Brave game
Time Limit: 1000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 6329 accepted submission (s): 4229
Problem description when I was a university student ten years ago, China introduced some film blockbusters from abroad every year. One of the movies was called "the player's game" (English name: zathura ), until now, I have been impressed by some of the Computer stunts in the movie.
Today, we chose the computer test as a brave choice. In this short term, we are talking about the game topic. Therefore, now everyone is playing the "multiplayer game", which is why I name this question.
Of course, in addition to being brave, I also hope to see "integrity". No matter what the test scores are, what I want to see is a real result. I believe everyone can do it ~
What is the first game to be played by beginners? It is very simple. It is defined as follows:
1. This game is a two-person game;
2. There are a pile of stones with N in total;
3. Take turns;
4. Each step can take 1... M stones;
5. the first party to win the light stones;
If both sides of the game are using the best strategy, please output which one can win.
The input data first contains a positive integer c (C <= 100), indicating that there are group C test data.
Each group of test data occupies one row and contains two integers n and M (1 <= n, m <= 1000). For the meanings of N and M, see the topic description.
Output: if the first person wins, output "first"; otherwise, output "second". The output of each instance occupies one row.
Sample Input
223 24 3
Sample output
firstsecond
Water game, classic Bashi game !!!
This is also the first question in my system to start learning the game !!!
When N = m + 1 is second win, but when n> m + 1 & n <2 * m, first can use the value to reach the value that was originally second win,
But now second is gone, so first will win, so you can get the rule by pushing it down !!!!
The AC code is as follows:
#include<stdio.h>int main(){ int t; scanf("%d",&t); while(t--) { int n,m; scanf("%d%d",&n,&m); if(n%(m+1)==0) printf("second\n"); else printf("first\n"); } return 0;}
