A. Jeff and Rounding time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jeff got 2n real numbers a1, a2, ..., a2n as a birt
重复至少 K 次且长度为 M 的模式 思路: 直接模拟 class Solution {public:bool containsPattern(vector<int>& arr, int m, int k) {int n = arr.size();for(int i = 0;i + m - 1 < n;i++) { //起点int cnt = 1;for(int j = i + m;j +
class Solution {public int countPrimes(int n) {int result=0;for(int i=2;i<n;i++){if(isPrimes(i)){result++;}}return result;}boolean isPrimes(int n){for(int i=2;i<=Math.sqrt(n);i++){if(n%i==0){return fa
题目 Description: Count the number of prime numbers less than a non-negative number, n. Credits: Special thanks to @mithmatt for adding this problem and creating all test cases. Hint: Let’s start
小a与"204"(C++) 点击做题网站链接 题目描述 小a非常喜欢204这个数字,因为′a′+′k′=204。 现在他有一个长度为n的序列,其中只含有2,0,4这三种数字 设ai为序列中第i个数,你需要重新排列这个数列,使得 ∑ i = 1 n ( a i − a i − 1 ) 2 \sum_{i=1}^n (a_i−a_{i−1})^2 ∑i=1n(ai−ai−1)2最大(公式的含