HDU 4371 Alice and Bob

Alice and Bob

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 691    Accepted Submission(s): 420

Problem Description

Alice and Bob are interested in playing games. One day, they invent a game which has rules below:
1. Firstly, Alice and Bob choose some random positive integers, and here we describe them as n, d₁, d₂,..., d_m.
2. 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 s₂which satisfies the condition that s₂ = s₁ + d_k 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 s_iwhich satisfies the condition that s_i = s_i-1 + d_k or s_i = s_i-1 - d_k , and s_i-2 < s_i <= n, 1 <= k <= m, i >= 3 at the same time.
3. 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,..., d_m.
T <= 100;
1 <= n <= 10⁶;
1 <= m <= 100;
1 <= d_k <= 10⁶, 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

27 137 22 6

 

Sample Output

Case #1: AliceCase #2: Bob

 

两个人轮流写数字，写不出来的那个人就算输了。

因为要是数列处于间隔上升趋势，那么只要每次都选择最小的数字，不给对手做减法的余地

#include <iostream>#include <stdio.h>#include <cstring>#include <string>#include <algorithm>#include <cmath>#define N 1000009using namespace std;int a[N];int n,m;int t;int main(){    while(~scanf("%d",&t))    {        for(int ca=1;ca<=t;ca++)        {        scanf("%d%d",&n,&m);        for(int i=1;i<=m;i++)        scanf("%d",&a[i]);        sort(a+1,a+1+m);        int s=0;        int num=1;        while(1)        {            num++;            if(s+a[1]>n)            {                break;            }            s+=a[1];        }        printf("Case #%d: ",ca);        if(num%2==0)        cout<<"Alice"<<endl;        else        cout<<"Bob"<<endl;        }    }    return 0;}