)return mex == 1 ? 1 : 0;for (int i = 0; i < mex; i++) {if ((state >> i & 1) == 0)return limit ? 0 : f[dep][state] = 0;}return limit ? 1 : f[dep][state] = 1;}if (!lead0 && !limit && f[dep][state] != -1) //不加lead0也行return f[dep][state];int up = limit ? nums[dep] : 9; //根据有无限制确定第dep位枚举上限int res = 0;for (int i = 0; i <= up; i++) {int fst = state | (1 << i);if (lead0 && i == 0) //如果有前导零，则无法传递fst = 0;if (i == mex && (lead0 & i == 0) == 0) //如果i等于mex（没前导零的情况），也是不成立的continue;res += dfs(dep + 1, fst, limit && i == up, lead0 && i == 0);}if (!limit && !lead0) //不加lead0也行f[dep][state] = res;return res;
}int get(int x) {memset(f, -1, sizeof f);nums.clear();if (!x) { //x为零无法进入while循环，特判return mex == 1 ? mex : 0;}while (x) {nums.push_back(x % 10);x /= 10;}reverse(nums.begin(), nums.end());return dfs(0, 0, 1, 1);
}void solve() {int a, b;cin >> a >> b;b += a;for (int i = 10; i >= 0; i--) {mex = i;int ans = get(b) - get(a - 1);if (ans) {cout << i << " " << ans << endl;return;}}
}int main() {cin.tie(nullptr);ios::sync_with_stdio(false);int _ = 1;cin >> _;while (_--) {solve();}return 0;
}
