Good Numbers The only difference between easy and hard versions is the maximum value of n.
You are given a positive integer number n. You really love good numbers so you want to find the smallest good number greater than or equal to n.
The positive integer is called good if it can be represented as a sum of distinct powers of 3 (i.e. no duplicates of powers of 3 are allowed).
For example:
30 is a good number: 30=3^ 3+3^1, 1 is a good number: 1=3^0, 12 is a good number: 12=3^ 2+3^ 1, but 2 is not a good number: you can’t represent it as a sum of distinct powers of 3 (2=3^ 0+3^ 0), 19 is not a good number: you can’t represent it as a sum of distinct powers of 3 (for example, the representations 19=3^ 2+3^ 2+3^ 0=3 ^2+3 ^1+3 ^1+3 ^1+3 ^0 are invalid), 20 is also not a good number: you can’t represent it as a sum of distinct powers of 3 (for example, the representation 20=3 ^2+3 ^2+3 ^0+3 ^0 is invalid). Note, that there exist other representations of 19 and 20 as sums of powers of 3 but none of them consists of distinct powers of 3.
For the given positive integer n find such smallest m (n≤m) that m is a good number.
You have to answer q independent queries.
Input The first line of the input contains one integer q (1≤q≤500) — the number of queries. Then q queries follow.
The only line of the query contains one integer n (1≤n≤10^18).
Output For each query, print such smallest integer m (where n≤m) that m is a good number.
Example
input 8 1 2 6 13 14 3620 10000 1000000000000000000 output 1 3 9 13 27 6561 19683 1350851717672992089
思路：定义一个sum=0，sum从3的0次方往后加， 当sum>=n的时候结束，然后sum依次减 之前累加的那些数，但仍要保证减过之后要>=n。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; typedef long long ll; int main() { int t; scanf("%d",&t); while(t--) { ll n; scanf("%lld",&n); ll sum=0,x=1; while(sum<=n) { sum+=x; x*=3; } while(x) { if(sum-x>=n) sum-=x; x/=3; } printf("%lld\n",sum); } return 0; }
