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 ).
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 ).
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.
10
1
2
3
4
5
6
7
8
9
1000
1
2
4
5
7
8
10
11
14
1666
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。
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)。
#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;
}
最后修改:2021 年 08 月 19 日 06 : 19 PM
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