HDU 4394 Digital Square

Digital Square

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 1882 Accepted Submission (s): 741

Problem Descriptiongiven An integer n,you should come up with the minimum nonnegativeInteger M.M meets the follow Condition:m2%10x=n (x=0,1,2,3 ...)

Inputthe first line have an integer T (t< = +), the number of test cases.
For each case, each line contains one integer n (0<= n <=109), indicating the given number.

Outputfor each case output the answer if it exists, otherwise print "None".

Sample Input332125

Sample OutputNone115

Source multi-university Training Contest 10 parsing: Set m = abc, i.e. m = 100*a+10*b+c, then M2 = 10000*a2 + 1000*2*a*b + 100* (b2+2* A*C) + 10*2*b*c + c2. It is easy to know that the bits of M2 are only related to the last 1 bits of M, M2 10 are only related to the last 2 bits of M, and the hundred M2 are only related to the last 3 bits of M ... We can start from the single-digit search, a bit to meet N. If there is a workable solution, find the smallest solution in this layer and output it, otherwise the output is "None".

1#include <cstdio>2#include <queue>3 using namespacestd;4 5 structnode{6     Long LongValue,place;7 };8 9 voidBFsLong LongN)Ten { One node tmp; ATmp.value =0; -Tmp.place =1; -Queue<node>Q; the Q.push (TMP); -     BOOLFindans =false; -     Long LongAns =0xFFFFFFFF; -      while(!Q.empty ()) { +TMP =Q.front (); - Q.pop (); + node now; ANow.place = tmp.place*Ten; at          for(inti =0; i<Ten; ++i) { -Now.value = tmp.value+i*Tmp.place; -             if(Now.value*now.value%now.place = = n%now.place) { -                 if(!Findans) - Q.push (now); -                 if(Now.value*now.value%now.place = = N && now.value<ans) { inFindans =true; -Ans =Now.value; to                 } +             } -         } the     } *     if(ans = =0xFFFFFFFF) $printf"none\n");Panax Notoginseng     Else -printf"%i64d\n", ans); the } +  A intMain () the { +     intT; -scanf"%d",&t); $      while(t--){ $         Long LongN; -scanf"%i64d",&n); - BFS (n); the     } -     return 0;Wuyi}

The code above finds a workable solution that needs to be compared to find the optimal solution, we can optimize it, change the queue to Priority_queue, let Priority_queue help us do this work, so that the first feasible solution to find is to satisfy the test instructions optimal solution.

1#include <cstdio>2#include <queue>3 using namespacestd;4 5 structnode{6     Long LongValue,place;7     BOOL operator< (Constnode& b)Const8     {9         returnValue>B.value;Ten     } One }; A  - voidBFsLong LongN) - { the node tmp; -Tmp.value =0; -Tmp.place =1; -Priority_queue<node>Q; + Q.push (TMP); -      while(!Q.empty ()) { +TMP =q.top (); A Q.pop (); at         if(Tmp.value*tmp.value%tmp.place = =N) { -printf"%i64d\n", tmp.value); -             return ; -         } - node now; -Now.place = tmp.place*Ten; in          for(inti =0; i<Ten; ++i) { -Now.value = tmp.value+i*Tmp.place; to             if(Now.value*now.value%now.place = = n%now.place) { + Q.push (now); -             } the         } *     } $printf"none\n");Panax Notoginseng } -  the intMain () + { A     intT; thescanf"%d",&t); +      while(t--){ -         Long LongN; $scanf"%i64d",&n); $ BFS (n); -     } -     return 0; the}

HDU 4394 Digital Square