题目网址: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=110064#problem/D Description Matt’s friend K.Bro is an ACMer. Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they ar…

K.Bro Sorting Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 2510 Accepted Submission(s): 1174 Problem Description Matt's friend K.Bro is an ACMer. Yesterday, K.Bro learnt an algorithm: Bubble…

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Matt’s friend K.Bro is an ACMer. Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble so…

Matt's friend K.Bro is an ACMer.Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong order. The process repeats until no swap is needed.Today, K.Bro comes up wi…

Description Matt’s friend K.Bro is an ACMer. Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong order. The process repeats until no swap is needed. Today, K.B…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5122 Problem Description Matt's friend K.Bro is an ACMer.Yesterday, K.Bro learnt an algorithm: Bubble sort. Bubble sort will compare each pair of adjacent items and swap them if they are in the wrong or…

K.Bro Sorting Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) Total Submission(s): 571 Accepted Submission(s): 300 Problem Description Matt's friend K.Bro is an ACMer. Yesterday, K.Bro learnt an algorithm: Bubble so…

K.Bro Sorting Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 3072    Accepted Submission(s): 1390 Problem Description Matt’s friend K.Bro is an ACMer. Yesterday, K.Bro learnt an algorithm: Bu…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5122 解题报告:定义一种排序算法,每一轮可以随机找一个数,把这个数与后面的比这个数小的交换,一直往后判断,直到后面没有比这个数小的,这样称为一轮,现在给定一个长度为n的序列,要你求,至少要经过多少轮,可以使这个序列成为有序的. 由于只能跟后面的数进行比较,所以我只要统计后面的数有比这个数小的数的个数就可以了.从后往前扫一遍,每次更新当前最小的. #include<cstdio> #include&…

思路: 模拟. 实现: #include <iostream> #include <cstdio> using namespace std; ], n, t; int main() { int temp; ; cin >> t; while (t--) { scanf("%d", &n); ; ; i < n; i++) { scanf("%d", &a[i]); } ; ; i >= ; i--) {…