Problem - D - Codeforces题意: 有一个长度为n的只包含0和1的a数组,分别对该数组的前i位排序 ( 1\leq i\leq n ) ,可以构成n个长度为n的01数组,将n个数组的第i位相加可以得到一个b数组,现在给你b数组让你求出…
OrEwA_ThePratham04 → Codeforces Submissions Showing 'N/A' Instead of Code — Need Help! _MyAngelGasai → If I don't reach E by this summer, I will shave my head. Cipesta... → The Hidden Impact of CP _kaitokid → Change the flag of Syria Ryuuji_Nakamura → Why is my code ...
B:http://codeforces.com/contest/1360/problem/B 题意: 将n个数分成两组,要求一组的最大值减去另一组的最小值的绝对值最小。 解析: 直接sort然后找最小间距 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; typedef long long ll; const...
打Codeforces Round #782 (Div. 2)时遇到的有趣的题,记录一下 D. Reverse Sort Sum 题意 对于一个长度为 n 的01数列 A ,我们定义 Bi 为将A 的前i 位排序后得到的数列,如 A=[0,1,1,0,1,0] 那么 B1=[0,1,1,0,1,0] B2=[0,1,1,0,1,0] B3=[0,1,1,0,1,0] B4=[0,0,...
· codeforces round 782 Div2 D. Reverse Sort Sum · 【做题笔记】CF1659D Reverse Sort Sum · 【笔记】构造题 阅读排行: · 一个基于 C# Unity 开发的金庸群侠传 3D 版,直呼牛逼! · 四十种AI编程工具,让你码字如飞 · 当数据爆炸遇上SQL Server:优化策略全链路解析 · LM Studio本地部署...
sort(a + L, a + R +1); }; intmain() { while(cin>> n) { memset(sum,0,sizeof(sum)); inttot =1<< n; for(inti =1; i <= tot; i++){ scanf("%I64d", a + i); } build(n,1, tot); intm, q; cin>> m;
sort(a+l,a+r+1); }intmain() { scanf("%d",&n);inty=(1<<n); Riep(y)scanf("%d",a+i); dfs(1,y,n);intq,x; scanf("%d",&q);while(q--) { scanf("%d",&x);while(x) { swap(sum[x][0],sum[x][1]); x--; ...
I recently scraped almost all of the submissions from Codeforces. Here I share all the source code and metadata (problem ID, submitter, language, verdict, etc.):https://mega.nz/folder/Sypi0BrS#iNbQXf3EwcjZbpwXRKHOnQ. The dataset contains at least 99.8% of the public submissions with ID...
pop(); if(u == v) break; } sort(c.begin(),c.end()); scc.push_back(c); } } bool check(int val) { for(int i = 1;i <= n + 5;i ++) g[i].clear(); for(int i = 1;i <= m;i ++){ g[eu[i]].push_back(ev[i]); if(ew[i] > val) continue; g[ev[i]]....
sort(a + L, a + R + 1); }; int main() { while (cin >> n) { memset(sum, 0, sizeof(sum)); int tot = 1 << n; for (int i = 1; i <= tot; i++){ scanf("%I64d", a + i); } build(n, 1, tot); int m, q; cin >> m; for (int i = 0; i < m; i+...