标签：byte dash time max out gre col ssis include
The only difference between easy and hard versions is the maximum value of nn.
You are given a positive integer number nn. You really love good numbers so you want to find the smallest good number greater than or equal to nn.
The positive integer is called good if it can be represented as a sum of distinct powers of 33 (i.e. no duplicates of powers of 33 are allowed).
For example:
Note, that there exist other representations of 1919 and 2020 as sums of powers of 33 but none of them consists of distinct powers of 33.
For the given positive integer nn find such smallest mm (n≤mn≤m) that mm is a good number.
You have to answer qq independent queries.
The first line of the input contains one integer qq (1≤q≤5001≤q≤500) — the number of queries. Then qqqueries follow.
The only line of the query contains one integer nn (1≤n≤10181≤n≤1018).
For each query, print such smallest integer mm (where n≤mn≤m) that mm is a good number.
8 1 2 6 13 14 3620 10000 1000000000000000000
1 3 9 13 27 6561 19683 1350851717672992089
//#include <bits/stdc++.h> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <algorithm> #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdio.h> #include <queue> #include <stack>; #include <map> #include <set> #include <string.h> #include <vector> #define ME(x , y) memset(x , y , sizeof(x)) #define SF(n) scanf("%d" , &n) #define rep(i , n) for(int i = 0 ; i < n ; i ++) #define INF 0x3f3f3f3f #define mod 998244353 #define PI acos(-1) using namespace std; typedef long long ll ; ll a[109] ; int main() { int t ; scanf("%d" , &t); while(t--) { ll n ; scanf("%lld" , &n); memset(a , 0 , sizeof(a)); int cnt = 0 ; ll tmp = n ; while(tmp) { a[cnt++] = tmp % 3 ; tmp /= 3 ; } int pos = -1 ; for(int i = 0 ; i < cnt ; i++) { if(a[i] >= 2) { pos = i ; a[i] = 0 ; a[i+1]++; } } if(pos != -1) { for(int i = 0 ; i < pos ; i++)//确保最小 { a[i] = 0 ; } } ll ans = 0 ; ll res = 1 ; for(int i = 0 ; i <= cnt ; i++)//注意取等 { ans += a[i] * res; res *= 3 ; } cout << ans << endl ; } return 0 ; }
标签：byte dash time max out gre col ssis include
原文地址：https://www.cnblogs.com/nonames/p/11749966.html
