current : 5 / 9
A. The beautiful values of the palace????
B. super_log????
C. Tsy's number 5
D. Robots
概率DP
E. K Sum
F. Greedy Sequence????
G. Quadrilateral
H. Holy Grail????
https://blog.csdn.net/TDD_Master/article/details/100374149
学习了贪心的一个写法。
我们先把洗衣顺序按时间排序。
最后的策略一定是，前 $j$ 个人手洗，之后的人用洗衣机洗。
我们枚举x 在$[1, y]$的值，每次只有后 y/x个人可能用洗衣机洗。
复杂度是一个调和级数。
// #pragma GCC optimize(2) // #pragma GCC optimize(3) // #pragma GCC optimize(4) #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> #include <unordered_set> // #include<bits/extc++.h> // using namespace __gnu_pbds; using namespace std; #define pb push_back #define fi first #define se second #define debug(x) cerr<<#x << " := " << x << endl; #define bug cerr<<"-----------------------"<<endl; #define FOR(a, b, c) for(int a = b; a <= c; ++ a) typedef long long ll; typedef long double ld; typedef pair<int, int> pii; typedef pair<ll, ll> pll; const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9+7; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } /**********showtime************/ const int maxn = 1e6+9; int t[maxn]; int main(){ int n,m; while(~scanf("%d%d", &n, &m)) { for(int i=1; i<=n; i++) scanf("%d", &t[i]); sort(t+1, t+1+n); for(int i=1; i<=m; i++) { int x = i, y = m; int cnt = min(n, y / x); ll ans = inff; ll bk = 0; for(int j=n; j>=(n - cnt + 1); j--) { ll t1 = 1ll * (n-j+1)*x + t[j]; bk = max(bk, t1); t1 = bk; if(j > 1) t1 = max(t1, 1ll*t[j-1] + y); ans = min(ans, t1); } if(i < m) printf("%lld ", ans); else printf("%lld\n", ans); } } return 0; }