HDU 3595 GG and MM(Every-SG)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 805    Accepted Submission(s): 367

Problem Description

GG and MM like playing a game since they are children. At the beginning of game, there are two piles of stones. MM chooses a pile of stones first, which has x stones, and then she can choose a positive number k and remove k*x stones out from the other pile of stones, which has y stones (I think all of you know that y>=k*x - -!). Then it comes the turn of GG, followed the rules above-mentioned as well. When someone can't remove any stone, then he/she loses the game, and this game is finished.
Many years later, GG and MM find this game is too simple, so they decided to play N games at one time for fun. MM plays first, as the same, and the one on his/her turn must play every unfinished game. Rules to remove are as same as above, and if someone cannot remove any stone (i.e., loses the last ending game), then he/she loses. Of course we can assume GG and MM are clever enough, and GG will not lose intentionally, O(∩_∩)O~

 

 

Input

The input file contains multiply test cases (no more than 100).
The first line of each test case is an integer N, N<=1000, which represents there are N games, then N lines following, each line has two numbers: p and q, standing for the number of the two piles of stones of each game, p, q<=1000(it seems that they are so leisure = =!), which represent the numbers of two piles of stones of every game.
The input will end with EOF.

 

 

Output

For each test case, output the name of the winner.

 

 

Sample Input

3 1 1 1 1 1 1 1 3 2

 

 

Sample Output

MM GG

 

 

Author

alpc95

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（16）——Host by NUDT

 

 

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zhengfeng   |   We have carefully selected several similar problems for you:   3600  3593  3599  3598  3594 

 

題意：一共有n個遊戲，每個遊戲有兩堆石子，一次移動能夠從大的那堆石子裏拿小的那堆石子的整數倍的石子。

只要是能夠操做的遊戲都要進行操做，不能進行操做的人負。

 

比較神的博弈

模型是Every-SG確定是沒問題，框架按套路寫就能夠

有一個比較顯然的結論

設兩個數爲$(x,y)$，那麼當$\frac{y}{x}>1$，此時先手必勝，由於先手可能經過控制倍數來控制接下來步數的奇偶性

 

#include<cstdio> #include<cstring> #include<algorithm> #include<iostream>
const int MAXN=1001; inline int read() { char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } int a[MAXN],b[MAXN],SG[MAXN][MAXN],step[MAXN][MAXN]; int GetSG(int x,int y) { if(x>y) std::swap(x,y); if(SG[x][y]!=-1) return SG[x][y]; if(!x||!y) return SG[x][y]=step[x][y]=0; int willx=y%x,willy=x; int k=y/x; if(k==1) { SG[x][y]=GetSG(willx,willy)^1; step[x][y]=step[willx][willy]+1; return SG[x][y]; } else { step[x][y]=GetSG(willx,willy)+step[willx][willy]+1; return SG[x][y]=1;//此時先手必勝 
 } } int main() { #ifdef WIN32 freopen("a.in","r",stdin); #else
    #endif memset(SG,-1,sizeof(SG)); int N; while(scanf("%d",&N)!=EOF) { int ans=0; for(int i=1;i<=N;i++) { int x=read(),y=read(); if(x>y) std::swap(x,y); GetSG(x,y); ans=std::max(ans,step[x][y]); } puts(ans%2?"MM":"GG"); } return 0; }