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 下面请你构造一个1∼n1∼n的排列,并且恰好调用k...
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...
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...
int n,k,a[100005]; void merge(int l,int r) { if(k<=0 || l+1>=r) { sort(a+l,a+r); return; } k-=2; int mid=(l+r)/2; merge(l,mid); merge(mid,r); } int main(void) { scanf("%d%d",&n,&k); if(k%2==0 || k>2*n) { printf("-1\n"); return 0; } ...
I am facing a runtime issue in my code.Can anyone help me with my code: My code link: https://pastebin.com/qCC4GsPS. Just check the merge and mergesort function in this link.#merge sort, #linked list -6 rsudhanshu138 4 years ago 0 ...
private void MergeSort(int[] array, int start, int end) // 归并排序 { if (start < end) { int mid = (start + end) / 2; MergeSort(array, start, mid); // 将前半部分排序 MergeSort(array, mid + 1, end); // 将后半部分排序 ...
merge(u, v + n); dsu.merge(u + n, v); } else { if(bel[u] > bel[v]) swap(u, v); cons.push_back({u, v}); } } vector<int> ok(k, 1); for(int i = 0; i < n; i++) { if(dsu.find(i) == dsu.find(i + n)) ok[bel[i]] = 0; } ll okCnt = 0; for...
x : fa[x] = find(fa[x]); } void merge(int x, int y) { if ((x = find(x)) != (y = find(y))) fa[x] = y; } int countComponent() { int res = 0; for (int i = 1; i <= n; i++) res += fa[i] == i; return res; } } using namespace DSU; void solve(...
网速题,一个字符查看是否在"Codeforces"里包含。 #include<bits/stdc++.h>usingnamespacestd;intmain(){freopen("test.in","r", stdin);inttt; cin >> tt; string s ="codeforces";while(tt -- ) {charc; cin >> c;boolf =false;for(autot : s)if(c == t) f =true; ...
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