Alice and Bob（博弈）

Alice and Bob

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB

Total submit users: 20, Accepted users: 10

Problem 11499 : No special judgement

Problem description

Alice and Bob are interested in playing games. One day, they invent a game which has rules below:   Firstly, Alice and Bob choose some random positive integers, and here we describe them as n,d1,d2,..,dm. Then they begin to write numbers alternately. At the first step, Alice has to write a “0”, here we let S1=0; Then, at the second step, Bob has to write a number S2 which satisfies the condition that S2=S1+dk and S2≤n, 1≤k≤m; From the third step, the person who has to write a number in this step must write a number Si which satisfies the condition that Si=Si-1+dk or Si= Si-1-dk , and Si-2 ＜ Si ≤n, 1≤k≤m, i≥3 at the same time. The person who can’t write a legal number at his own step firstly lose the game. Here’s the question, please tell me who will win the game if both Alice and Bob play the game optimally.

Input

At the beginning, an integer T indicating the total number of test cases. Every test case begins with two integers n and m, which are described before. And on the next line are m positive integers d1,d2,..,dm. T≤100   1≤n≤10^6 1≤m≤100 1≤dk≤10^6,1≤k≤m

Output

For every test case, print “Case #k: ” firstly, where k indicates the index of the test case and begins from 1. Then if Alice will win the game, print “Alice”, otherwise “Bob”.

Sample Input

2 7 1 3 7 2 2 6

Sample Output

Case #1: Alice Case #2: Bob

 

ps：http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11499&courseid=136

【博弈：】

今天终于用到你了！

题意：Alice 和 Bob又玩游戏了，这次是这样的，看谁先超过给定的数，超过的人就输了；首先Alice

先输入s1=0；（相当于Bob优先）然后假设第 i 次的数字是 S[i] ，那么第 i+1 次的数字 S[i+1]= S[i]+d[k]或者 S[i]-d[k]，条件是 S[i+1]<= n && S[i-1]比赛的时候没有“或者 S[i]-d[k]”这句话，不过这也没关系。

思路：我当时是这么想的，如果我来选，那么我肯定选最小的啊，这样离n才远，所以就选出dk里面的最小的，然后看看n/min(dk)是奇数呢还是偶数，如果是奇数Bob就win，偶数Alice就win；

 

加了红色的那句呢，也可以这么理解，如果我没错加min(dk),那么为了满足S[i-1]

 

#include
int main()
{
    int T,n,m,ans,tp,Case;
    scanf("%d",&T);
    for(Case=1;Case<=T;Case++)
    {
        scanf("%d%d",&n,&m);
        int min=999999999;
        for(int i=0;i)
        {
            scanf("%d",&ans);
            if(ans<min)
                min=ans;
        }
        tp=(n/min)%2;
        printf("Case #%d: ",Case);
        if(tp==0)
            printf("Alice\n");
        else
            printf("Bob\n");
    }
    return 0;
}

 再不理解就想想那么直接想最后一步，什么时候就结束呢？自然是s(i)+min > n 时就输了

 

so。。你懂了。

 

 

 

 

 

 

 

 

再不懂的话。。。。

 

 

 

 

 

 

 

 

 

只能恕我能力低微，，，解释不清楚了。

转载于:https://www.cnblogs.com/yuyixingkong/p/3959494.html