Codeforces Round #595 (Div. 3) C2. Good Numbers (hard version)

感觉这种方法比较巧妙，且易于理解。贪心求，从最大位遍历，能减就减

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define debug(x, y) cout <<x<<" "<<y<<endl
#define nl cout << endl
#define inf 0x3f3f3f3f
const int maxn = 2e5+10;
ll ma, d;
void init()
{
    d = 1;
    ma = 0;
    while(ma < 1e18) {
        ma += d;
        d *= 3;
    }
}
int main()
{
    int t;
    init();
    scanf("%d", &t);
    while(t--) {
        ll n;
        scanf("%I64d", &n);
        ll temp = d;
        ll ans = ma;
        while(temp)
        {
            if(ans-temp >= n) ans -= temp;
            temp /= 3;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}

第二种方法
这种方法就是先转成三进制，找位上为2的数，把之前的变成0，后一位加一，有限次数内一定能完成找到需要的数。

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
#define debug(x, y) cout <<x<<" "<<y<<endl
#define nl cout << endl
#define inf 0x3f3f3f3f
const int maxn = 2e5+10;
const int N = 2e5+10;

int main()
{
    int t;
    scanf("%d", &t);
    while(t--) {
        ll n;vector<int> pre;
        scanf("%I64d", &n);
        while(n)
        {
            pre.push_back(n%3);
            n /= 3;
        }
        pre.push_back(0);
        pre.push_back(0);
        for(int i = 0; i <= 60; ++i)
        {
            int ans = -1;
            for(int j = 0; j < (int)pre.size(); ++j)
                if(pre[j]==2) ans = j;
            if(ans == -1) break;
            for(int j = 0; j <= ans; ++j) pre[j] = 0;
            pre[ans+1]++;
        }
        ll ans = 0;
        ll d = 1;
        for(int i = 0; i < (int)pre.size(); ++i) {
            ans += d*pre[i];
            d *= 3;
        }
        printf("%lld\n", ans);
    }
    return 0;
}