welcome to 浩·C's blog

//思路：枚举y:[k+1,2k]#include#includeusing namespace std;int main() { int k; while(scanf("%d", &k) == 1 && k) { vector X, Y; for(int y = k+1; y <= k*2; y++) { // 1/k = 1/x + 1/y =>

Given a sequence of integers S = {S1, S2, . . . , Sn}, you should determine what is the value of themaximum positive product involving consecutive terms of S. If you cannot find a positive sequence,

Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N,

DesertTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 1126    Accepted Submission(s): 803Problem DescriptionA tourist gets lost

美团外卖的品牌代言人袋鼠先生最近正在进行音乐研究。他有两段音频，每段音频是一个表示音高的序列。现在袋鼠先生想要在第二段音频中找出与第一段音频最相近的部分。具体地说，就是在第二段音频中找到一个长度和第一段音频相等且是连续的子序列，使得它们的 difference 最小。两段等长音频的 difference 定义为：difference = SUM(a[i] - b[i])2 (1 ≤