Luogu2022 有趣的数-二分答案+数位DP

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

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define ll long long
 using namespace std;

 ll m, rank, sum[][][];
 int b[], len, a[], len2;

 ll dfs(int pos, int pos2, bool cmp, bool lim, bool lead) {
     if (pos == )
         return (cmp || pos2 > ) && (!lead);
     if (!lim && !lead && sum[pos][pos2][cmp] != -)
         return sum[pos][pos2][cmp];
     ll tmp = , up = lim ? a[pos] : ;
     for (int i = ; i <= up; ++i) {
         bool nxcmp = !(lead && i == ) && i < b[pos2];
         nxcmp |= cmp;
         if (!cmp && i > b[pos2])
             continue;
         tmp += dfs(pos - , pos2 - ((lead && i == ) ?  : ), nxcmp, lim && i == a[pos], lead && i == );
     }
     if (!lim && !lead)
         sum[pos][pos2][cmp] = tmp;
     return tmp;
 }

 ll check(ll n) {
     len = ;
     while (n)
         a[++len] = n % , n /= ;
     ll re = dfs(len, len2, false, true, true);
     return re;
 }

 int main()
 {
     memset(sum, -, sizeof (sum));
     scanf("%lld%lld", &m, &rank);
     len2 = ;
     ll l = rank > m ? rank : m, r = 7e17, ans = ;
     while (m)
         b[++len2] = m % , m /= ;
     for (; l <= r;) {
         ll mid = (l + r) >> , tmp = check(mid);
         if (tmp == rank - ) {
             ans = mid; r = mid - ;
         }
         else if (tmp < rank - ) l = mid + ;
         else r = mid - ;
     }
     printf("%lld\n", ans);
 }