Good Numbers (hard version) CodeForces - 1249C2

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

题意：
输出大于输入数据所给的最小数（3的n次方且n不能重复的数）

思路：
将输入数据转为三进制数，从高位往低位找到第一个为2的位置，再从第一个为2的位置向高位找到第一个为0的位置，将这个为0的位置的值改为1并舍去所有低位的数，这个数即是我们要输出的数。

代码：

#include <stdio.h>
#include <string.h>
long long a[500];
long long good(long long b,long long c)
{
    long long s=1;
    while(c)
    {
        if(c%2==1)
            s*=b;
        b*=b;
        c=c/2;
    }
    return s;
}
int main(void)
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        long long n;
        scanf("%lld",&n);
        long long i=0,j,k,s=n;
        while(n>0)
        {
            a[i++]=n%3;
            n/=3;
        }
        j=i-1;
        for(; j>=0; j--)
        {
            if(a[j]==2)
                break;
        }
        if(j==-1)
            printf("%lld\n",s);
        else
        {
            k=j+1;
            for(; k<=i-1; k++)
            {
                if(a[k]==0)
                {
                    a[k]=1;
                    break;
                }
            }
            long long s1=0;
            if(k==i)
                a[i]=1;
            for(j=k; j<=i; j++)
            {
                if(a[j]==1)
                    s1+=good(3,j);
            }
            printf("%lld\n",s1);
        }
    }
    return 0;
}