chp居然没来。。然后我跟zmc鼓捣了一下午。。出了3题 有一道貌似是DP水题。。然后我俩都不会dp。。A题水题不说了，E题博弈论模板。。然后我俩也都不会博弈论。。现场找的模板。。F题无脑BFS。。一开始看成计算几何了没敢做sad。。

A题：

Post Robot

Apple bananaiPad lemon ApplepiSony
233
Tim cook is doubi from Apple
iPhoneipad
iPhone30 is so biiiiiiig Microsoft
makes good App.

MAI MAI MAI!
MAI MAI MAI!
MAI MAI MAI!
SONY DAFA IS GOOD!
MAI MAI MAI!
MAI MAI MAI!
MAI MAI MAI!

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
using namespace std;
#define LL long long
char s[1000010];
int main()
{
	while(gets(s))
	{
		int len=strlen(s);
		for(int i=0;i<len;i++)
			if(!strncmp(s+i,"Apple",5))
			puts("MAI MAI MAI!");
		    else if(!strncmp(s+i,"iPhone",6))
			puts("MAI MAI MAI!");
			else if(!strncmp(s+i,"iPod",4))
			puts("MAI MAI MAI!");
			else if(!strncmp(s+i,"iPad",4))
			puts("MAI MAI MAI!");
			else if(!strncmp(s+i,"Sony",4))
			puts("SONY DAFA IS GOOD!");
	}
	return 0;
}

Game

1
1
2
1 1
3
1 2 3

Win
Lose
Lose

博弈论 尼姆博奕（Nimm Game）

指的是这样的一个博弈游戏，目前有任意堆石子，每堆石子个数也是任意的，双方轮流从中取出石子，规则如下：
1)每一步应取走至少一枚石子；每一步只能从某一堆中取走部分或全部石子；
2)如果谁取到最后一枚石子就胜。
也就是尼姆博弈（Nimm Game）。

结论：假设有n堆物品，每堆物品有a[i]件，若a[0]^a[1]^a[2]^........a[n-1]!=0 先手必胜，否则必败。

#include <cstdio>
using namespace std;
int main()
{
	int n,t,x;
	while(~scanf("%d",&n))
	{
		scanf("%d",&t);--n;
		while(n--)
		{
			scanf("%d",&x);t^=x;
		}
		if(t)puts("Win");
		else puts("Lose");
	}
	return 0;
}

Dice

There are 2 special dices on the table. On each face of the dice, a distinct number was written. Consider a₁.a₂,a₃,a₄,a₅,a₆ to be numbers written on top face, bottom face, left face, right face, front face and back face of dice A. Similarly, consider b₁.b₂,b₃,b₄,b₅,b₆ to be numbers on specific faces of dice B. It’s guaranteed that all numbers written on dices are integers no smaller than 1 and no more than 6 while a_i ≠ a_j and b_i ≠ b_j for all i ≠ j. Specially, sum of numbers on opposite faces may not be 7.

At the beginning, the two dices may face different(which means there exist some i, a_i ≠ b_i). Ddy wants to make the two dices look the same from all directions(which means for all i, a_i = b_i) only by the following four rotation operations.(Please read the picture for more information)


Now Ddy wants to calculate the minimal steps that he has to take to achieve his goal.

1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 5 6 4 3
1 2 3 4 5 6
1 4 2 5 3 6

0
3
-1

六维BFS。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
struct node
{
	int a[7],step;
};
bool vis[7][7][7][7][7][7];
int s[7],e[7];
void bfs()
{
	queue <node> Q;
	node t,f;
	for(int i=1;i<7;i++)
	t.a[i]=s[i];t.step=0;
	vis[t.a[1]][t.a[2]][t.a[3]][t.a[4]][t.a[5]][t.a[6]]=1;
	Q.push(t);
	while(!Q.empty())
	{
		f=Q.front();Q.pop();
		if(f.a[1]==e[1]&&f.a[2]==e[2]&&f.a[3]==e[3]&&f.a[4]==e[4]&&f.a[5]==e[5]&&f.a[6]==e[6])
		{
			printf("%d\n",f.step);
			return ;
		}
		t.step=f.step+1;
		//left
		t.a[1]=f.a[4];t.a[2]=f.a[3];t.a[3]=f.a[1];t.a[4]=f.a[2];t.a[5]=f.a[5];t.a[6]=f.a[6];
		if(!vis[t.a[1]][t.a[2]][t.a[3]][t.a[4]][t.a[5]][t.a[6]])
		{
			vis[t.a[1]][t.a[2]][t.a[3]][t.a[4]][t.a[5]][t.a[6]]=1;
			Q.push(t);
		}
		//right
		t.a[1]=f.a[3];t.a[2]=f.a[4];t.a[3]=f.a[2];t.a[4]=f.a[1];t.a[5]=f.a[5];t.a[6]=f.a[6];
		if(!vis[t.a[1]][t.a[2]][t.a[3]][t.a[4]][t.a[5]][t.a[6]])
		{
			vis[t.a[1]][t.a[2]][t.a[3]][t.a[4]][t.a[5]][t.a[6]]=1;
			Q.push(t);
		}
		//front
		t.a[1]=f.a[6];t.a[2]=f.a[5];t.a[3]=f.a[3];t.a[4]=f.a[4];t.a[5]=f.a[1];t.a[6]=f.a[2];
		if(!vis[t.a[1]][t.a[2]][t.a[3]][t.a[4]][t.a[5]][t.a[6]])
		{
			vis[t.a[1]][t.a[2]][t.a[3]][t.a[4]][t.a[5]][t.a[6]]=1;
			Q.push(t);
		}
		//back
		t.a[1]=f.a[5];t.a[2]=f.a[6];t.a[3]=f.a[3];t.a[4]=f.a[4];t.a[5]=f.a[2];t.a[6]=f.a[1];
		if(!vis[t.a[1]][t.a[2]][t.a[3]][t.a[4]][t.a[5]][t.a[6]])
		{
			vis[t.a[1]][t.a[2]][t.a[3]][t.a[4]][t.a[5]][t.a[6]]=1;
			Q.push(t);
		}
	}
    puts("-1");
}
int main()
{
	while(~scanf("%d",&s[1]))
	{
		memset(vis,0,sizeof(vis));
		for(int i=2;i<7;i++)
			scanf("%d",&s[i]);
		for(int i=1;i<7;i++)
			scanf("%d",&e[i]);
		bfs();
	}
	return 0;
}

acm_icpc网络赛第三站：西安赛区