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M[16],X=16,W,k; | |
main(){ | |
T(system("stty cbreak")); | |
puts(W&1?"WIN":"LOSE"); | |
} | |
K[]={2,3,1}; | |
s(f,d,i,j,l,P){ | |
for(i=4;i--;) | |
for(j=k=l=0;k<4;) | |
j<4?P=M[w(d,i,j++)],W|=P>>11,l*P&&(f?M[w(d,i,k)]=l<<(l==P):0,k++),l=l?P?l-P?P:0:l:P:(f?M[w(d,i,k)]=l:0,++k,W|=2*!l,l=0); | |
} | |
w(d,i,j){ | |
return d?w(d-1,j,3-i):4*i+j; | |
} | |
T(i){ | |
for(i=X+rand()%X;M[i%X]*i;i--); | |
i?M[i%X]=2<<rand()%2:0; | |
for(W=i=0;i<4;)s(0,i++); | |
for(i=X,puts("\e[2J\e[H");i--;i%4||puts("")) | |
printf(M[i]?"%4d|":" |",M[i]); | |
W-2||read(0,&k,3)|T(s (1,K[(k>>X)%4])); | |
} | |
//[2048] reformated | |
Original Version : https://gist.github.com/justecorruptio/9967738 | |
by : Jay Chan |
I don't understand, why read(0,&k,3), the arrow keys only need 2 bytes.
The reason why 3 characters are read is that arrow keys generate 3-byte sequences.
Key | bytes | Hex value |
---|---|---|
arrow left | 0x1b, '[', 'D' | 0x1b5b44 |
arrow right | 0x1b, '[', 'C' | 0x1b5b43 |
arrow up | 0x1b, '[', 'A' | 0x1b5b41 |
arrow down | 0x1b, '[', 'B' | 0x1b5b42 |
When the characters are read, the 3-byte sequence actually is reversed.
So for instance, in the last case ('arrow down'), k is assigned 0x425b1b.
k is then used to decide which of the K-array value to use.
k is right-shifted by 16 bits, reducing it to 0x42, so we have K [ 0x042 % 4 ], which is equivalent to K[2], the 3rd element of K[].
Thus, the statement after the read (0, &k, 3) evaluates to calling s (1, 1).
I wonder why K[] does not have 4 values, one for each arrow key.
What is this mate i dont understand this code can u pls help me? I think w,a,s,d doesnt work for example i put w but numbers doesnt went up why?
This passes gcc and clang without a warning.
Simplified the three functions T(), w() and s(); the arguments weren't passed in from outside, but were later initialised.
Unrolled the for-loops, sticking to the "1 statement per line" rule.
Added random seed.