B. Godsend（Round #429 (Div. 2)）

最新推荐文章于 2017-11-19 18:30:36 发布

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本文介绍了一个基于数组的游戏，两位玩家轮流移除具有特定和的子序列。文章详细解释了游戏规则，并通过示例展示了如何判断哪位玩家将在双方都采用最优策略的情况下获胜。

Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?

Input
First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.

Next line contains n integers a1, a2, …, an (0 ≤ ai ≤ 109).

Output
Output answer in single line. “First”, if first player wins, and “Second” otherwise (without quotes).

Examples
input
4
1 3 2 3
output
First
input
2
2 2
output
Second
Note
In first sample first player remove whole array in one move and win.

In second sample first player can’t make a move and lose.

如果刚开始所有数和为奇数，则第一个人全拿完赢了。如果和为偶数且不存在奇数，则第一个人什么也不拿，第二个人全拿，第二个人赢，如果存在奇数，第一个人赢

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int a,sum=0,flag=1;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a);
            sum=sum+a;
            if(a%2==1)flag=0;
        }
        if(sum%2==1)
        {
            printf("First\n");
        }
        else
        {
            if(flag)
            {
                printf("Second\n");
            }
            else
            {
                printf("First\n");
            }
        }
    }
    return 0;
}