D. Reverse Sort Sum(思维+构造) Problem - D - Codeforcescodeforces.com/contest/1659/problem/D 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≤i≤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数组让你求出原始的a数组。 思路: 对...
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,...
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+...
The first line of input contains a single integern(0 ≤ n ≤ 20). The second line of input contains2nspace-separated integersa[1], a[2], ..., a[2n] (1 ≤ a[i] ≤ 109), the initial array. The third line of input contains a single integerm(1 ≤...
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 int maxn=1e3...
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]]....
count where the 1st element is from left and the 2nd element is from right. If after solving the subproblems on left and right, these two subarrays are also in sorted order, then we can count the cross pair in linear time by using two pointers. This can be embedded into merge sort. ...
[i].p;44}45sort(N,N+n,cmp1);46sort(M,M+n,cmp2);47intexm=0;48for(inti=0;i<n;i++)49{50exm+=N[i].p;51if(N[i].init-c*exm>0)52{53ans1+=N[i].init-c*exm;54}55}56exm=0;57for(inti=0;i<n;i++)58{59exm+=M[i].p;60if(M[i].init-c*exm>0)61{62ans2+=(...