Your task is to calculate the minimum number of operations described above to sort the permutation pp in ascending order (i. e. transform pp so that p1<p2<⋯<pnp1<p2<⋯<pn). The first line contains a single integer tt (1≤t≤1041≤t≤104)— the number of test cases. The fir...
C. Min Max Sort Problem - C - Codeforcescodeforces.com/contest/1792/problem/C 分析: 若n为偶数,那么操作的时候一定是操作(1,n),(2,n−1),(3,n−2),⋯这种成对的数. 为了保证最终有序我们一定是从中间的数对向两边操作. 并且有一个性质,对(x,y)操作之后,所有的(x−1,y+1),(x...
for(inti=1;i<=n;i++)scanf("%d",a+i),pq.push(a[i]); sort(a+1,a+n+1); intmaxx=a[n]+m; intminn=0; for(inti=1;i<=m;i++) { intcur=pq.top();pq.pop(); pq.push(cur+1); } for(inti=1;i<=n;i++) { minn=max(minn,pq.top());pq.pop(); } printf("%d %d\...
AC代码: 1#include<iostream>2#include<algorithm>3usingnamespacestd;4constintmaxn =107;5intvis[maxn];6intn,m;7intmain()8{9cin>>n>>m;10for(inti=1;i<=n;i++)11{12intx,y;13cin>>x>>y;14for(inti=x;i<y;i++)15{16vis[i]++;17}18if(y==m)vis[y]++;19}20intok =0;21for...
Limited Sorting Use: Although Heap Sort exists, it’s generally outperformed by Merge Sort & Quick Sort in practice. No Ordered Traversal: Unlike BSTs, you cannot traverse elements in sorted order efficiently. 4. When to Use Heaps in Codeforces?
codeforces 839A 水题 题意 分糖果 每次最多8个 多出来的可以遗留到下一天 注意下一天少于8个要清0; 代码: #include<bits/stdc++.h> using namespace std; int n,k; int q,b,ok; int a; long long sum; int main() { cin>>n>>k;
Usual algorithm: Dynamic programming (refer to this blog :Codeforces blog) My (maybe-wrong) greedy idea: Root the tree at arbitrary node, let say node 1. Find the distances of other nodes to node 1 (means find the height of each node). Sort the nodes according to their distances to no...
'\n':' '); continue; } sort(a,a+n,greater<int>()); for (i = 0; i < n; i++) { for (j = n-1; j > i; j--) { LLI diff = -abs(a[j]-a[j-1]-c); if (j < n-1) { diff -= abs(a[j+1]-a[j]-c); diff += abs(a[j+1]-a[j-1]-c); } if (i ...
D - Maxflow E - MinCostFlow F - Convolution G - SCC H - Two SAT I - Number of Substrings J - Segment Tree K - Range Affine Range Sum L - Lazy Segment Tree AtCoder Typical Contest 001 https://atcoder.jp/contests/atc001
else if(x1<=b&&ma[i].y<=a&&ma[j].x<=a)sum=max(sum,m); else if(x4<=a&&ma[i].x<=b&&ma[j].x<=b)sum=max(sum,m); else if(x4<=b&&ma[i].x<=a&&ma[j].x<=a)sum=max(sum,m); else if(x2<=a&&ma[i].y<=b&&ma[j].y<=b)sum=max(sum,m); ...