2014
02-12

# Snake

Snake is a popular game , and I believe most of us had played it . The Original game is that you can control a snake to eat the magic bean and after the snake eat one magic bean , the length of the snake’s body will get longer .But today we are talking about a new game. These are the rules of the new Snake Game :
1. The length of the snake’s body won’t change even though it eat a magic bean.
2. Some pairs of the beans have a relation , that is, one of them can not be eaten until another one had been eaten . We call the latter “the key bean” . For example , if A can’t be eaten until B had been eaten ,we say “B is the key bean of A”. (That means when A can’t be eaten , the snake can not move into the grid where A is.)
3. The snake could not move to a wall or its body.Befor it move,it will chooses an adjacent vacant square of its head,which is neither a stone nor occupied by its body.
Figure 1 and figure2 shows how the snake move

Figure 1

Figure 2

The first line contain a integer T (T <= 10).Followed by T cases. Each case contain five parts.
The first part: six integers ,H,W,L,K,R,N,(H <= 20 , W <= 20 , L <= 8 , K <= 7 ) means the height of the map , the width of the map , the length of the snake’s body, the number of the magic beans . the number of the relations , the number of the wall respectively.
The second part: L lines , each line contain two integer hi ,wi, indicating the original position of each block of snake’s body, from B1(h1,w1) to BL(hL,wL) orderly, where 1<=hi<=H, and 1<=wi<=W,1<=i<=L.
The third part: K lines ,each line contain two integer hi ,wi , indicating the position of each magic bean , from MB1(h1,w1) to MBK(hK,wK) orderly, where 1<=hi<=H, and 1<=wi<=W,1<=i<=K.
The fourth part : R lines , each line contain two integer A ,B means “A is the key bean of B ”. The A and B may appear several times , but “one bean will have only one key bean”.
The fifth part: N lines , each line contain two integer hi ,wi , indicating the position of each wall , from W1(h1,w1) to WN(hN,wN) orderly, where 1<=hi<=H, and 1<=wi<=W,1<=i<=N.

The first line contain a integer T (T <= 10).Followed by T cases. Each case contain five parts.
The first part: six integers ,H,W,L,K,R,N,(H <= 20 , W <= 20 , L <= 8 , K <= 7 ) means the height of the map , the width of the map , the length of the snake’s body, the number of the magic beans . the number of the relations , the number of the wall respectively.
The second part: L lines , each line contain two integer hi ,wi, indicating the original position of each block of snake’s body, from B1(h1,w1) to BL(hL,wL) orderly, where 1<=hi<=H, and 1<=wi<=W,1<=i<=L.
The third part: K lines ,each line contain two integer hi ,wi , indicating the position of each magic bean , from MB1(h1,w1) to MBK(hK,wK) orderly, where 1<=hi<=H, and 1<=wi<=W,1<=i<=K.
The fourth part : R lines , each line contain two integer A ,B means “A is the key bean of B ”. The A and B may appear several times , but “one bean will have only one key bean”.
The fifth part: N lines , each line contain two integer hi ,wi , indicating the position of each wall , from W1(h1,w1) to WN(hN,wN) orderly, where 1<=hi<=H, and 1<=wi<=W,1<=i<=N.

1
8 9
5 2
1 8
5 2
6 2
6 3
6 4
6 5
4 2
2 6
2 1
2 5
3 5
4 4
4 5
4 6
5 6
5 7
6 7

21

#include<cstring>
#include<cstdio>
#include<iostream>
using namespace std;

int cal(int x,int y)
{
int ans=1;
while(y)
{
if(y&1)
ans=(ans*x)%100;
y>>=1;
x=(x*x)%100;
}
return ans;
}
int rr[25];
int main()
{
__int64 n,nn;
for(int i=1; i<=20; ++i)
{
rr[i]=cal(2,i+20-1);
rr[i]=rr[i]*rr[i]+rr[i];
rr[i]%=100;
}
while(scanf("%I64d",&n)&&n!=0)
{
long long count=0;
while(n--)
{
count++;
scanf("%I64d",&nn);
printf("Case %d: ",count);
if(nn==1) printf("2\n");
else if(nn==2) printf("6\n");
else
{
nn=(nn+19)%20+1;
printf("%d\n",rr[nn]);
}
}
printf("\n");
}
return 0;
}