2019南京网络赛
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👌
I. Washing clothes
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; }
李超树
https://www.cnblogs.com/pkgunboat/p/11449031.html
// #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; ll a[maxn]; ll ans[maxn]; struct node{ int flag; int k, b; ll cal(int x) { return 1ll*k * x + b; } }tree[maxn << 2]; void build(int le, int ri ,int rt) { tree[rt].flag = 0; if(le == ri) return ; int mid = (le + ri) >> 1; build(le, mid, rt<<1); build(mid+1, ri, rt<<1|1); } void update(int k, int b, int le, int ri, int rt) { if(tree[rt].flag == 0) { tree[rt].flag = 1; tree[rt].k = k; tree[rt].b = b; return; } ll l1 = tree[rt].cal(le); ll r1 = tree[rt].cal(ri); ll l2 = 1ll*k * le + b; ll r2 = 1ll*k * ri + b; if(l1 >= l2 && r1 >= r2) return; if(l1 <= l2 && r1 <= r2) { tree[rt].k = k; tree[rt].b = b; return; } double pos = 1.0 * (b - tree[rt].b) / (tree[rt].k - k); int mid = (le + ri) >> 1; if(l1 >= l2) { if(pos <= mid) { update(tree[rt].k, tree[rt].b, le, mid, rt<<1); tree[rt].k = k; tree[rt].b = b; } else { update(k, b, mid+1, ri, rt<<1|1); } } else { if(pos <= mid) { update(k, b, le, mid, rt<<1); } else { update(tree[rt].k, tree[rt].b, mid+1, ri, rt<<1|1); tree[rt].k = k; tree[rt].b = b; } } } ll query(int pos, int le, int ri, int rt) { if(tree[rt].flag == 0) return 0; ll res = tree[rt].cal(pos); if(le == ri) return res; int mid = (le + ri) >> 1; if(pos <= mid) res = max(res,query(pos, le, mid, rt<<1)); else res = max(res, query(pos, mid+1, ri, rt<<1|1)); return res; } int main(){ int n,m; while(~scanf("%d%d", &n, &m)) { for(int i=1; i<=n; i++) scanf("%lld", &a[i]); sort(a+1, a+1+n); build(1, m, 1); update(1, a[n], 1, m, 1); a[0] = -m; int p = n-1; for(int x=m; x>=1; x--){ ll res = max(a[p] + m, query(x, 1, m, 1)); // cout<<x<<" , " << res<<endl; while(p >= 1) { int tmp = max(a[p-1] + m, max(a[p] + x * (n - p + 1), query(x, 1, m, 1))); if(tmp <= res) { res = tmp; update(n - p + 1, a[p], 1, m, 1); p--; } else break; } ans[x] = res; } for(int i=1; i<m; i++) printf("%lld ", ans[i]); printf("%lld\n", ans[m]); } return 0; }
skr
