12400-3, 2, 1, 0 (mathematical thinking Questions & amp; high precision or table)

 

 

Ideas:

 

According to this idea, this sequence is actually unique, as shown in the following code: (1000 digits. For more information about the code, see the final part of the article)

3233232323222233222323222322322222332322232332333223232332222233232232233232223322232222233322223233323332333332323332333233333332233233222232223222222322322333322322222332323222322333232222222232223322332333223232233232222222323333323323232232333322333332223223322232323333232333332323233333323322223332223322222222232223332332332333233333233232333333233322223322222333333223232223232232223323232323333222323333332222333332222323323332323233332233323233233232332232223323223323222222323322333333323232322232222333233332232232233233332232223322233233232222222323333322322222222322323322333233333222323222222223222223233333223323232333223233222333232233333222233333322332322233332222333222323232223223233323223232332222332323232233333223333232232323322222233222333332223222233322323233233233232332323332222222332323322232332233323222222322333223322323322332232332222333222233333323332322232232232222233232332323332232222233323332233323223222232232232232332222333332332333333332223233333323323232223232

The O (1) Complexity code is as follows:

 

/*0.015s*/#include
 
  #include
  
   const int mx = 1000;const char s[mx + 1] = 233232323222233222323222322322222332322232332333223232332222233232232233232223322232222233322223233323332333332323332333233333332233233222232223222222322322333322322222332323222322333232222222232223322332333223232233232222222323333323323232232333322333332223223322232323333232333332323233333323322223332223322222222232223332332332333233333233232333333233322223322222333333223232223232232223323232323333222323333332222333332222323323332323233332233323233233232332232223323223323222222323322333333323232322232222333233332232232233233332232223322233233232222222323333322322222222322323322333233333222323222222223222223233333223323232333223233222333232233333222233333322332322233332222333222323232223223233323223232332222332323232233333223333232232323322222233222333332223222233322323233233233232332323332222222332323322232332233323222222322333223322323322332232332222333222233333323332322232232232222233232332323332232222233323332233323223222232232232232332222333332332333333332223233333323323232223232;int cnt[2][mx];int main(){for (int i = 1; i < mx; ++i){++cnt[s[mx - i - 1] - '2'][i];cnt[0][i] += cnt[0][i - 1];cnt[1][i] += cnt[1][i - 1];}int n, m, k;while (~scanf(%d%d%d, &n, &m, &k)){if (k && n >= cnt[1][k] && m >= cnt[0][k]) puts(s + mx - 1 - k);else puts(Impossible.);}return 0;}
  
 

Code for table creation: (enter 999 999 999)

 

 

/* 1.802 s */import java. io. bufferedInputStream; import java. math. bigInteger; import java. util. *; public class Main {static partition cin = new partition (new BufferedInputStream (System. in); public static void main (String [] args) {int N, M, K, kk, I; BigInteger MASK, ten, cur; char ans []; while (cin. hasNextInt () {N = cin. nextInt (); M = cin. nextInt (); K = cin. nextInt (); if (K = 0) {System. out. println (Impossible .); continue;} if (K> M + N) {System. out. println (Impossible .); continue;} MASK = BigInteger. valueOf (2 ). pow (K ). subtract (BigInteger. ONE); // MASK = 2 ^ k-1ans = new char [K]; kk = K; ten = BigInteger. ONE; cur = BigInteger. ZERO; for (I = 0; I <K; ++ I) {if (cur. and (BigInteger. ONE. shiftLeft (I )). equals (BigInteger. ZERO) {// The I-bit of the cur binary is 0ans [-- kk] = '2'; if (-- M <0) {System. out. println (Impossible .); break;} cur = cur. add (ten. multiply (BigInteger. valueOf (2); if (cur. compareTo (MASK)> 0) {cur = cur. and (MASK); // modulo to prevent the cur from slowing down the operation speed too much} else {// The I-bit of the cur binary is 1ans [-- kk] = '3 '; if (-- N <0) {System. out. println (Impossible .); break;} cur = cur. add (ten. multiply (BigInteger. valueOf (3); if (cur. compareTo (MASK)> 0) {cur = cur. and (MASK); // modulo to prevent the cur from slowing down the computation speed.} ten = ten. multiply (BigInteger. valueOf (10);} if (I = K) System. out. println (ans );}}}