51Nod 1010 contains only the number of factor 2 3 5 Label:none

The factor of K contains only 2 3 5. The first 10 numbers that meet the criteria are: 2,3,4,5,6,8,9,10,12,15. All such k make up a sequence s, and now give a number n, to find the smallest number of >= given in S. For example: n = the minimum number of >= 13 in 13,s is 15, so output 15. Input

Line 1th: A number t that represents the number of numbers that are later used as input tests. (1 <= T <= 10000) 2-t + 1 lines: 1 numbers per line n (1 <= n <= 10^18)

Output

A total of T-lines, 1 digits per line, and the smallest of the output >= n contains only the number of factors 2 3 5.

Input example

518133577

Output example

28153680

Code

1#include <iostream>2#include <cstdlib>3#include <cstring>4#include <cstdio>5#include <algorithm>6#include <queue>7#include <map>8#include <vector>9 #definell Long LongTen #defineINF 1500000000000000000 One using namespacestd;  A  - structcmp{BOOL operator(ll A,ll b) {returnA>b;}}; -Priority_queue<ll,vector<ll>,cmp>Q; theMap<ll,int>m; -  -ll a[1000005],m,p; -  + voidInit_ () { -Q.push (1); +      while(Q.top () <INF) { All x=q.top (); Q.pop (); ata[p++]=x; -         if(!m.count (2*x)) {Q.push (2*X); m[2*x]=1;} -         if(!m.count (3*x)) {Q.push (3*X); m[3*x]=1;} -         if(!m.count (5*x)) {Q.push (5*X); m[5*x]=1;} -     } -      in     //cout<<p<<endl; -scanf"%d",&M); to } +  - voidWork () { the      while(m--){ * ll X; $scanf"%lld",&x);Panax Notoginseng          -printf"%lld\n", *lower_bound (A +1, a+p+1, x)); the     } + } A  the intMain () { + //freopen ("01.in", "R", stdin); -      $ Init_ (); $ Work (); -      -Fclose (stdin); fclose (stdout);return 0; the}

Make a priority queue, generate larger numbers with existing numbers

Map de-weight, lower_bound two-point acceleration

INF less hit a 0 and unhappy

Time is OK, I think the main question is the time, because the counter bacteria that the number in the data range only 1w+

51Nod 1010 contains only the number of factor 2 3 5 Label:none