Solution
我好像写了一个非常有趣的解法233，
我们可以用数位$DP$ 算出比$N$小的数中 字典序比 $X$ 小的数有多少个， 再和 $rank$进行比较。
由于具有单调性， 显然可以二分答案。
Code
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define ll long long 5 using namespace std; 6 7 ll m, rank, sum[40][40][3]; 8 int b[40], len, a[40], len2; 9 10 ll dfs(int pos, int pos2, bool cmp, bool lim, bool lead) { 11 if (pos == 0) 12 return (cmp || pos2 > 23) && (!lead); 13 if (!lim && !lead && sum[pos][pos2][cmp] != -1) 14 return sum[pos][pos2][cmp]; 15 ll tmp = 0, up = lim ? a[pos] : 9; 16 for (int i = 0; i <= up; ++i) { 17 bool nxcmp = !(lead && i == 0) && i < b[pos2]; 18 nxcmp |= cmp; 19 if (!cmp && i > b[pos2]) 20 continue; 21 tmp += dfs(pos - 1, pos2 - ((lead && i == 0) ? 0 : 1), nxcmp, lim && i == a[pos], lead && i == 0); 22 } 23 if (!lim && !lead) 24 sum[pos][pos2][cmp] = tmp; 25 return tmp; 26 } 27 28 ll check(ll n) { 29 len = 0; 30 while (n) 31 a[++len] = n % 10, n /= 10; 32 ll re = dfs(len, len2, false, true, true); 33 return re; 34 } 35 36 int main() 37 { 38 memset(sum, -1, sizeof (sum)); 39 scanf("%lld%lld", &m, &rank); 40 len2 = 23; 41 ll l = rank > m ? rank : m, r = 7e17, ans = 0; 42 while (m) 43 b[++len2] = m % 10, m /= 10; 44 for (; l <= r;) { 45 ll mid = (l + r) >> 1, tmp = check(mid); 46 if (tmp == rank - 1) { 47 ans = mid; r = mid - 1; 48 } 49 else if (tmp < rank - 1) l = mid + 1; 50 else r = mid - 1; 51 } 52 printf("%lld ", ans); 53 }