White cattle off season May 16 A stone's attendance problems (Game)

Links: https://ac.nowcoder.com/acm/contest/949/A
Source: Cattle-off network

Time limit: C / C ++ 1 second, 2 seconds languages other
space restrictions: C / C ++ 262144K, other languages 524288K
64bit the IO the Format: LLD%

Title Description

Small stone male and play games, a total of  $\mathrm{n\; number\; of\; n,\; respectively \; .\; 1\; ~\; n\; 1~n.\; They\; take\; turns\; fetching,\; stone\; upper\; hand.\; For\; each\; round\; access,\; you\; must\; select\; the\; remaining\; number\; in\; any\; of\; a\; number\; of\; the\; X-\; the\; X-,\; but\; also\; take\; away \; the\; X-\; ,\; \lfloor \; the\; X-/\; 2\; \rfloor \; ,\; \lfloor \; \lfloor \; the\; X-/\; 2\; \rfloor \; /\; 2\; \rfloor \; ...\; If\; a\; number\; does\; not\; exist,\; stop\; access\; (not\; a\; number\; not\; take).\; Who\; caught\; the\; last\; number,\; whoever\; lost.\; Small\; stone\; wanted\; to\; know\; if\; I\; can\; win.\; If\; the\; stone\; win,\; output \; "Shi"\; ,\; otherwise\; output \; "Yang"\; (the\; quotation\; marks\; are\; not\; output).}$

Enter a description:

A total line, enter a number  $\mathrm{n-\; n-.}$

Output Description:

A total line, the output "Shi" or "Yang" (without the quotes output).

Example 1

Entry

1

Export

That

Explanation

Stone removed only $\mathrm{1\; 1\; Xiaoyang\; win.}$

Example 2

Entry

2

Export

Shi

Explanation

If the stone removed $\mathrm{1\; 1,\; a\; small\; sun\; can\; only\; take\; away\; 2\; 2\; stone\; win.}$

Remarks:

$\mathrm{1\; \le \; the\; n-\; \le \; 1E9\; ideas:\; game\; theory\; problem,\; because\; the\; stone\; is\; the\; upper\; hand,\; it\; will\; definitely\; be\; a\; stone\; taken\; away\; and\; would\; be\; taken\; away\; in\; the\; first\; round,\; assuming\; n>\; 1,\; and\; stone\; to\; lose,\; then\; for\; all\; stone\; will\; lose\; of\; the\; first\; round\; stone\; if\; you\; take\; a\; small\; positive\; becomes\; equivalent\; to\; the\; upper\; hand,\; it\; will\; lose\; a\; small\; positive,\; so\; in\; addition\; to\; n\; =\; 1\; than\; losing\; a\; small\; positive,\; stone\; will\; win\; the\; code:}$

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 int n;
 6 
 7 int main()
 8 {
 9     cin >> n;
10     if(n == 1)  cout << "Yang";
11     else cout << "Shi";
12     return 0;
13 }