题目描述

Polycarp doesn't like integers that are divisible by $3$ or end with the digit $3$ in their decimal representation. Integers that meet both conditions are disliked by Polycarp, too.

Polycarp starts to write out the positive (greater than $0$ ) integers which he likes: $1, 2, 4, 5, 7, 8, 10, 11, 14, 16, \dots$ . Output the $k$ -th element of this sequence (the elements are numbered from $1$ ).

输入格式

The first line contains one integer $t$ ( $1 \le t \le 100$ ) — the number of test cases. Then $t$ test cases follow.

Each test case consists of one line containing one integer $k$ ( $1 \le k \le 1000$ ).

输出格式

For each test case, output in a separate line one integer $x$ — the $k$ -th element of the sequence that was written out by Polycarp.

输入输出样例

输入 #1

10
1
2
3
4
5
6
7
8
9
1000

输出 #1

1
2
4
5
7
8
10
11
14
1666

分析

题意：给定一个序列，包含除了 $3$ 的倍数和个位是 $3$ 数之外的全部正整数，求序列的第 $k$ 项。

• 判断一个数是否是 $3$ 的倍数，只需要检查这个数 $\mod 3$ 的结果，如果为 $0$ 则说明是 $3$ 的倍数，反之则不是 $3$ 的倍数。
• 判断一个数的个位是否为 $3$，只需要检查这个数 $\mod 10$ 的结果，得到的运算结果即为这个数的个位。

于是，我们可以从 $1$ 开始从小到大扫描正整数，检查当前的数字是否满足要求，如果满足要求，则累加当前序列项数，判断是否满足条件输出即可。

时间复杂度 $O(k)$。

代码

#include<bits/stdc++.h>
using namespace std;
int t,k;
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&k);
        int now=1,x=0;
        for(int i=1;i<=k;i++){
            x++;
            while(x%3==0||x%10==3)x++;
        }
        printf("%d\n",x);
    }
    return 0;
}