結果

問題 No.2167 Fibonacci Knapsack
ユーザー 👑 hos.lyric
提出日時 2023-05-21 11:26:24
言語 D
(dmd 2.109.1)
結果
AC  
実行時間 9 ms / 2,000 ms
コード長 2,792 bytes
コンパイル時間 995 ms
コンパイル使用メモリ 132,000 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-06-22 17:57:18
合計ジャッジ時間 2,154 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 21
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import std.conv, std.functional, std.range, std.stdio, std.string;
import std.algorithm, std.array, std.bigint, std.bitmanip, std.complex, std.container, std.math, std.mathspecial, std.numeric, std.regex, std.typecons
    ;
import core.bitop;
class EOFException : Throwable { this() { super("EOF"); } }
string[] tokens;
string readToken() { for (; tokens.empty; ) { if (stdin.eof) { throw new EOFException; } tokens = readln.split; } auto token = tokens.front; tokens
    .popFront; return token; }
int readInt() { return readToken.to!int; }
long readLong() { return readToken.to!long; }
real readReal() { return readToken.to!real; }
bool chmin(T)(ref T t, in T f) { if (t > f) { t = f; return true; } else { return false; } }
bool chmax(T)(ref T t, in T f) { if (t < f) { t = f; return true; } else { return false; } }
int binarySearch(alias pred, T)(in T[] as) { int lo = -1, hi = cast(int)(as.length); for (; lo + 1 < hi; ) { const mid = (lo + hi) >> 1;
    (unaryFun!pred(as[mid]) ? hi : lo) = mid; } return hi; }
int lowerBound(T)(in T[] as, T val) { return as.binarySearch!(a => (a >= val)); }
int upperBound(T)(in T[] as, T val) { return as.binarySearch!(a => (a > val)); }
void main() {
enum LIM = 90;
auto F = new long[LIM];
F[0] = 0;
F[1] = 1;
foreach (i; 2 .. LIM) {
F[i] = F[i - 1] + F[i - 2];
}
debug {
writeln("F = ", F);
writeln(cast(real)(F[$ - 1]));
}
try {
for (; ; ) {
const numCases = readInt;
foreach (caseId; 0 .. numCases) {
const N = readInt;
const W = readLong;
auto A = new long[N + 2];
foreach (i; 2 .. N + 2) {
A[i] = readLong;
}
alias Entry = Tuple!(long, "x", long, "cost");
Entry[] crt = [Entry(0, 0)];
foreach_reverse (i; 2 .. N + 2) {
/*
will remain: F[2] + ... + F[i-1] = F[i+1] - 2
{x, x + F[i]} -> {x, x + F[i], x + 2 F[i]}
2 F[i] > F[i+1] - 2
{x, x + F[i+1]} -> {x, x + F[i], x + F[i+1], x + F[i+2]}
F[i+1] > F[i+1] - 2
*/
assert(crt.length == 1 ||
(crt.length == 2 && (crt[1].x == crt[0].x + F[i] || crt[1].x == crt[0].x + F[i+1])));
Entry[] nxt = crt.dup;
foreach (e; crt) {
if (e.cost + A[i] <= W) {
nxt ~= Entry(e.x + F[i], e.cost + A[i]);
}
}
nxt.sort;
const x0 = nxt[$ - 1].x - (F[i+1] - 2);
crt = [];
foreach (e; nxt) if (e.x >= x0) {
if (crt.empty || crt[$ - 1].x < e.x) {
crt ~= e;
}
}
debug {
writeln(crt);
}
}
writeln(crt[$ - 1].x);
}
}
} catch (EOFException e) {
}
}
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