PAT basic 1030
来源:互联网 发布:linux 终端切换 编辑:程序博客网 时间:2024/06/13 19:22
#include <iostream>#include <algorithm>#include <vector>using namespace std;int main() { int n; long long p; cin >> n >> p; if (n == 0) { cout << n; return 0; } vector<long long int> a(n); for (int i = 0; i < n; i++) cin >> a[i]; sort(a.begin(), a.end()); int result = 1; for (int i = 0; i <= n - 2; i++) { //轮扫 for (int j = i + result; j <= n - 1; j++) { if (a[j] > a[i] * p) break; result = max(result, j - i + 1); } } cout << result; return 0;}
阅读全文
0 0
- PAT basic 1030
- PAT Basic Level (1026~1030)
- PAT Basic
- Pat(Basic Level)Practice--1030(完美数列)
- PAT(basic level) 1030 完美数列(25)
- PAT (Basic Level) Practise
- PAT Basic 1001
- PAT Basic 1002
- PAT Basic 1005
- PAT Basic 1006
- PAT Basic 1007
- PAT Basic 1008
- PAT Basic 1009
- PAT Basic 1010
- pat basic level 1016
- pat basic level 1018
- pat basic level 1019
- PAT(basic level)题解
- PAT basic 1028
- JAVA语言(五)
- eclipse找不到tools.jar下的内容的解决方法
- maven 介绍
- PAT basic 1029
- PAT basic 1030
- 机器学习集训营---第四周总结
- Java菜鸟教程 一些简单的练习
- Educational Codeforces Round 26 C. Two Seals
- 瞬时态、持久态、脱管态
- python 學習的一些心得分享(基礎)
- Java 异常总结
- Unity3D
- DNS (域名系统)