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;
}
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