NOIP模拟3
三元
构造题，显然第一列要填 \(n\) 个 \(0、n\) 个\(1\) 和\(n\)个\(2\)，所以就直接把 $ 0、1、2$ 按顺序放好
将\(0、1、2\)开头的串分为三块
题目要求字典序最大的串字典序最小，所以就尽可能地在\(2\)后面放\(0\)，然而都放\(0\)显然会重复，所以就在最后一列按照\(0、1、2\)的顺序从上往下放，然而\(n\)大于\(3\)的时候又会重复，所以要把前面的一个\(0\)换成\(1\)，还不够就把\(0\)换成\(2\)，再不够就继续往前一列换，为了方便替换的时候依然按照\(0、1、2\)的顺序放，就会发现每一列都有一个循环，而每一列的循环的长度都是\(3^{L-i+1}\)(\(L\)是串的长度，\(i\)是第几列)， 因此可以直接按照循环放，而为了保证每一列\(0、1、2\)的出现次数都为\(n\)，应在上面的块中改变\(0、1、2\)的顺序，改为\(2、0、1\) 和\(1、2、0\)即可，不理解可以手模一下
code
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<cmath>
using namespace std;
#define Miuna printf ("dai suki...")
#define mizuki signed
#define love (1209 & 322 & 901)
namespace LRH
{
template <typename Shiodome_miuna> inline void in (Shiodome_miuna &x)
{
x = 0;
char f = 0; char ch;
while (!isdigit (ch = getchar ())) if (ch == '-') f = 1;
do
{
x = (x << 1) + (x << 3) + (ch ^ 48);
} while (isdigit (ch = getchar ()));
if (f) x = -x;
}
int stk[20], tp;
template <typename Shiodome_miuna> inline void ot (Shiodome_miuna x, int f = 2)
{
if (x < 0) putchar ('-'), x = -x;
do
{
stk[++ tp] = x % 10;
x /= 10;
} while (x);
while (tp)
{
putchar (stk[tp --] | 48);
}
if (f == 1) putchar ('\n');
if (!f) putchar (' ');
}
}using namespace LRH;
typedef long long ll;
const int maxn = 5e4 + 10;
int n, L;
int a[3 * maxn][20];
int up[20];
mizuki main ()
{
freopen ("three.in", "r", stdin);
freopen ("three.out", "w", stdout);
in (n), in (L);
int nn = n << 1, nnn = n * 3;
for (int i = 1; i <= L; ++ i) up[i] = pow (3, L - i);
for (int i = 1; i <= n; ++ i) a[i][1] = 0;
for (int i = n + 1; i <= nn; ++ i) a[i][1] = 1;
for (int i = nn + 1; i <= nnn; ++ i) a[i][1] = 2;
for (int l = 2; l <= L; ++ l)
{
for (int i = 1; i <= n; ++ i)
{
int tmp = i % (3 * up[l]);
if (tmp == 0) tmp = 3 * up[l];
int kk = (tmp - 1) / up[l];
if (kk == 0) a[i][l] = 2;
else if (kk == 1) a[i][l] = 0;
else if (kk == 2) a[i][l] = 1;
}
}
for (int l = 2; l <= L; ++ l)
{
for (int i = n + 1; i <= nn; ++ i)
{
int k = i - n;
int tmp = k % (3 * up[l]);
if (tmp == 0) tmp = 3 * up[l];
if ((tmp - 1) / up[l] == 0) a[i][l] = 1;
else if ((tmp - 1) / up[l] == 1) a[i][l] = 2;
else if ((tmp - 1) / up[l] == 2) a[i][l] = 0;
}
}
for (int l = 2; l <= L; ++ l)
{
for (int i = nn + 1; i <= nnn; ++ i)
{
int k = i - nn;
int tmp = k % (3 * up[l]);
if (tmp == 0) tmp = 3 * up[l];
if ((tmp - 1) / up[l] == 0) a[i][l] = 0;
else if ((tmp - 1) / up[l] == 1) a[i][l] = 1;
else if ((tmp - 1) / up[l] == 2) a[i][l] = 2;
}
}
for (int i = 1; i <= nnn; ++ i)
{
for (int l = 1; l <= L; ++ l)
{
ot (a[i][l]);
}
putchar ('\n');
}
return love;
}
鹅国
非正解，是\(O(nk^3)\)的，\(c_i\)是\(c_{i-1}\)的倍数，也就是说\(c_i\)是可以替换成几个\(c_{i-1}\)的，于是就可以想\(DP\)了
设\(f_{i, J,k}\)和\(g_{i, J, k}\)表示后\(i\)个币一共用了\(J\)个，用了\(k\)个第\(i\)种币的最大\(/\)小重量
可以枚举将几个第\(i\)种币转成第\(i - 1\)种币，设转化\(s\)个，一个\(i\)种币可以转成\(t\)个\(i-1\)种币，则有转移
\(f_{i, J, k} +s\times w_i+(s\times t)\times w_{i-1}→ f_{i-1,J+s\times t-s, s\times t}\)
发现初始状态不好设
因为我们是倒着枚举\(i\)，所以可以尽量先选面值大的币，于是就可以处理出来初始状态来
然而直接这样内存会炸，所以用滚动数组，所以要在转移过程中将初始状态加进去，具体见代码
code
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
using namespace std;
#define Miuna printf ("dai suki...")
#define mizuki signed
#define love (1209 & 322 & 901)
namespace LRH
{
template <typename Shiodome_miuna> inline void in (Shiodome_miuna &x)
{
x = 0;
char f = 0; char ch;
while (!isdigit (ch = getchar ())) if (ch == '-') f = 1;
do
{
x = (x << 1) + (x << 3) + (ch ^ 48);
} while (isdigit (ch = getchar ()));
if (f) x = -x;
}
int stk[20], tp;
template <typename Shiodome_miuna> inline void ot (Shiodome_miuna x, int f = 2)
{
if (x < 0) putchar ('-'), x = -x;
do
{
stk[++ tp] = x % 10;
x /= 10;
} while (x);
while (tp)
{
putchar (stk[tp --] | 48);
}
if (f == 1) putchar ('\n');
if (!f) putchar (' ');
}
}using namespace LRH;
typedef long long ll;
#define int ll
const int maxn = 1e3 + 10;
const int inf = 4e18;
int n, k;
ll ans1, ans2, p;
struct miuna
{
ll c, w;
}a[65];
ll f[2][1010][1010], g[2][1010][1010];
int v[65];
mizuki main ()
{
freopen ("goose.in", "r", stdin);
freopen ("goose.out", "w", stdout);
in (n), in (k), in (p);
for (int i = 1; i <= n; ++ i) in (a[i].c), in (a[i].w);
if (p % a[1].c) return (puts ("-1"), love);
memset (f, 127, sizeof (f)), memset (g, 128, sizeof (g));
ll tmp = p;
for (int i = n; i >= 1; -- i) v[i] = tmp / a[i].c, tmp %= a[i].c;
f[0][v[n]][v[n]] = v[n] * a[n].w, g[0][v[n]][v[n]] = v[n] * a[n].w;
int op = 0;
for (int i = n; i >= 2; -- i)
{
op ^= 1;
int t = a[i].c / a[i - 1].c;
for (int J = 0; J <= k; ++ J)
{
for (int K = 0; K <= J; ++ K)
{
if (f[op ^ 1][J][K] >= inf && g[op ^ 1][J][K] <= -inf) continue;
for (int s = 0; s <= K; ++ s)
{
int o = s * t + v[i - 1];
if (J + o - s > k) break;
f[op][J + o - s][o] = min (f[op][J + o - s][o], f[op ^ 1][J][K] - s * a[i].w + o * a[i - 1].w);
g[op][J + o - s][o] = max (g[op][J + o - s][o], g[op ^ 1][J][K] - s * a[i].w + o * a[i - 1].w);
}
f[op ^ 1][J][K] = inf, g[op ^ 1][J][K] = -inf;
}
}
}
ll ans1 = inf, ans2 = -inf;
for (int i = v[1]; i <= k; ++ i) ans1 = min (ans1, f[op][k][i]), ans2 = max (ans2, g[op][k][i]);
if (ans1 == inf) return puts ("-1"), love;
ot (ans1, 0), ot (ans2, 1);
return love;
}
楼盘
离谱\(DP\)
我这么\(J\)显然是不会的
矩形
对于每个\(i\)，可以先单调栈预处理出来以它为最小值的区间
设所有矩形的面积排好序后组成的集合为\(A\)
二分求\(A_L\)，要求有多少个矩形面积小于等于\(mid\)，下面直接贺过来题解
\(f(s)\)为面积小于等于\(s\)的矩形个数
接下来对于每个\(h_i\)求出\(len\times h_i\)大于\(A_L\)最小的\(len\)
然后求出这个\(len\)可以用几次，塞进一个按\(len\times h_i\)排序的小根堆里，每次输出之后将下一个不小于它的数塞进去
code
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<set>
using namespace std;
#define Miuna printf ("dai suki...")
#define mizuki signed
#define love (1209 & 322 & 901)
namespace LRH
{
template <typename Shiodome_miuna> inline void in (Shiodome_miuna &x)
{
x = 0;
char f = 0; char ch;
while (!isdigit (ch = getchar ())) if (ch == '-') f = 1;
do
{
x = (x << 1) + (x << 3) + (ch ^ 48);
} while (isdigit (ch = getchar ()));
if (f) x = -x;
}
int stk[20], tp;
template <typename Shiodome_miuna> inline void ot (Shiodome_miuna x, int f = 2)
{
if (x < 0) putchar ('-'), x = -x;
do
{
stk[++ tp] = x % 10;
x /= 10;
} while (x);
while (tp)
{
putchar (stk[tp --] | 48);
}
if (f == 1) putchar ('\n');
if (!f) putchar (' ');
}
}using namespace LRH;
typedef long long ll;
const int maxn = 3e5 + 10;
#define int ll
int n, L, R;
int h[maxn], li[maxn], ri[maxn], a[maxn], b[maxn];
ll S(int x){return x <= 0 ? 0 : x * (x + 1) / 2;}
ll f (int w, int l, int r) {return S (w) - S (w - l - 1) - S (w - r - 1) + S (w - l - r - 2);}
ll ck (ll x)
{
ll res = 0;
for (int i = 1; i <= n; ++ i)
{
int k = x / h[i];
res += f (min (k, ri[i] - li[i] + 1), a[i], b[i]);
}
return res;
}
struct miuna
{
int id, s, c, w;
friend bool operator < (miuna a, miuna b)
{
return a.s > b.s;
}
};
ll cnt (int i, ll len)
{
if (li[i] + len - 1 <= i) return min (ri[i] - i + 1, len);
return min (ri[i] - (li[i] + len - 1) + 1, i - li[i] + 1);
}
priority_queue<miuna>q;
int st[maxn], top;
mizuki main ()
{
freopen ("rectangle.in", "r", stdin);
freopen ("rectangle.out", "w", stdout);
in (n);
for (int i = 1; i <= n; ++ i) in (h[i]);
for (int i = 1; i <= n; ++ i)
{
while (top && h[st[top]] > h[i]) ri[st[top]] = i - 1, top --;
st[++ top] = i;
}
while (top) ri[st[top --]] = n;
for (int i = n; i >= 1; -- i)
{
while (top && h[st[top]] >= h[i]) li[st[top]] = i + 1, top --;
st[++ top] = i;
}
while (top) li[st[top --]] = 1;
for (int i = 1; i <= n; ++ i)
{
a[i] = i - li[i];
b[i] = ri[i] - i;
}
in (L), in (R);
ll l = 0, r = 4e18, ansl;
while (l <= r)
{
ll mid = (l + r) >> 1;
if (ck (mid) < L) l = mid + 1, ansl = mid;
else r = mid - 1;
}
ansl ++;
ll up = min (R, ck (ansl));
for (ll i = L; i <= up; ++ i) ot (ansl, 0);
ansl ++;
L = up + 1;
for (int i = 1; i <= n; ++ i)
{
int w = ansl / h[i];
if (w > (ri[i] - li[i] + 1)) continue;
q.push ({i, w * h[i], cnt (i, w), w});
}
while (L <= R && !q.empty ())
{
miuna tmp = q.top ();
q.pop ();
tmp.c --;
if (tmp.c > 0) q.push (tmp);
else if (tmp.w + 1 <= (ri[tmp.id] - li[tmp.id] + 1)) q.push ({tmp.id, (tmp.w + 1) * h[tmp.id], cnt (tmp.id, tmp.w + 1), tmp.w + 1});
if (tmp.s < ansl) continue;
ot (tmp.s, 0);
L ++;
}
return love;
}
