Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Alice and Bob are very smart guys and they like to play all kinds of games in their spare time. the most amazing thing is that they always find the best strategy, and that's why they feel bored again and again. They just invented a new game, as they usually did.

The rule of the new game is quite simple. At the beginning of the game, they write down N random positive integers, then they take turns (Alice first) to either:

1. Decrease a number by one.
2. Erase any two numbers and write down their sum.

Whenever a number is decreased to 0, it will be erased automatically. the game ends when all numbers are finally erased, and the one who cannot play in his(her) turn loses the game.

Here's the problem: Who will win the game if both use the best strategy? Find it out quickly, before they get bored of the game again!

Input 

The first line contains an integer T (1T4000), indicating the number of test cases.

Each test case contains several lines.

The first line contains an integer N (1N50).

The next line contains N positive integers A₁...A_N (1A_i1000), represents the numbers they write down at the beginning of the game.

Output 

For each test case in the input, print one line: `Case #X: Y', where X is the test case number (starting with 1) and Y is either "Alice" or "Bob".

Sample Input 

3
3
1 1 2
2
3 4
3
2 3 5

Sample Output 

Case #1: Alice
Case #2: Bob
Case #3: Bob

---

就兩種操作，取走一個，或者合併兩堆。沒有辦法動作者輸。

根據經驗，先將等效盤面壓縮成每堆只有 1 個 2 個 3 個 4 個。

計算這四種可能的堆數，進行狀態轉移即可。

#include <stdio.h>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
int dp[55][55][55][55] = {};
int used[55][55][55][55] = {};
int dfs(int A, int B, int C, int D) {
    if(A == 0 && B == 0 && C == 0 && D == 0)
        return 0;
    if(used[A][B][C][D] == 2)
        puts("EEEEEEEE");
    if(used[A][B][C][D])
        return dp[A][B][C][D];
    used[A][B][C][D] = 2;
    int &val = dp[A][B][C][D];
    val = 0;
    if(A >= 1)
        val |= !dfs(A-1, B, C, D);
    if(A >= 2)
        val |= !dfs(A-2, B+1, C, D);
    if(B >= 1)
        val |= !dfs(A+1, B-1, C, D);
    if(B >= 2)
        val |= !dfs(A, B-2, C, D+1);
    if(C >= 1)
        val |= !dfs(A, B+1, C-1, D);
    if(C >= 2)
        val |= !dfs(A, B, C-2, D+1);
    if(D >= 1)
        val |= !dfs(A, B, C+1, D-1);
    if(D >= 2)
        val |= !dfs(A, B, C, D-1);
    if(A >= 1 && B >= 1)
        val |= !dfs(A-1, B-1, C+1, D);
    if(A >= 1 && C >= 1)
        val |= !dfs(A-1, B, C-1, D+1);
    if(A >= 1 && D >= 1)
        val |= !dfs(A-1, B, C+1, D-1);
    if(B >= 1 && C >= 1)
        val |= !dfs(A, B-1, C, D);
    if(B >= 1 && D >= 1)
        val |= !dfs(A, B-1, C, D);
    if(C >= 1 && D >= 1)
        val |= !dfs(A, B, C, D-1);
    used[A][B][C][D] = 1;
    return val;
}
int main() {
    int testcase, cases = 0;
    int n, i;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d", &n);
        int x;
        int A = 0, B = 0, C = 0, D = 0;
        for(i = 0; i < n; i++) {
            scanf("%d", &x);
            if(x == 1)          A++;
            else if(x == 2)     B++;
            else if(x%2 == 1)   C++;
            else                D++;
        }
        printf("Case #%d: %s\n", ++cases, dfs(A, B, C, D) ? "Alice" : "Bob");
    }
    return 0;
}