AtCoder Regular Contest 137

题目链接

AtCoder Regular Contest 137

C - Distinct Numbers

Problem Statement

You are given a non-negative integer sequence of length $N: A=\left(A_{1}, A_{2}, \cdots, A_{N}\right)$. Here, all elements of $A$ are pairwise distinct.
Alice and Bob will play a game. They will alternately play a turn, with Alice going first. In each turn, the player does the operation below.

• Choose the largest element in $A$ at the moment, and replace it with a smaller non-negative integer. Here, all elements in $A$ must still be pairwise distinct after this operation.
  The first player to be unable to do the operation loses. Determine the winner when both players play optimally.
  Constraints
• $2 \leq N \leq 3 \times 10^{5}$
• $0 \leq A_{1}<A_{2}<\cdots<A_{N} \leq 10^{9}$
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A1 A2 ... AN

Output

If Alice wins, print Alice; if Bob wins, print Bob.

Sample Input 1

2
2 4

Sample Output 1

Alice

In Alice's first turn, she may replace $4$ with $0$, $1$, or $3$. If she replaces $4$ with $0$ or $1$, Bob's move in the next turn will make Alice unable to do the operation and lose. On the other hand, if she replaces $4$ with $3$, she can win regardless of Bob's subsequent moves. Thus, Alice wins in this input.

Sample Input 2

3
0 1 2

Sample Output 2

Bob

解题思路

博弈论

• 如果 $a[n]-a[n-1]\geq 2$，先手必胜
  证明：如果存在一种方案将 $a[n]$ 变为小于 $a[n-1]$ 的数必胜，有两种情况：1.小于 $a[n-1]$ 的数都不能选，则先手可以选择将 $a[n]$ 变为 $a[n-1]+1$ 使后手无数可选；2.有小于 $a[n-1]$ 的数可选，则先手选择一个小于 $a[n-1]$ 使其必胜。否则所有将 $a[n]$ 变为小于 $a[n-1]$ 的数必败，先手可以选择将 $a[n]$ 变为 $a[n-1]+1$ 使后手处于必败局面

• $a[n]-a[n-1]=1$
  为了不使后手处于必胜局面，先手要保证每一次操作后最大值与次最大值之间相差为 $1$，每次操作最大值都要减小，最终必败局面为 $0,1,\dots,n-1$，即 $a[n]-(n-1)$ 的奇偶性决定最后的胜者

• 时间复杂度：$O(1)$

代码

// Problem: C - Distinct Numbers
// Contest: AtCoder - AtCoder Regular Contest 137
// URL: https://atcoder.jp/contests/arc137/tasks/arc137_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// %%%Skyqwq
#include <bits/stdc++.h>
 
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
 
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
 
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
 
template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}

const int N=300005;
int n,a[N];
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    bool f=0;
    if(a[n]-a[n-1]>=2)f=1;
    else if((a[n]-(n-1))%2)f=1;
    puts(f?"Alice":"Bob");
    return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/16032554.html
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