HDU-2509-Be the Winner,博弈题~~水过~~

Be the Winner

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

http://acm.hdu.edu.cn/showproblem.php?pid=2509

Problem Description

Let's consider m apples divided into n groups. Each group contains no more than 100 apples, arranged in a line. You can take any number of consecutive apples at one time.

For example "@@@" can be turned into "@@" or "@" or "@ @"(two piles). two people get apples one after another and the one who takes the last is 

the loser. Fra wants to know in which situations he can win by playing strategies (that is, no matter what action the rival takes, fra will win).

 

Input

You will be given several cases. Each test case begins with a single number n (1 <= n <= 100), followed by a line with n numbers, the number of apples in each pile. There is a blank line between cases.

 

Output

If a winning strategies can be found, print a single line with "Yes", otherwise print "No".

 

Sample Input

2
2 2
1
3

 

Sample Output

No
Yes

 

Source

ECJTU 2008 Autumn Contest

刚做完南理工的校外镜像赛，就在HUD上看到这个题，我嘞个坑啊，简直和最强战舰一模一样，除了输入输出（https://icpc.njust.edu.cn/Contest/749/H/），怪不得这么多人过了；

题意应该很好懂吧，n堆苹果，每次从一堆中拿任意个，最后拿的输，问先拿者是否能赢？？？？？

题解请看：http://blog.csdn.net/nyist_tc_lyq/article/details/51180906-我的另一篇博客，题意都是一样的，那有具体的思路题解；

AC代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=100+10;
int a[N];
int main()
{
    int n,i,j,x;
    while(~scanf("%d",&n))
    {
        x=j=0;
        int f;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            x^=a[i];
            if(a[i]>=2)
                j++;
        }
        if(x!=0)
        {
            if(j==0)
                f=1;
            else
                f=2;
        }
        else
        {
            if(j==0)
                f=2;
            else
                f=1;
        }
        if(f==1)
            printf("No\n");
        else
            printf("Yes\n");
    }
    return 0;
}