There are two seagulls playing a very peculiar game. First they line up N unit squares in a line, all originally colored white.
Let L be the length of the longest continuous sub-segment of white unit squares. Let P be any prime that satisfies the condition that . The player can choose any P if it satisfies the conditions but if there is no value of P that does, then P is set to 1.
The current player then proceeds to choose any continuous sub-segment of white unit squares of length P and paint them black. The player who can’t make a move loses.
If both players play optimally and the first player starts, which player is going to win the game?
Input
The first line of input is T – the number of test cases.
Each test case contains a line with a single integer N (1 ≤ N ≤ 107).
Output
For each test case, output on a single line "first" (without quotes) if the first player would win the game, or "second" if the second would win.
Example
2
2
5
second
first
题意:
每次可以取一段白色连续区间内p个,把白的染成黑色。不能染色的玩家失败。
整段区间内,最长的白色连续区间的长度为L,p是小于等于L/2的任意质数,如果没有质数,p可以取1;
思路:
手推,2,3是先手必败,1是先手必胜。
大于3的偶数,在中间取2,使留下来的两段长度相等。
大于3的奇数,在中间取3,使留下来的两段长度相等。
然后就像nim博弈只有两堆石子那样,跟着第二个人取,就可以坐等胜利了。
![](http://img.e-com-net.com/image/info8/b8d97b5613f94ed2ba791cad57d0b2ed.gif)
![](http://img.e-com-net.com/image/info8/2f88dd3f1cd145f59c0e47b51acdbd4b.gif)
#includeint main() { int T; scanf("%d",&T); while(T--){ int n; scanf("%d",&n); if(n==2||n==3){printf("second\n");} else printf("first\n"); } return 0; }