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hdu 2689 超级大水题....两种代码都过了,开始以为n^2会tle,后来竟然过了...汗 注意下cin写在while里面,就可以了 #include <iostream> using namespace std; int main() { int n,i,j,a[1010]; while(cin>>n) { for(i=1; i<=n; i++) cin>>a[i]; int cnt = 0; for(i=1; i<=n; i++) for(j=i+…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2689 Sort it Problem Description You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it …
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2689 题目分析:求至少交换多少次可排好序,可转换为逆序对问题. 用冒泡排序较为简单,复杂度较大~~ 也可用归并排序,复杂度O(lognn), 统计个数后复杂都不变. /* Sort it Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2660…
Sort it Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4672    Accepted Submission(s): 3244 Problem Description You want to processe a sequence of n distinct integers by swapping two adjacent s…
Problem Description You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need. For example, 1 2 3 5 4, we only need one operation : s…
Sort it Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4679    Accepted Submission(s): 3250 Problem Description You want to processe a sequence of n distinct integers by swapping two adjacent s…
Sort it You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need. For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.…
Tree Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2877    Accepted Submission(s): 883 Problem Description There are N (2<=N<=600) cities,each has a value of happiness,we consider two cities…
(一)树状数组的概念 如果给定一个数组,要你求里面所有数的和,一般都会想到累加.但是当那个数组很大的时候,累加就显得太耗时了,时间复杂度为O(n),并且采用累加的方法还有一个局限,那就是,当修改掉数组中的元素后,仍然要你求数组中某段元素的和,就显得麻烦了.所以我们就要用到树状数组,他的时间复杂度为O(lgn),相比之下就快得多.下面就讲一下什么是树状数组: 一般讲到树状数组都会少不了下面这个图: 下面来分析一下上面那个图看能得出什么规律: 据图可知:c1=a1,c2=a1+a2,c3=a3,c4…
找素数本来是很简单的问题,但当数据变大时,用朴素思想来找素数想必是会超时的,所以用素数筛法. 素数筛法 打表伪代码(用prime数组保存区间内的所有素数): void isPrime() vis[]数组清零://vis[]数组用于标记是否已被检验过 prime[]数组全赋初值false://prime[]数组从下标0开始记录素数 for i = 2 to MAXN (i++) if 数i未被检验过 prime[tot++]=i; for j = i*i to MAXN (j+=i) //j是i的…