穷酸博主没有bz权限号， 也不会去$poi$官网， 在某咕刷的$poi$，按照某咕的难度排序刷， poi~
$Luogu3572 PTA-Little Bird$
单调队列， 队列内按照 步数为第一关键字递增， 高度为第二关键字递减
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define R register 6 #define N 1000005 7 using namespace std; 8 9 int n, k, Q; 10 int d[N], f[N], q[N]; 11 12 inline int read() { 13 int X = 0, p = 1; char c = getchar(); 14 for (; c > '9' || c < '0';c = getchar()) 15 if (c == '-') p = -1; 16 for (; c >= '0' && c <= '9'; c = getchar()) 17 X = X * 10 + c - '0'; 18 return X * p; 19 } 20 21 int main() 22 { 23 n = rd; 24 for (R int i = 1; i <= n; ++i) d[i] = rd; 25 Q = rd; 26 for (; Q; Q--) { 27 k = rd; 28 int l = 1, r = 1; 29 q[l] = 1; 30 for (R int i = 2; i <= n; ++i) { 31 while (l <= r && q[l] < i - k) l++; 32 f[i] = f[q[l]] + (d[q[l]] <= d[i] ? 1 : 0); 33 while (l <= r && (f[q[r]] > f[i] || (f[q[r]] == f[i] && d[q[r]] <= d[i]))) r--; 34 q[++r] = i; 35 } 36 printf("%d\n", f[n]); 37 } 38 }
$Luogu3566KLO-Bricks$
贪心， 优先用右端点$ed$的颜色的来填， 如果$ed$用光了或者上一个颜色是$ed$， 则选取剩下来个数最多的去填
个数最多可以用线段树维护
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define N 1000005 6 using namespace std; 7 typedef pair<int, int > P; 8 9 int n, m, cnt[N], st, ed, a[N]; 10 11 int read() { 12 int X = 0, p = 1; char c = getchar(); 13 for (; c > '9' || c < '0'; c = getchar()) 14 if (c == '-') p = -1; 15 for (; c >= '0' && c <= '9'; c = getchar()) 16 X = X * 10 + c - '0'; 17 return X * p; 18 } 19 20 void up(int &A, int B) { 21 if (A < B) A = B; 22 } 23 24 P cmax(P A, P B) { 25 return A > B ? A : B; 26 } 27 28 namespace SegT { 29 #define lson nd << 1 30 #define rson nd << 1 | 1 31 #define mid ((l + r) >> 1) 32 P Max[N << 2]; 33 34 void pushup(int nd) { 35 Max[nd] = cmax(Max[lson], Max[rson]); 36 } 37 38 void build(int l, int r, int nd) { 39 if (l == r) { 40 Max[nd] = P(cnt[l], l); return; 41 } 42 build(l, mid, lson); build(mid + 1, r, rson); 43 pushup(nd); 44 } 45 46 void modify(int c, int l, int r, int nd) { 47 if (l == r) { 48 Max[nd].first-- ; return; 49 } 50 if (c <= mid) modify(c, l, mid, lson); 51 else modify(c, mid + 1, r, rson); 52 pushup(nd); 53 } 54 55 P query(int L, int R, int l, int r, int nd) { 56 if (L > R) return P(0, 0); 57 if (L <= l && r <= R) return Max[nd]; 58 P tmp = P(0, 0); 59 if (mid >= L) 60 tmp = cmax(tmp, query(L, R, l, mid, lson)); 61 if (mid < R) 62 tmp = cmax(tmp, query(L, R, mid + 1, r, rson)); 63 return tmp; 64 } 65 }using namespace SegT; 66 67 int main() 68 { 69 m = rd; st = rd; ed = rd; 70 for (int i = 1; i <= m; ++i) 71 n += (cnt[i] = rd); 72 cnt[st] --; cnt[ed] --; 73 a[1] = st; a[n] = ed; 74 build(1, m, 1); 75 for (int i = 2; i < n; ++i) { 76 if (cnt[ed] && a[i - 1] != ed) { 77 a[i] = ed; 78 cnt[a[i]]--; 79 modify(a[i], 1, m, 1); 80 } else { 81 P tmp = query(1, a[i - 1] - 1, 1, m, 1); 82 tmp = cmax(tmp, query(a[i - 1] + 1, m, 1, m, 1)); 83 if (tmp.first == 0) return printf("0"), 0; 84 a[i] = tmp.second; 85 cnt[a[i]]--; 86 modify(a[i], 1, m, 1); 87 } 88 } 89 if (a[n - 1] == a[n]) 90 return printf("0"), 0; 91 printf("%d", a[1]); 92 for (int i = 2; i <= n; ++i) 93 printf(" %d", a[i]); 94 return 0; 95 }
$Luogu3575DOO-Around the world$
双指针扫一遍， 找到从 $i$ 往前最多能飞多少格。 并记录步数和起始位置。 当起始位置和当前位置相距不少于N， 则更新答案。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define N 2000005 6 #define R register 7 using namespace std; 8 9 int n, m, f[N], head[N], maxn, a[N]; 10 11 int read() { 12 int X = 0, p = 1; char c = getchar(); 13 for (; c > '9' || c < '0'; c = getchar()) 14 if (c == '-') p = -1; 15 for (; c >= '0' && c <= '9'; c = getchar()) 16 X = X * 10 + c - '0'; 17 return X * p; 18 } 19 20 void cmax(int &A, int B) { 21 if (A < B) A = B; 22 } 23 24 void cmin(int &A, int B) { 25 if (A > B) A = B; 26 } 27 28 void work(int d) { 29 if (d < maxn) { 30 puts("NIE"); return; 31 } 32 int ans = n; 33 for (R int i = n + 1, j = 1; i <= 2 * n; ++i) { 34 while (a[i] - a[j] > d) j++; 35 f[i] = f[j] + 1, head[i] = head[j]; 36 if (i - head[i] >= n) cmin(ans, f[i]); 37 } 38 printf("%d\n", ans); 39 } 40 41 int main() 42 { 43 n = rd; m = rd; 44 for (R int i = 1; i <= n; ++i) 45 a[i] = a[i + n] = rd, 46 cmax(maxn, a[i]), 47 head[i] = i; 48 for (R int i = 1; i <= 2 * n; ++i) 49 a[i] += a[i - 1]; 50 for (R int i = 1; i <= m; ++i) work(rd); 51 }
$Luogu3565Hotels$
枚举根节点为中心， 在根的不同儿子的子树中找$3$个相同深度的点的方案数。 时间复杂度$O(N^2)$，
不过好像用长链剖分可以做到$O(N)$。。。老年选手也懒得去学了唔
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define N 5005 6 #define ll long long 7 #define R register 8 using namespace std; 9 10 int head[N], tot, dep[N], g[N], n; 11 ll ans, f[N][3]; 12 13 struct edge { 14 int nxt, to; 15 }e[N << 1]; 16 17 inline int read() { 18 int X = 0, p = 1; char c = getchar(); 19 for (; c > '9' || c < '0'; c = getchar()) 20 if (c == '-') p = -1; 21 for (; c >= '0' && c <= '9'; c = getchar()) 22 X = X * 10 + c - '0'; 23 return X * p; 24 } 25 26 void add(int u, int v) { 27 e[++tot].to = v; 28 e[tot].nxt = head[u]; 29 head[u] = tot; 30 } 31 32 void cmax(int &A, int B) { 33 if (A < B) A = B; 34 } 35 36 void dfs(int u, int fa) { 37 g[dep[u]] ++; 38 for (R int i = head[u]; i; i = e[i].nxt) { 39 int nt = e[i].to; 40 if (nt == fa) continue; 41 dep[nt] = dep[u] + 1; 42 dfs(nt, u); 43 } 44 } 45 46 int main() 47 { 48 n = rd; 49 for (int i = 1; i < n; ++i) { 50 int u = rd, v = rd; 51 add(u, v); add(v, u); 52 } 53 for (R int rt = 1; rt <= n; ++rt) { 54 dep[rt] = 1; 55 memset(f, 0, sizeof(f)); 56 for (R int i = head[rt]; i; i = e[i].nxt) { 57 memset(g, 0, sizeof(g)); 58 dep[e[i].to] = 2; 59 dfs(e[i].to, rt); 60 for (R int j = 2; j; --j) 61 for (R int k = 1; k <= n; ++k) 62 f[k][j] += f[k][j - 1] * g[k]; 63 for (R int k = 1; k <= n; ++k) 64 f[k][0] += g[k]; 65 } 66 for (R int i = 1; i <= n; ++i) 67 ans += f[i][2]; 68 } 69 printf("%lld\n", ans); 70 }
$Luogu3580Freight$
列车肯定是 分批， 一批过去再回来， 才轮到下一批。
然后就开始$DP$, 设$F[i]$ 为前 $i$ 辆列车回来的最早时间， 同时为了防止不同列车同时出发， 令$t[i] = max(t[i], t[i-1] + 1)$
有状态转移方程：
$f[i] \ = \ min( \ max(f[j] + i -j-1, t[i])+2 \times S + i - j - 1)$
移项得：
$f[i] \ = \ 2 \times S + 2 \times i -1 \ + \ min( \ max(f[j] - j - 1, t[i] - i)-j)$
由于 $f[j] - j-1$ 和 $t[i] - i$ 是不下降的， 所以用单调队列找到第一个分界点， 并且队列 除第一位 按照 $f[j]-2 \times j - 1$ 递增
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define N 1000005 6 #define R register 7 #define ll long long 8 using namespace std; 9 10 int n, t[N], q[N], S; 11 ll f[N]; 12 13 inline int read() { 14 int X = 0, p = 1; char c = getchar(); 15 for (; c > '9' || c < '0'; c = getchar()) 16 if (c == '-') p = -1; 17 for (; c >= '0' && c <= '9'; c = getchar()) 18 X = X * 10 + c - '0'; 19 return X * p; 20 } 21 22 inline int cmax(int A, int B) { 23 return A > B ? A : B; 24 } 25 26 template<typename T> 27 inline void down(T &A, T B) { 28 if (A > B) A = B; 29 } 30 31 int main() 32 { 33 n = rd; S = rd; 34 for (R int i = 1; i <= n; ++i) 35 t[i] = cmax(rd, t[i - 1] + 1); 36 int l = 1, r = 1; 37 for (R int i = 1; i <= n; ++i) { 38 while (l < r && f[q[l + 1]] - q[l + 1] - 1<= t[i] - i) l++; 39 f[i] = 1LL * t[i] + i - q[l] + 2 * S - 1; 40 if (l < r) down(f[i], f[q[l + 1]] - 2 * q[l + 1] - 1 + 2 * S + 2 * i - 1); 41 while (l < r && f[q[r]] - 2 * q[r] >= f[i] - 2 * i) r--; 42 q[++r] = i; 43 } 44 printf("%lld\n", f[n]); 45 }
$Luogu3574FarmCraft$
贪心排序+树形DP
我们可以递归求出每颗子树所需要的最大时间，
子节点到父节点的转移方程：
$f[u] \ = \ max(2 \times \sum\limits_{k=1}^{i-1}sz[k] + f[i]+1)$ ， 父节点 $f[u]$ 初始为 $a[u]$
根据贪心， 把子树按照$2 \times sz[i]-f[i]$进行排序（证明挺简单的， 邻项交换 就可以证出）
由于根节点是要最后安装电脑的， $f[1] = max(f[1], a[1] + 2 \times (sz[1] - 1) )$
最后输出$f[1]$， 复杂度$O(NlogN)$
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 #define N 500005 6 #define rd read() 7 using namespace std; 8 9 int head[N], tot, n, f[N], sz[N], a[N]; 10 vector<int> v[N]; 11 12 struct edge { 13 int nxt, to; 14 }e[N << 1]; 15 16 inline int read() { 17 int X = 0, p = 1; char c = getchar(); 18 for (; c > '9' || c < '0'; c = getchar()) 19 if (c == '-') p = -1; 20 for (; c >= '0' && c <= '9'; c = getchar()) 21 X = X * 10 + c - '0'; 22 return X * p; 23 } 24 25 void add(int fr, int to) { 26 e[++tot].to = to; 27 e[tot].nxt = head[fr]; 28 head[fr] = tot; 29 } 30 31 int cmp(int A, int B) { 32 return 2 * sz[A] - f[A] < 2 * sz[B] - f[B]; 33 } 34 35 void dfs(int u, int fa) { 36 for (int i = head[u]; i; i = e[i].nxt) { 37 int nt = e[i].to; 38 if (nt == fa) continue; 39 dfs(nt, u); 40 v[u].push_back(nt); 41 } 42 } 43 44 void cmax(int &A, int B) { 45 if (A < B) A = B; 46 } 47 48 void dp(int u) { 49 sz[u] = 1; 50 f[u] = a[u]; 51 for (int i = 0, up = v[u].size(); i < up; ++i) { 52 int nt = v[u][i]; 53 dp(nt); 54 } 55 sort(v[u].begin(), v[u].end(), cmp); 56 for (int i = 0, up = v[u].size(); i < up; ++i) { 57 int nt = v[u][i]; 58 cmax(f[u], 2 * sz[u] + f[nt] - 1); 59 sz[u] += sz[nt]; 60 } 61 } 62 63 int main() 64 { 65 n = rd; 66 for (int i = 1; i <= n; ++i) a[i] = rd; 67 for (int i = 1; i < n; ++i) { 68 int fr = rd, to = rd; 69 add(fr, to); add(to, fr); 70 } 71 dfs(1, 0); dp(1); 72 cmax(f[1], 2 * (n - 1) + a[1]); 73 printf("%d\n", f[1]); 74 }
$Luogu3570Criminals$
挺早前就做了的题。。。看不惯当时的毒瘤代码， 空格有时有， 有时没有。
就是一个链表乱搞， 就不贴代码丢脸了（捂脸）
$Luogu3567Couriers$
主席树模板题
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 using namespace std; 6 7 const int N = 1e5; 8 9 int lson[N * 100], rson[N * 100], sum[N * 100]; 10 int a[N * 10], n, m, root[N * 10], b[N * 10], tot, cnt; 11 12 int read() { 13 int X = 0, p = 1; char c = getchar(); 14 for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1; 15 for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0'; 16 return X * p; 17 } 18 19 int fd(int x) { 20 return lower_bound(b + 1, b + 1 + tot, x) - b; 21 } 22 23 void build0(int &o, int l, int r) { 24 o = ++cnt; 25 if(l == r) return; 26 int mid = (l + r) >> 1; 27 build0(lson[o], l, mid); 28 build0(rson[o], mid + 1, r); 29 } 30 31 void build(int last, int &o, int pos, int l, int r) { 32 o = ++cnt; 33 sum[o] = sum[last] + 1; 34 lson[o] = lson[last]; 35 rson[o] = rson[last]; 36 if(l == r) 37 return; 38 int mid = (l + r) >> 1; 39 if(pos <= mid) 40 build(lson[last], lson[o], pos, l, mid); 41 else 42 build(rson[last], rson[o], pos, mid + 1, r); 43 } 44 45 int query(int last, int now, int k, int l, int r) { 46 if(l == r) 47 return (2 * (sum[now] - sum[last]) > k) ? l : 0; 48 int mid = (l + r) >> 1; 49 if(2 * (sum[lson[now]] - sum[lson[last]]) > k) 50 return query(lson[last], lson[now], k, l, mid); 51 else 52 return query(rson[last], rson[now], k, mid + 1, r); 53 } 54 55 int main() 56 { 57 n = rd; m = rd; 58 for(int i = 1; i <= n; ++i) 59 b[i] = a[i] = rd; 60 sort(b + 1, b + 1 + n); 61 tot = unique(b + 1, b + 1 + n) - b - 1; 62 for(int i = 1; i <= n; ++i) 63 build(root[i - 1], root[i], fd(a[i]), 1, tot); 64 for(int i = 1; i <= m; ++i) { 65 int l = rd, r = rd, ans; 66 ans = query(root[l - 1], root[r], r - l + 1, 1, tot); 67 printf("%d\n", b[ans]); 68 } 69 }
$Luogu3572Ant colony$
倍增LCA+二分查找
$f[i][j]$ 表示 $i$ 的 $2^j$ 级祖先， $g[i][j]$ 表示从i 到 $f[i][j]$ 要分支成几份。 另外$g[i][j]$ 可能非常大， 最高约是 $3$的几万次方， 所以如果$g[i][j] \times k > a[m]$，直接赋为$-1$
然后就可以倍增求要几个分支，设分支数为 $x$， 则最后被吃掉的蚂蚁数就是 数组a中 $k \times x<= <=(k + 1) \times -1$的个数， 二分查找即可。
时间复杂度为$O(NlogN)$， (做POI习惯性卡常数了
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define N 1000005 6 #define ll long long 7 #define R register 8 using namespace std; 9 10 const int base = 20; 11 12 int n, m, k, tot, st, ed; 13 int head[N], a[N], f[N][base], d[N], leaf[N], cnt, dep[N]; 14 ll g[N][base], ans; 15 bool in[N]; 16 17 struct edge { 18 int nxt, to; 19 }e[N << 1]; 20 21 inline int read() { 22 int X = 0, p = 1; char c = getchar(); 23 for (; c > '9' || c < '0'; c = getchar()) 24 if (c == '-') p = -1; 25 for (; c >= '0' && c <= '9'; c = getchar()) 26 X = X * 10 + c - '0'; 27 return X * p; 28 } 29 30 inline void add(int u, int v) { 31 e[++tot].to = v; 32 e[tot].nxt = head[u]; 33 head[u] = tot; 34 } 35 36 void dfs(int u, bool flag) { 37 in[u] = flag; 38 for (R int i = head[u]; i; i = e[i].nxt) { 39 R int nt = e[i].to; 40 if (nt == f[u][0]) continue; 41 f[nt][0] = u; 42 g[nt][0] = (d[nt] - 1) ? d[nt] - 1 : 1; 43 dep[nt] = dep[u] + 1; 44 if ((u == e[1].to && nt == e[2].to) || (nt == e[1].to && u == e[2].to)) 45 dfs(nt, true); 46 else dfs(nt, flag); 47 } 48 } 49 50 inline int cal(int x) { 51 int maxn = upper_bound(a + 1, a + 1 + m, 1LL * (k + 1) * x - 1) - a, 52 minn = lower_bound(a + 1, a + 1 + m, k * x) - a; 53 return maxn - minn; 54 } 55 56 int LCA(R int x, R int y) { 57 ll res = 1; 58 if (dep[x] < dep[y]) swap(x, y); 59 for (R int j = base - 1; ~j; --j) 60 if (dep[f[x][j]] >= dep[y]) { 61 res *= g[x][j]; 62 if (res * k > a[m] || res < 0) return -1; 63 x = f[x][j]; 64 } 65 if (x == y) { 66 res *= g[x][0]; 67 if (res * k > a[m] || res < 0) return -1; 68 else return res; 69 } 70 for (R int j = base - 1; ~j; --j) 71 if (f[x][j] != f[y][j]) { 72 res *= g[x][j]; 73 res *= g[y][j]; 74 if (res * k > a[m] || res < 0) return -1; 75 x = f[x][j]; 76 y = f[y][j]; 77 } 78 res *= g[x][1] * g[y][0]; 79 if (res * k > a[m] || res < 0) return -1; 80 return res; 81 } 82 83 void work(int x) { 84 R int tmp = LCA(x, in[x] ? ed : st); 85 if (tmp != -1) ans += 1LL * cal(tmp) * k; 86 } 87 88 int main() 89 { 90 n = rd; m = rd; k = rd; 91 for (R int i = 1; i <= m; ++i) 92 a[i] = rd; 93 for (R int i = 1; i < n; ++i) { 94 int u = rd, v = rd; 95 add(u, v); add(v, u); 96 d[u] ++; d[v] ++; 97 } 98 dep[1] = 1; 99 dfs(1, false); 100 g[1][0] = (d[1] - 1) ? d[1] - 1 : 1; 101 sort(a + 1, a + 1 + m); 102 if (dep[e[1].to] > dep[e[2].to]) st = e[2].to, ed = e[1].to; 103 else st = e[1].to, ed = e[2].to; 104 for (R int i = 1; i <= n; ++i) 105 if (d[i] == 1) leaf[++cnt] = i; 106 for (R int j = 1; j < base; ++j) 107 for (R int i = 1; i <= n; ++i) { 108 f[i][j] = f[f[i][j - 1]][j - 1], 109 g[i][j] = g[i][j - 1] * g[f[i][j - 1]][j - 1]; 110 if (g[i][j] * k > a[m]) g[i][j] = -1; 111 else if (g[i][j] < 0) g[i][j] = -1; 112 } 113 for (R int i = 1; i <= cnt; ++i) 114 work(leaf[i]); 115 printf("%lld\n", ans); 116 }
$Luogu3564Salad Bar$
树状数组
若$s[i]=='j'$， 则赋为 $-1$， 若为$'p'$， 则赋为 $1$。 定义 $sum$为前缀和。
则一个字符串 $[j,i]$ 从前往后 的$1$的个数 不少于 $-1$的个数， 则满足 $sum[j-1] <= sum[k]$ $(j<=k<=i)$ 恒成立。
根据这一点， 若要知道最多往后到哪里， 只需往后找到第一个 $i$，$sum[i]>sum[j]$。 单调栈$O(N)$即可维护。 往前同理可得。
我们设从 $i$往后最多到 $R[i]$， 往前最多到 $L[i]$。 一个字符串 $[j,i]$ 满足条件 必须使 $R[j]>=i && L[i]<=j$
我们只需要先把 $i$ 按照 $R[i]$从小往大排序，用树状数组维护即可。
时间复杂度$O(NlogN)$
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rg register 5 #define rd read() 6 #define N 1000005 7 using namespace std; 8 9 int a[N], b[N], n, pre[N], nxt[N]; 10 int st[N], tp, Max[N], ans; 11 char s[N]; 12 13 struct node { 14 int id, L, R; 15 bool operator < (const node &A) const { 16 return R < A.R; 17 } 18 }t[N]; 19 20 int read() { 21 int X = 0, p = 1; char c = getchar(); 22 for (; c > '9' || c < '0'; c = getchar()) 23 if (c == '-') p = -1; 24 for (; c >= '0' && c <= '9'; c = getchar()) 25 X = X * 10 + c - '0'; 26 return X * p; 27 } 28 29 void init() { 30 for (rg int i = 1; i <= n + 1; ++i) { 31 int pr = 0; 32 while (tp) { 33 if (b[st[tp]] >= b[i]) tp--; 34 else { 35 pr = st[tp]; break; 36 } 37 } 38 pre[i] = pr; 39 st[++tp] = i; 40 } 41 tp = 0; 42 for (rg int i = n; ~i; --i) { 43 int nt = n + 1; 44 while (tp) { 45 if (a[st[tp]] >= a[i]) tp--; 46 else { 47 nt = st[tp]; break; 48 } 49 } 50 nxt[i] = nt; 51 st[++tp] = i; 52 } 53 } 54 55 inline int cmin(int A, int B) { 56 return A > B ? B : A; 57 } 58 59 inline int cmax(int A, int B) { 60 return A > B ? A : B; 61 } 62 63 inline void up(int &A, int B) { 64 if (A < B) A = B; 65 } 66 67 int lowbit(int x) { 68 return x & -x; 69 } 70 71 void add(int x, int val) { 72 for (; x <= n; x += lowbit(x)) 73 up(Max[x], val); 74 } 75 76 int query(int x) { 77 int res = 0; 78 for (; x; x -= lowbit(x)) 79 up(res, Max[x]); 80 return res; 81 } 82 83 int main() 84 { 85 n = rd; 86 scanf("%s", s + 1); 87 for (rg int i = 1; i <= n; ++i) 88 a[i] = a[i - 1] + (s[i] == 'p' ? 1 : -1); 89 for (rg int i = n; i; --i) 90 b[i] = b[i + 1] + (s[i] == 'p' ? 1 : -1); 91 init(); 92 for (rg int i = 1; i <= n; ++i) 93 pre[i] = pre[i + 1] + 1; 94 for (rg int i = n; i; --i) 95 nxt[i] = nxt[i - 1] - 1; 96 for (rg int i = 1; i <= n; ++i) { 97 t[i].id = i; 98 t[i].L = pre[i]; 99 t[i].R = nxt[i]; 100 } 101 sort(t + 1, t + 1 + n); 102 for (rg int i = 1, j = 1; i <= n; ++i) { 103 while (j <= n && j <= t[i].R) add(pre[j], j), j++; 104 int res = query(t[i].id); 105 up(ans, res - t[i].id + 1); 106 } 107 printf("%d\n", ans); 108 }
$Luogu3579Solar Panels$
整除分块枚举。。。
真的没有想到会这么简单。
$\left\lfloor\dfrac{1}{2}\right\rfloor$
要使一个数 $p$ 满足 条件， 则 存在$x, y$， $a<=x \times p<=b\ \&\&\ c<=y \times p <=d$
把$p$ 除掉 则
$$\left\lceil\dfrac{a}{p}\right\rceil <=x <=\left\lfloor\dfrac{b}{p}\right\rfloor$$
$$\left\lceil\dfrac{c}{p}\right\rceil <=y <=\left\lfloor\dfrac{d}{p}\right\rfloor$$
把向上取整变为向下取整：
$\left\lfloor\dfrac{a+p-1}{p}\right\rfloor <= \left\lfloor\dfrac{b}{p}\right\rfloor$
$\left\lfloor\dfrac{b+p-1}{p}\right\rfloor <= \left\lfloor\dfrac{d}{p}\right\rfloor$
然后就变成了 ：
$\left\lfloor\dfrac{a-1}{p}\right\rfloor < \left\lfloor\dfrac{b}{p}\right\rfloor$
$\left\lfloor\dfrac{b-1}{p}\right\rfloor < \left\lfloor\dfrac{d}{p}\right\rfloor$
最后整除分块。 只需按照 $b/p$和$d/p$ 相同时进行分类。 这样能使 $b/p$ 和 $d/p$相等的同时 $c/p$ 和 $d/p$尽量小， 更可能满足条件
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define rd read() 5 #define R register 6 using namespace std; 7 8 inline int read() { 9 int X = 0, p = 1; char c = getchar(); 10 for (; c > '9' || c < '0'; c = getchar()) 11 if (c == '-') p = -1; 12 for (; c >= '0' && c <= '9'; c = getchar()) 13 X = X * 10 + c - '0'; 14 return X * p; 15 } 16 17 inline void cmax(int &A, int B) { 18 if (A < B) A = B; 19 } 20 21 inline int cmin(int A, int B) { 22 return A > B ? B : A; 23 } 24 25 void work() { 26 int ans = 1; 27 int a = rd - 1, b = rd, c = rd - 1, d = rd; 28 for (R int i = 1, j = 1, up = cmin(b, d); i <= up; i = j + 1) { 29 j = cmin(b / (b / i), d / (d / i)); 30 if (b / j > a / j && d / j > c / j) cmax(ans, j); 31 } 32 printf("%d\n", ans); 33 } 34 35 int main() 36 { 37 int n = rd; 38 for (; n; --n) work(); 39 }