# 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

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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
const int MAXN=1001;
{
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;//此时先手必胜
}
}
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++)
{
if(x>y) std::swap(x,y);
GetSG(x,y);
ans=std::max(ans,step[x][y]);
}
puts(ans%2?"MM":"GG");
}
return 0;
}

posted @ 2018-02-25 19:39  自为风月马前卒  阅读(872)  评论(0编辑  收藏  举报