I am not good in Explaining but still I'm trying, Hope you will get it. Let's consider Arrays A and B and resultant array C (merged array). The element from the first array will be appended to C until the element of the second array is larger, For example lets take first Testcase...
CodeForces 873D Merge Sort 构造 分治 题意 给出一个归并排序的算法mergesortmergesort,如果对于当前区间[l,r)[l,r)是有序的,则函数直接返回。 否则会分别调用mergesort(l,mid)mergesort(l,mid)和mergesort(mid,r)mergesort(mid,r),其中mid=⌊l+r2⌋mid=⌊l+r2⌋ 最后合并左右两个子区间 下面请...
codeforces 873 D. Merge Sort(分治) 题目链接:http://codeforces.com/contest/873/problem/D 题解:这题挺简单的,除了一开始算作是调用到一次,然后每次执行操作时都会调用2次,所以最多调用几次就很好算了,而且只有奇数调用次数才合理。然后就是类似分治的思想,每次dfs二分过去,发现调用次数不够就交换mid和mid-1...
In the third example all integers in the given array are distinct, so it will not change. 题意:给你一组数,找到其中最小的数(最少有2个或以上),将最左边的数删除,左数第二的*2。然后重复操作 例如[3,4,1,2,2,1,1] → [3,4,2,2,2,1] → [3,4,4,2,1] → [3,8,2,1][3,4...
Sometimes I encounter a type of range queries that I don't know how to do using segment tree, so I use a merge sort tree instead. It can answer queries in O(log2n)O(log2n). I decided to share this because it can be useful to many in contests. First, we know that a node in...
Educational Codeforces Round 30-D-Merge Sort(瞎搞) 题意:给你n和k,问你能否构造出一个长为n的全排列,使得按照归并排序的执行规则,正好调动k次归并排序函数使得原排列变成递增排列。 题解:我们知道对于递减的排列,需要调用归并函数最多次,因此我们可以初始化数组为递减全排列,然后调用归并函数即可,每次让k减2(...
14 changes: 7 additions & 7 deletions 14 docs/basic/stl-sort.md Original file line numberDiff line numberDiff line change @@ -102,13 +102,13 @@ std::sort(da + 1, da + 1 + 10, cmp); // 使用 cmp 函数进行比较,从大 严格弱序的要求: 1. $x \not< x$ (非自反性) 2. 若$...
int c=0; sort(a,a+n); for(int i=0;i<n;i++){ if(a[i]!=a[i+1]){ c++; } } cout<<n-c<<endl; } Binary file added BIN +60.2 KB Is_your_horseshoe_on_the_other_hoof.exe Binary file not shown. 16 changes: 16 additions & 0 deletions 16 calculating.cpp Original file ...
Dragons http://codeforces.com/problemset/problem/230/A Dragons time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Kirito is stuck on a level of the MMORPG he is playing now. To move on in the game, he's got to defeat alln...
847B - Preparing for Merge Sort 思路:前面的排序的最后一个一定大于后面的排序的最后一个。所以判断要不要开始新的排序只要拿当前值和上一个排序最后一个比较就可以了。 代码: #include<bits/stdc++.h>usingnamespacestd;#definell long long#definepb push_back#definemem(a,b) memset(a,b,sizeof(a))...