結果
問題 | No.121 傾向と対策:門松列(その2) |
ユーザー |
|
提出日時 | 2017-03-25 11:32:21 |
言語 | D (dmd 2.109.1) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,323 bytes |
コンパイル時間 | 1,086 ms |
コンパイル使用メモリ | 128,688 KB |
実行使用メモリ | 153,700 KB |
最終ジャッジ日時 | 2024-06-12 18:33:51 |
合計ジャッジ時間 | 12,870 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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ファイルパターン | 結果 |
---|---|
other | AC * 4 TLE * 1 -- * 4 |
ソースコード
import std.stdio, std.array, std.string, std.conv, std.algorithm;import std.typecons, std.range, std.random, std.math, std.container;import std.numeric, std.bigint, core.bitop, core.stdc.stdio;void main() {auto N = readln.chomp.to!int;auto A = readln.split.map!(to!int).array;auto B = A.dup.sort().uniq.array;int[int] compressor;foreach (i, e; enumerate(B)) {compressor[e] = i.to!int;}foreach (i; 0..N) {A[i] = compressor[A[i]];}auto M = B.length.to!int - 1;auto st_left = new SegmentTree(M+1);auto st_right = new SegmentTree(M+1);auto st_same = new SegmentTree(M+1);foreach (i; 1..N) {st_right.add(A[i], 1);}long ans = 0;foreach (i; 1..N-1) {st_right.add(A[i], -1);st_left.add(A[i-1], 1);auto same = st_same.sum(A[i-1], A[i-1]);st_same.add(A[i-1], -same);st_same.add(A[i-1], st_left.sum(A[i-1], A[i-1]) * st_right.sum(A[i-1], A[i-1]));if (A[i] > 0) {ans += st_left.sum(0, A[i]-1) * st_right.sum(0, A[i]-1) - st_same.sum(0, A[i]-1);}if (A[i] < M) {ans += st_left.sum(A[i]+1, M) * st_right.sum(A[i]+1, M) - st_same.sum(A[i]+1, M);}}ans.writeln;}class SegmentTree {long[] table;int size;this(int n) {assert(bsr(n) < 29);size = 1 << (bsr(n) + 2);table = new long[](size);fill(table, 0);}void add(int pos, long num) {return add(pos, num, 0, 0, size/2-1);}void add(int pos, long num, int i, int left, int right) {table[i] += num;if (left == right)return;auto mid = (left + right) / 2;if (pos <= mid)add(pos, num, i*2+1, left, mid);elseadd(pos, num, i*2+2, mid+1, right);}long sum(int pl, int pr) {return sum(pl, pr, 0, 0, size/2-1);}long sum(int pl, int pr, int i, int left, int right) {if (pl > right || pr < left)return 0;else if (pl <= left && right <= pr)return table[i];elsereturnsum(pl, pr, i*2+1, left, (left+right)/2) +sum(pl, pr, i*2+2, (left+right)/2+1, right);}}