题目链接：CodeForces - 1249C1
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
7 1 2 6 13 14 3620 10000Output
1 3 9 13 27 6561 19683题目大意：找到>=n的最小的数，这个数是Good number。
也就是将这个数分解成3的幂的和（但是幂不能相等）。。。
30=3^ 3+3^1（1和3不等）；YES
2=3^ 0+3^ 0（0==0）；NO
思路：增加一位最高位设为0，用来避免当前位数的3进制数无法满足要求，然后从最高位向下寻找，如果为0或者1，跳过；遇到2，说明不满足要求，我们要增大数字，令该位置及更低位的数字都为0，并向上寻找，如果为2或者1，就令其为0；否则遇到0（遇到的第一个0改为1就结束），令其为1；这样得到的就是满足条件的最小值。
#include<stdio.h> #include<string.h> #include<math.h> int a[110]; int r; void solve(int x) { r=0; while(x) { a[r++]=x%3; x/=3; } } int main() { int t; scanf("%d",&t); while(t--) { memset(a,0,sizeof a); int n; scanf("%d",&n); solve(n); a[r]=0; for(int i=r; i>=0; i--) { if(a[i]==2) { for(int j=i+1; j<=r; j++) { if(a[j]==1) a[j]=0; else if(a[j]==0) { a[j]=1; break; } } for(int k=i; k>=0; k--) a[k]=0; break; } } int ans=0; for(int i=r; i>=0; i--) { ans+=a[i]*pow(3,i); } printf("%d\n",ans); } return 0; }
