F. Digital Patterns：

题目描述：

 

思路解析：

要求在一个方块中，任意相邻的方块中他的透明度系数不能相同，这样的方块称为趣味性方块，问这样的方块有多少种。

那么我们可以相当，假设 a1 == a2, 那么每一列的第一行元素都等于第二行元素。  假设b1==b2，那么每一行的第一列元素都等于第二列元素。那么可以想到 a1 != a2, a3!=a2, b1 != b2, b2 != b3，那么就可以组成3*3的趣味性方块，并且其中2*2和1*1方块皆为趣味性方块。那么通过上述分析知道，只有出现相邻元素时，他们便无法组成趣味性方块。那么我们就可以将整个a数组和b数组分为多个区间块。

那么可以发现最后可以利用八个树状数组来维护a b c d这四个数组信息。

代码实现：

import java.util.*;import java.io.*;public class Main {public static void main(String[] args) throws IOException {int t = 1;while (t > 0) {solve();t--;}w.flush();w.close();}static TreeSet<Integer> set[] = new TreeSet[2];static Fenwick[][] fen = new Fenwick[2][4];static int N;static long ans;public static void solve() throws IOException{int n = f.nextInt(); int m = f.nextInt();int q = f.nextInt();ans = 0;for (int i = 0; i < Math.min(n, m); i++) {ans += (long) (n - i) * (m - i);}long[] a = new long[n+1];long[] b = new long[m+1];N = Math.max(n, m);for (int i = 0; i < n; i++) {a[i] = f.nextInt();}for (int i = 0; i < m; i++) {b[i] = f.nextInt();}for (int i = 0; i < 2; i++) {set[i] = new TreeSet<>();for (int j = 0; j < 4; j++) {fen[i][j] = new Fenwick(N+1);}}set[0].add(0); set[0].add(n); set[1].add(0); set[1].add(m);fen[0][0].add(n, 1);fen[0][1].add(n, n);fen[0][2].add(n, S1(n));fen[0][3].add(n, S2(n));fen[1][0].add(m, 1);fen[1][1].add(m, m);fen[1][2].add(m, S1(m));fen[1][3].add(m, S2(m)); // 初始化for (int i = n-1; i > 0; i--) {a[i] -= a[i-1];if (a[i] == 0){insert(0, i);}}for (int i = m-1; i > 0; i--) {b[i] -= b[i-1];if (b[i] == 0){insert(1, i);}}w.println(ans);for (int i = 0; i < q; i++) {int t = f.nextInt(); int l = f.nextInt() - 1; int r = f.nextInt(); int x = f.nextInt();if (x == 0) continue;if (t == 1){if (l > 0){if (a[l] == 0){earse(0, l);}a[l] += x;if (a[l] == 0){insert(0, l);}}if (r < n){if (a[r] == 0){earse(0, r);}a[r] -= x;if (a[r] == 0){insert(0, r);}}}else {if (l > 0){if (b[l] == 0){earse(1, l);}b[l] += x;if (b[l] == 0){insert(1, l);}}if (r < m){if (b[r] == 0){earse(1, r);}b[r] -= x;if (b[r] == 0){insert(1, r);}}}w.println(ans);}}public static void add(int t, int x, int coef){ans +=(long) coef * x * fen[t ^ 1][2].sum(x); // n <= m ans += (long) coef * fen[t ^ 1][3].sum(x);// n <= mans += (long) coef * S2(x) * fen[t ^ 1][0].rangesum(x, N+1); // n > mans +=(long) coef * S1(x) * fen[t ^ 1][1].rangesum(x, N+1);fen[t][0].add(x, coef);fen[t][1].add(x, (long) coef *x);fen[t][2].add(x, coef * S1(x));fen[t][3].add(x, coef * S2(x));}public static void insert(int t, int i){int l = set[t].floor(i);int r = set[t].ceiling(i);set[t].add(i);add(t, r - l, -1);add(t, i - l, 1);add(t, r - i, 1);}public static void earse(int t, int i){set[t].remove(i);int l = set[t].floor(i);int r = set[t].ceiling(i);add(t, r - l, 1);add(t, i - l, -1);add(t, r - i, -1);}public static class Fenwick{int n;long[] a;public Fenwick(int n){this.n = n;a = new long[n];}public void add(int x, long val){for(int i = x+1; i <= n; i += i & -i) a[i-1] += val;}public long sum(int x){long res = 0;for (int i = x; i >= 1; i-= i & -i) {res += a[i-1];}return res;}public long rangesum(int l, int r){return sum(r) - sum(l);}}public static long S1(int x) {return (long) x * (x + 1) / 2;}public static long S2(int x) {return (long) x * (x + 1) * (2L * x + 1) / 6 - (long) x * x * (x+1L) / 2;}static PrintWriter w = new PrintWriter(new OutputStreamWriter(System.out));static Input f = new Input(System.in);static class Input {public BufferedReader reader;public StringTokenizer tokenizer;public Input(InputStream stream) {reader = new BufferedReader(new InputStreamReader(stream), 32768);tokenizer = null;}public String next() throws IOException{while (tokenizer == null || !tokenizer.hasMoreTokens()) {tokenizer = new StringTokenizer(reader.readLine());}return tokenizer.nextToken();}public int nextInt() throws IOException {return Integer.parseInt(next());}public long nextLong() throws IOException {return Long.parseLong(next());}public Double nextDouble() throws IOException {return Double.parseDouble(next());}}
}