Section 4.1 4.2

省选准备/assets/day4/CF359D.cpp

@@ -0,0 +1,49 @@
+#include <cstdio>
+#include <cctype>
+inline int min(int a, int b) { return a < b ? a : b; }
+inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); }
+inline void read(int &x)
+{
+    int ch = x = 0;
+    while (!isdigit(ch = getchar()));
+    for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + ch - '0';
+}
+int n, a[1 << 20 | 1], Log[1 << 20 | 1];
+int M[1 << 20 | 1][20], G[1 << 20 | 1][20];
+struct
+{
+    int c, a[1 << 20 | 1], r;
+}ans;
+inline bool can(int l, int r)
+{
+    int k = Log[r - l + 1];
+    return min(M[l][k], M[r - (1 << k) + 1][k]) == gcd(G[l][k], G[r - (1 << k) + 1][k]);
+}
+bool check(int len)
+{
+    int cnt = 0;
+    for (int i = 0; i + len < n; i++)
+        if (can(i, i + len))
+            ans.a[cnt++] = i;
+    if (cnt) ans.r = len, ans.c = cnt;
+    return cnt;
+}
+int main()
+{
+    read(n);
+    Log[0] = -1;
+    for (int i = 1; i <= n; i++) Log[i] = Log[i >> 1] + 1;
+    for (int i = 0; i < n; i++) read(a[i]);
+    for (int i = 0; i < n; i++) M[i][0] = G[i][0] = a[i];
+    for (int j = 1; j <= Log[n]; j++)
+        for (int i = 0; i + (1 << j) - 1 < n; i++)
+            M[i][j] = min(M[i][j - 1], M[i + (1 << (j - 1))][j - 1]),
+            G[i][j] = gcd(G[i][j - 1], G[i + (1 << (j - 1))][j - 1]);
+    int L = 0, R = n, m;
+    while (L < R)
+        check(m = (L + R) >> 1) ? L = m + 1 : R = m;
+    printf("%d %d\n", ans.c, ans.r);
+    for (int i = 0; i < ans.c; i++)
+        printf("%d ", ans.a[i] + 1);
+    return 0;
+}

省选准备/assets/day4/FenwickTree.cpp

@@ -0,0 +1,13 @@
+void add(int x, int v)
+{
+    for (; x <= n; x += lowbit(x))
+        A[x] += v;
+}
+
+void sum(int x)
+{
+    int res = 0;
+    for (; x; x -= lowbit(x))
+        res += A[x];
+    return res;
+}

省选准备/assets/day4/poj1990.cpp

@@ -0,0 +1,42 @@
+#include <algorithm>
+#include <cstdio>
+#define lowbit(x) ((x) & -(x))
+const int N = 20005;
+void add(int *arr, int pos, int val)
+{
+    for (; pos < N; pos += lowbit(pos))
+        arr[pos] += val;
+}
+int query(int *arr, int pos)
+{
+    int ans = 0;
+    for (; pos; pos -= lowbit(pos))
+        ans += arr[pos];
+    return ans;
+}
+int sum[2][N];
+struct cow
+{
+    int pos, vol;
+    bool operator<(const cow &rhs) const { return vol < rhs.vol; }
+} cows[N];
+int main()
+{
+    int n;
+    scanf("%d", &n);
+    for (int i = 1; i <= n; i++)
+        scanf("%d%d", &cows[i].vol, &cows[i].pos);
+    std::sort(cows + 1, cows + n + 1);
+    long long ans = 0;
+    for (int i = 1; i <= n; i++)
+    {
+        long long a = query(sum[0], cows[i].pos), b = query(sum[1], cows[i].pos);
+        ans += (cows[i].pos * a - b + query(sum[1], 20000) - b -
+                (i - 1 - a) * cows[i].pos) *
+               cows[i].vol;
+        add(sum[0], cows[i].pos, 1);
+        add(sum[1], cows[i].pos, cows[i].pos);
+    }
+    printf("%lld", ans);
+    return 0;
+}

省选准备/assets/day4/poj2299.cpp

@@ -0,0 +1,42 @@
+#include <algorithm>
+#include <cstdio>
+#include <cstring>
+#include <stdint.h>
+#define lowbit(x) ((x) & -(x))
+const int N = 500005;
+int sum[N];
+int query(int x)
+{
+    int ans = 0;
+    for (; x; x -= lowbit(x)) ans += sum[x];
+    return ans;
+}
+void update(int x, int y)
+{
+    for (; x <= N; x += lowbit(x)) sum[x] += y;
+}
+struct abcd
+{
+    int val, pos;
+    bool operator<(const abcd &rhs) const { return val < rhs.val; }
+} nodes[N];
+int map[N];
+int main()
+{
+    int n;
+    while (~scanf("%d", &n) && n)
+    {
+        memset(sum, 0, sizeof(sum));
+        for (int i = 1; i <= n; i++) scanf("%d", &nodes[i].val), nodes[i].pos = i;
+        std::sort(nodes + 1, nodes + n + 1);
+        for (int i = 1; i <= n; i++) map[nodes[i].pos] = i;
+        int64_t ans = 0;
+        for (int i = 1; i <= n; i++)
+        {
+            update(map[i], 1);
+            ans += i - query(map[i]);
+        }
+        printf("%lld\n", ans);
+    }
+    return 0;
+}

省选准备/assets/day4/poj2352.cpp

@@ -0,0 +1,30 @@
+#include <cstdio>
+#include <cstring>
+#define lowbit(x) ((x) & -(x))
+const int maxn = 1 << 15 | 1;
+int a[maxn], n, m, level[maxn];
+void update(int pos, int x)
+{
+    for (; pos < maxn; pos += lowbit(pos)) a[pos] += x;
+}
+int query(int pos)
+{
+    int ans = 0;
+    for (; pos; pos -= lowbit(pos)) ans += a[pos];
+    return ans;
+}
+int main()
+{
+    scanf("%d", &n);
+    memset(level, 0, sizeof(level));
+    memset(a, 0, sizeof(a));
+    for (int i = 0, x, y; i < n; ++i)
+    {
+        scanf("%d%d", &x, &y);
+        level[query(++x)]++;
+        update(x, 1);
+    }
+    for (int i = 0; i < n; ++i)
+        printf("%d\n", level[i]);
+    return 0;
+}

省选准备/assets/day4/poj3264.cpp

@@ -0,0 +1,29 @@
+#include <cstdio>
+inline int min(int a, int b) { return a < b ? a : b; }
+inline int max(int a, int b) { return a > b ? a : b; }
+const int N = 50010;
+const int LogN = 16;
+int minT[LogN][N], maxT[LogN][N], Log[N];
+int main()
+{
+    Log[0] = -1;
+    for (int i = 1; i < N; i++) Log[i] = Log[i >> 1] + 1;
+    int m, n;
+    scanf("%d%d", &n, &m);
+    for (int i = 1, x; i <= n; i++)
+    {
+        scanf("%d", &x);
+        minT[0][i] = maxT[0][i] = x;
+    }
+    for (int j = 1; j < LogN; j++)
+        for (int i = 1; i + (1 << j) - 1 <= n; i++)
+            minT[j][i] = min(minT[j - 1][i], minT[j - 1][i + (1 << (j - 1))]),
+            maxT[j][i] = max(maxT[j - 1][i], maxT[j - 1][i + (1 << (j - 1))]);
+    for (int i = 1, x, y, z; i <= m; i++)
+    {
+        scanf("%d%d", &x, &y);
+        z = Log[y - x + 1];
+        printf("%d\n", max(maxT[z][x], maxT[z][y - (1 << z) + 1]) - min(minT[z][x], minT[z][y - (1 << z) + 1]));
+    }
+    return 0;
+}

省选准备/省选基础算法.tex

@@ -284,15 +284,13 @@ Manacher 模板题
 \paragraph{中国剩余定理}
 	若$m_1,m_2,m_3,\ldots,m_r$是两两互素的正整数，则同余方程组
 	\begin{equation}
-	\left\{
-	\begin{aligned}
+	\left\{\begin{aligned}
 		& x\equiv a_1\pmod{m_1}\\
 		& x\equiv a_2\pmod{m_2}\\
 		& x\equiv a_3\pmod{m_3}\\
 		& \cdots\\
 		& x\equiv a_r\pmod{m_r}\\
-	\end{aligned}
-	\right.
+	\end{aligned}\right.
 	\end{equation}
 	有$\mod M=\prod\limits_1^r m$的唯一解，即为中国剩余定理。\\
 	若$n=pq$且$gcd(p,q)=1$，那么$x\mod p,x\mod q$的值确定后，$x\mod n$的值也会随之确定。
@@ -330,10 +328,40 @@ Manacher 模板题
 	$$gcd(a,n)=1\Rightarrow Ord_n(a)|\varphi(n)$$
 	$$\mbox{特别的，}p\in\mathbb{P},gcd(a,p)=1\Rightarrow Ord_p(a)|p-1\mbox{，显然}\varphi(p)=p-1$$
 \paragraph{原根} $n\in\mathbb{N_+},a\in\mathbb{Z},Ord_n(a)=\varphi(n)$，则称$a$为模$n$的一个原根，由阶的定义可知原根和$n$必然互质。
-\paragraph{求质数$p$的原根算法}
-	暴力从小到大枚举$g\in\mathbb{N_+}$，$\forall a\in\{x\;|\;x|p-1,x\in\mathbb{P}\},g^{\frac{p-1}{a}}\not\equiv1\pmod{p}$
+\paragraph{求质数$p$的原根算法} 暴力从小到大枚举$g\in\mathbb{N_+}$，$\forall a\in\{x\;|\;x|p-1,x\in\mathbb{P}\},g^{\frac{p-1}{a}}\not\equiv1\pmod{p}$
 \codeinput[Primitive Root]{assets/day3/proot.cpp}
 \paragraph{练习题}
 \subparagraph{\href{http://poj.org/problem?id=1284}{POJ1284} - Primitive Roots} 如果$n\in\mathbb{N_+}$有一个原根，那么$n$一共有$\varphi(\varphi(n))$个不同余的原根
 \codeinput[Primitive Roots]{assets/day3/poj1284.cpp}
+\section{day4 数据结构}
+\subsection{树状数组} 在$\lg{n}$的时间内更新或查询$\sum\limits_{i=1}^n a_i$
+\paragraph{lowbit} \verb|lowbit(x) = x & -x|
+\paragraph{两个操作} $\lg{n}$更新或查询
+\codeinput[Fenwick tree]{assets/day4/FenwickTree.cpp}
+\paragraph{练习题}
+\subparagraph{\href{http://poj.org/problem?id=2352}{POJ2352} - Stars} 。。。
+\codeinput[Stars]{assets/day4/poj2352.cpp}
+\subparagraph{\href{http://poj.org/problem?id=2299}{POJ2299} - Ultra-QuickSort} 求逆序对数量，不过配图是啥玩意？马桶橛子？
+\codeinput[Ultra-QuickSort]{assets/day4/poj2299.cpp}
+\begin{center}\includegraphics[height=10cm]{assets/day4/poj2299.jpg}\end{center}
+\subparagraph{\href{http://poj.org/problem?id=1990}{POJ1990} - MooFest} 杀光奶牛问题就会得到解决
+\codeinput[MooFest]{assets/day4/poj1990.cpp}
+\subsection{Sparse Table} 处理区间最值，即 RMQ（Range Minimum Query）问题。
+\paragraph{预处理} 计算一个数组$f$，使$f[i][j]=min[i,i+2^j)$，然后你就可以开始倍增了。
+	\begin{eqnarray}
+	f[i][j]=\left\{\begin{array}{lr}
+		a[i] & j=0 \\
+		min(f[i][j-1],f[i+2^{j-1}][j-1]) & j>0
+		\end{array}\right.
+	\end{eqnarray}
+\paragraph{询问}
+	考虑一个询问区间$[x,y)$的最小值的询问操作。\\
+	我们可以求出满足$2^i\le y-x < 2^{i+1}$的$i$，即$\lfloor\log_2{y-x}\rfloor$，这样我们可以用两个长度为$2^i$的小区间覆盖询问的大区间。\\
+	而长度为$2^i$的小区间的最小值在预处理时已经求出，于是区间$[x,y)$的最小值为$min\{f[x][i],f[y-2^i][i]\}$，由于是求最值，区间有交集也没关系。
+\paragraph{练习题}
+\subparagraph{\href{http://poj.org/problem?id=3264}{POJ3264} - Balanced Lineup}
+	给你一个长度为$n$的序列$a[N]$，询问$Q$次，每次输出$[L,R]$区间最大值与最小值的差是多少，果真这种简化了的题面真是清晰易懂。
+\codeinput[Balanced Lineup]{assets/day4/poj3264.cpp}
+\subparagraph{\href{http://codeforces.com/problemset/problem/359/D}{CF359D} - Pair of Numbers} 首先要知道gcd有“最值的性质”，然后二分暴力。
+\codeinput[Pair of Numbers]{assets/day4/CF359D.cpp}
 \end{document}