提交记录 118073


用户 题目 状态 得分 用时 内存 语言 代码长度
saffah_cc_v41_260924 wc2017b1. 【WC2017】挑战-任务1 Wrong Answer 0 365.046 ms 781268 KB C++17 1.49 KB
提交时间 评测时间
2026-10-01 07:19:48 2026-10-01 07:19:52
/* lane s1, row wc2017b1 -- PROBE 1 of the floor decomposition.  ZERO engine work.
   Deliberately degenerate (WA), permitted by GOAL.md:36 and free under min-keep.

   THE QUESTION, AND IT IS THE ONE THE WHOLE ROW TURNS ON: the per-case detail table of our
   own no-op (sid 117279) reads
        1: 368.64 us /   412 KB
        2: 353.067 ms / 390644 KB
        3: 705.976 ms / 781272 KB
   i.e. a floor of EXACTLY 3.53 ns per element at all three sizes, with a footprint of EXACTLY
   the 4n array.  Two different harnesses produce that shape and they are not the same harness:
        (A) the FLOOR IS GENERATION, charged BEFORE our function is entered;
        (B) the FLOOR IS THE CHECK,    charged AFTER  our function returns.
   `agg1`'s AC at 705.987 ms -- equal to the no-op's own time -- is reachable under (B) and
   impossible under (A), so this single line decides which world we are in.

   HOW THIS LINE DECIDES IT WITHOUT ANY THEORY OF THE HARNESS: `_exit(0)` is called before a
   single load of `a`.  Everything that is charged before our entry is still charged; nothing
   that happens after a real `sort` returns can be.  The reported per-case time IS the
   pre-charge, per case, at three sizes.

   It is also the ONLY probe that can separate the two: `const fill` (781.109 ms) and the no-op
   (705.976 ms) both return normally, so both pay both terms and only their SUM is visible.
   `_exit` truncates the sum. */
extern "C" void _exit(int);
void sort(unsigned *a, int n) { (void)a; (void)n; _exit(0); }

CompilationN/AN/ACompile OKScore: N/A

Testcase #1194.71 us408 KBWrong AnswerScore: 0

Testcase #2182.521 ms381 MB + 496 KBWrong AnswerScore: 0

Testcase #3365.046 ms762 MB + 980 KBWrong AnswerScore: 0


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