📄 usaco 2_1_5 hamming codes 题解_leokan的blog.mht
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<DIV class=3Dtit>USACO 2.1.5 Hamming Codes =CC=E2=BD=E2</DIV>
<DIV class=3Ddate>2008=C4=EA01=D4=C228=C8=D5 =D0=C7=C6=DA=D2=BB =
13:25</DIV>
<TABLE style=3D"TABLE-LAYOUT: fixed">
<TBODY>
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<TD>
<DIV class=3Dcnt>
<H2>USACO 2.1.5 Hamming Codes</H2>
<DIV class=3Dt_msgfont>Hamming Codes<BR>Rob Kolstad <BR>Given N, =
B, and D:=20
Find a set of N codewords (1 <=3D N <=3D 64), each of length =
B bits (1=20
<=3D B <=3D 8), such that each of the codewords is at least =
Hamming=20
distance of D (1 <=3D D <=3D 7) away from each of the other =
codewords.=20
The Hamming distance between a pair of codewords is the number of =
binary=20
bits that differ in their binary notation. Consider the two =
codewords=20
0x554 and 0x234 and their differences (0x554 means the hexadecimal =
number=20
with hex digits 5, 5, and 4): <BR><BR> =
=20
0x554 =3D 0101 0101 0100<BR> 0x234 =
=3D 0010=20
0011 0100<BR>Bit differences: xxx xx<BR><BR>Since five =
bits=20
were different, the Hamming distance is 5. <BR><BR>PROGRAM NAME:=20
hamming<BR>INPUT FORMAT<BR>N, B, D on a single line <BR><BR>SAMPLE =
INPUT=20
(file hamming.in)<BR>16 7 3<BR><BR>OUTPUT FORMAT<BR>N codewords, =
sorted,=20
in decimal, ten per line. In the case of multiple solutions, your =
program=20
should output the solution which, if interpreted as a base 2^B =
integer,=20
would have the least value. <BR><BR>SAMPLE OUTPUT (file =
hamming.out)<BR>0=20
7 25 30 42 45 51 52 75 76<BR>82 85 97 102 120 =
127<BR><BR><BR><BR>Hamming=20
Codes <BR>=BA=A3=C3=F7=C2=EB<BR><BR>Rob Kolstad<BR><BR>=D2=EB by =
Felicia Crazy<BR><BR>=B8=F8=B3=F6 N=A3=ACB =BA=CD=20
D=A3=BA=D5=D2=B3=F6 N =B8=F6=B1=E0=C2=EB=A3=A81 <=3D N <=3D =
64=A3=A9=A3=AC=C3=BF=B8=F6=B1=E0=C2=EB=D3=D0 B =CE=BB=A3=A81 <=3D B =
<=3D =
8=A3=A9=A3=AC=CA=B9=B5=C3=C1=BD=C1=BD=B1=E0=C2=EB=D6=AE=BC=E4=D6=C1=C9=D9=
=D3=D0 D=20
=
=B8=F6=B5=A5=CE=BB=B5=C4=A1=B0=BA=A3=C3=F7=BE=E0=C0=EB=A1=B1=A3=A81 =
<=3D D <=3D =
7=A3=A9=A1=A3=A1=B0=BA=A3=C3=F7=BE=E0=C0=EB=A1=B1=CA=C7=D6=B8=B6=D4=D3=DA=
=C1=BD=B8=F6=B1=E0=C2=EB=A3=AC=CB=FB=C3=C7=B5=C4=B6=FE=BD=F8=D6=C6=B1=ED=CA=
=BE=B7=A8=D6=D0=B5=C4=B2=BB=CD=AC=B6=FE=BD=F8=D6=C6=CE=BB=B5=C4=CA=FD=C4=BF=
=A1=A3=BF=B4=CF=C2=C3=E6=B5=C4=C1=BD=B8=F6=B1=E0=C2=EB=20
0x554 =BA=CD 0x234 =D6=AE=BC=E4=B5=C4=C7=F8=B1=F0=A3=A80x554 =
=B1=ED=CA=BE=D2=BB=B8=F6=CA=AE=C1=F9=BD=F8=D6=C6=CA=FD=A3=AC=C3=BF=B8=F6=CE=
=BB=C9=CF=B7=D6=B1=F0=CA=C7 5=A3=AC5=A3=AC4=A3=A9=A3=BA =
<BR><BR> =20
0x554 =3D 0101 0101 0100<BR> =20
0x234 =3D 0010 0011 0100<BR> =
=B2=BB=CD=AC=B5=C4=B6=FE=BD=F8=D6=C6=CE=BB:=20
xxx =
xx<BR><BR>=D2=F2=CE=AA=D3=D0=CE=E5=B8=F6=CE=BB=B2=BB=CD=AC=A3=AC=CB=F9=D2=
=D4=A1=B0=BA=A3=C3=F7=BE=E0=C0=EB=A1=B1=CA=C7 5=A1=A3 <BR><BR>PROGRAM =
NAME:=20
hamming<BR>INPUT FORMAT<BR>=D2=BB=D0=D0=A3=AC=B0=FC=C0=A8 N, B, =
D=A1=A3 <BR><BR>SAMPLE INPUT (file=20
hamming.in)<BR>16 7 3<BR><BR>OUTPUT FORMAT<BR>N =
=B8=F6=B1=E0=C2=EB=A3=A8=D3=C3=CA=AE=BD=F8=D6=C6=B1=ED=CA=BE=A3=A9=A3=AC=D2=
=AA<SPAN=20
class=3Dt_tag=20
=
href=3D"tag.php?name=3D%C5%C5%D0%F2">=C5=C5=D0=F2</SPAN>=A3=AC=CA=AE=B8=F6=
=D2=BB=D0=D0=A1=A3=C8=E7=B9=FB=D3=D0=B6=E0=BD=E2=A3=AC=C4=E3=B5=C4=B3=CC=D0=
=F2=D2=AA=CA=E4=B3=F6=D5=E2=D1=F9=B5=C4=BD=E2=A3=BA=BC=D9=C8=E7=B0=D1=CB=FC=
=BB=AF=CE=AA=20
2^B =
=BD=F8=D6=C6=B5=C4=CA=FD=A3=AC=CB=FC=B5=C4=D6=B5=D2=AA=D7=EE=D0=A1=A1=A3 =
<BR><BR>SAMPLE OUTPUT (file hamming.out)<BR>0 7 25 30 42=20
45 51 52 75 76<BR>82 85 97 102 120 127</DIV>
<P></P>
<HR>
<P></P>
<P><STRONG>USACO 2.1.5 Hamming =
Codes<BR>=CC=E1=BD=BB=B4=CE=CA=FD=A3=BA2=B4=CE</STRONG></P>
=
<P><STRONG>=D5=E2=CC=E2=BF=C9=D2=D4=D6=B1=BD=D3=CB=D1=CB=F7=A3=AC=B9=D8=D3=
=DA=C5=D0=B6=CF=B6=FE=BD=F8=D6=C6=B2=BB=CD=AC=B5=C4=B8=F6=CA=FD=BF=C9=D2=D4=
=D5=E2=C3=B4=D7=F6=A1=A3=CF=C8=D3=C3x xor=20
=
y=B5=C3=B5=BDxy=D6=AE=BC=E4=C4=C4=BC=B8=B8=F6=CA=FD=B2=BB=CD=AC=A3=AC=C8=BB=
=BA=F3=CD=B3=BC=C6=D5=E2=B8=F6=CA=FD=D3=D0=B6=E0=C9=D9=B8=F61.=D3=C3x =
and=20
=
-x=B5=C3=B5=BD=D7=EE=D3=D2=B1=DF=B5=C4=D2=BB=B8=F61=A3=AC=C8=BB=BA=F3=BC=F5=
=C8=A5=CB=FC=A3=AC=D6=D8=B8=B4=D5=E2=B8=F6=D6=B1=B5=BD=CB=FC=CE=AA0.=BB=B9=
=D3=D0=D2=BB=D6=D6=B7=BD=B7=A8=CA=C7=B7=D6=D2=B1=B7=A8=B5=C3=B5=BD=A3=BA<=
/STRONG></P><STRONG>
<HR>
</STRONG>
<P><FONT =
size=3D2><STRONG>=CE=D2=BD=E2=CA=CD=B5=C3=BF=CF=B2=BB=C8=E7m67=C5=A3=B5=C4=
=BA=C3=A3=AC=D2=FD=D3=C3=CB=FC=B5=C4=B0=C9=A1=A3=D2=BB=CF=C2<FONT=20
=
color=3D#3366ff>=C0=B6=C9=AB=D7=D6=CC=E5=B2=BF=B7=D6</FONT>=D7=AA=D7=D4m6=
7=B5=C4blog</STRONG></FONT></P>
<P><FONT size=3D2><FONT style=3D"BACKGROUND-COLOR: #ffffff"><FONT=20
=
color=3D#3366ff>=BC=C6=CB=E3=B6=FE=BD=F8=D6=C6=D6=D0=B5=C41=B5=C4=B8=F6=CA=
=FD<BR> =20
=
=CD=AC=D1=F9=BC=D9=C9=E8x=CA=C7=D2=BB=B8=F632=CE=BB=D5=FB=CA=FD=A1=A3=BE=AD=
=B9=FD=CF=C2=C3=E6=CE=E5=B4=CE=B8=B3=D6=B5=BA=F3=A3=ACx=B5=C4=D6=B5=BE=CD=
=CA=C7=D4=AD=CA=FD=B5=C4=B6=FE=BD=F8=D6=C6=B1=ED=CA=BE=D6=D0=CA=FD=D7=D61=
=B5=C4=B8=F6=CA=FD=A1=A3=B1=C8=C8=E7=A3=AC=B3=F5=CA=BC=CA=B1x=CE=AA131452=
0=A3=A8=CD=F8=D3=D1=D7=A5=BF=F1=A3=BA=C4=DC=B2=BB=C4=DC=BB=BB=D2=BB=B8=F6=
=CA=FD=B0=A1=A3=A9=A3=AC=C4=C7=C3=B4=D7=EE=BA=F3x=BE=CD=B1=E4=B3=C9=C1=CB=
9=A3=AC=CB=FC=B1=ED=CA=BE1314520=B5=C4=B6=FE=BD=F8=D6=C6=D6=D0=D3=D09=B8=F6=
1=A1=A3<BR></FONT></FONT></FONT></P>
<DIV class=3DUBBPanel>
<DIV class=3DUBBContent><PRE><CODE class=3Dpascal><FONT =
style=3D"BACKGROUND-COLOR: #ffffff" color=3D#3366ff size=3D2>x :=3D (x =
<SPAN class=3Dkeyword>and</SPAN> $<SPAN class=3Dnumber>55555555</SPAN>) =
+ ((x <SPAN class=3Dkeyword>shr</SPAN> <SPAN class=3Dnumber>1</SPAN>) =
<SPAN class=3Dkeyword>and</SPAN> $<SPAN class=3Dnumber>55555555</SPAN>); =
x :=3D (x <SPAN class=3Dkeyword>and</SPAN> $<SPAN =
class=3Dnumber>33333333</SPAN>) + ((x <SPAN class=3Dkeyword>shr</SPAN> =
<SPAN class=3Dnumber>2</SPAN>) <SPAN class=3Dkeyword>and</SPAN> $<SPAN =
class=3Dnumber>33333333</SPAN>);=20
x :=3D (x <SPAN class=3Dkeyword>and</SPAN> $<SPAN =
class=3Dnumber>0</SPAN>F0F0F0F) + ((x <SPAN class=3Dkeyword>shr</SPAN> =
<SPAN class=3Dnumber>4</SPAN>) <SPAN class=3Dkeyword>and</SPAN> $<SPAN =
class=3Dnumber>0</SPAN>F0F0F0F);=20
x :=3D (x <SPAN class=3Dkeyword>and</SPAN> $<SPAN =
class=3Dnumber>00</SPAN>FF00FF) + ((x <SPAN class=3Dkeyword>shr</SPAN> =
<SPAN class=3Dnumber>8</SPAN>) <SPAN class=3Dkeyword>and</SPAN> $<SPAN =
class=3Dnumber>00</SPAN>FF00FF);=20
x :=3D (x <SPAN class=3Dkeyword>and</SPAN> $<SPAN =
class=3Dnumber>0000</SPAN>FFFF) + ((x <SPAN class=3Dkeyword>shr</SPAN> =
<SPAN class=3Dnumber>16</SPAN>) <SPAN class=3Dkeyword>and</SPAN> $<SPAN =
class=3Dnumber>0000</SPAN>FFFF); </FONT></CODE></PRE></DIV></DIV>
<P><BR><FONT style=3D"BACKGROUND-COLOR: #ffffff" color=3D#3366ff=20
size=3D2> =20
=
=CE=AA=C1=CB=B1=E3=D3=DA=BD=E2=CB=B5=A3=AC=CE=D2=C3=C7=CF=C2=C3=E6=BD=F6=CB=
=B5=C3=F7=D5=E2=B8=F6=B3=CC=D0=F2=CA=C7=C8=E7=BA=CE=B6=D4=D2=BB=B8=F68=CE=
=BB=D5=FB=CA=FD=BD=F8=D0=D0=B4=A6=C0=ED=B5=C4=A1=A3=CE=D2=C3=C7=C4=C3=CA=FD=
=D7=D6211=A3=A8=CE=D2=C3=C7=B0=E0=C4=B3MM=B5=C4=C9=FA=C8=D5=A3=A9=C0=B4=BF=
=AA=B5=B6=A1=A3211=B5=C4=B6=FE=BD=F8=D6=C6=CE=AA11010011=A1=A3<BR><BR></F=
ONT><FONT=20
size=3D2><FONT style=3D"BACKGROUND-COLOR: #ffffff"><FONT=20
color=3D#3366ff><SPAN>+---+---+---+---+---+---+---+---+<BR>| 1 | 1 =
| 0 | 1 |=20
0 | 0 | 1 | 1 | =20
=
<---=D4=AD=CA=FD<BR>+---+---+---+---+---+---+---+---+<BR>| =
1=20
0 | 0 1 | 0 =
0 =20
| 1 0 | =20
=
<---=B5=DA=D2=BB=B4=CE=D4=CB=CB=E3=BA=F3<BR>+-------+-------+-------+-=
------+<BR>| =20
0 0 1 1 | 0 0 1=20
0 | =20
=
<---=B5=DA=B6=FE=B4=CE=D4=CB=CB=E3=BA=F3<BR>+---------------+---------=
------+<BR>| =20
0 0 0 0 0 1 0 1 =20
| =20
=
<---=B5=DA=C8=FD=B4=CE=D4=CB=CB=E3=BA=F3=A3=AC=B5=C3=CA=FD=CE=AA5<BR>+=
-------------------------------+</SPAN><BR><BR> =20
=
=D5=FB=B8=F6=B3=CC=D0=F2=CA=C7=D2=BB=B8=F6=B7=D6=D6=CE=B5=C4=CB=BC=CF=EB=A1=
=A3=B5=DA=D2=BB=B4=CE=CE=D2=C3=C7=B0=D1=C3=BF=CF=E0=C1=DA=B5=C4=C1=BD=CE=BB=
=BC=D3=C6=F0=C0=B4=A3=AC=B5=C3=B5=BD=C3=BF=C1=BD=CE=BB=C0=EF1=B5=C4=B8=F6=
=CA=FD=A3=AC=B1=C8=C8=E7=C7=B0=C1=BD=CE=BB10=BE=CD=B1=ED=CA=BE=D4=AD=CA=FD=
=B5=C4=C7=B0=C1=BD=CE=BB=D3=D02=B8=F61=A1=A3=B5=DA=B6=FE=B4=CE=CE=D2=C3=C7=
=BC=CC=D0=F8=C1=BD=C1=BD=CF=E0=BC=D3=A3=AC10+01=3D11=A3=AC00+10=3D10=A3=AC=
=B5=C3=B5=BD=B5=C4=BD=E1=B9=FB=CA=C700110010=A3=AC=CB=FC=B1=ED=CA=BE=D4=AD=
=CA=FD=C7=B04=CE=BB=D3=D03=B8=F61=A3=AC=C4=A94=CE=BB=D3=D02=B8=F61=A1=A3=D7=
=EE=BA=F3=D2=BB=B4=CE=CE=D2=C3=C7=B0=D10011=BA=CD0010=BC=D3=C6=F0=C0=B4=A3=
=AC=B5=C3=B5=BD=B5=C4=BE=CD=CA=C7=D5=FB=B8=F6=B6=FE=BD=F8=D6=C6=D6=D01=B5=
=C4=B8=F6=CA=FD=A1=A3=B3=CC=D0=F2=D6=D0=C7=C9=C3=EE=B5=D8=CA=B9=D3=C3=C8=A1=
=CE=BB=BA=CD=D3=D2=D2=C6=A3=AC=B1=C8=C8=E7=B5=DA=B6=FE=D0=D0=D6=D0$333333=
33=B5=C4=B6=FE=BD=F8=D6=C6=CE=AA00110011001100....=A3=AC=D3=C3=CB=FC=BA=CD=
x=D7=F6and=D4=CB=CB=E3=BE=CD=CF=E0=B5=B1=D3=DA=D2=D42=CE=AA=B5=A5=CE=BB=BC=
=E4=B8=F4=C8=A1=CA=FD=A1=A3shr=B5=C4=D7=F7=D3=C3=BE=CD=CA=C7=C8=C3=BC=D3=B7=
=A8=D4=CB=CB=E3=B5=C4=CF=E0=CD=AC=CA=FD=CE=BB=B6=D4=C6=EB=A1=A3</FONT></F=
ONT></FONT></P><FONT=20
size=3D2><FONT style=3D"BACKGROUND-COLOR: #ffffff"><FONT=20
color=3D#3366ff><STRONG>
<HR>
</STRONG></FONT></FONT></FONT>
=
<P><STRONG>=D5=E2=CC=E2=B5=C4=CA=E4=B3=F6=D5=E6=C2=E9=B7=B3...=CE=AA=B4=CB=
wa=C1=CB=D2=BB=B4=CE=A1=A3</STRONG></P>
<P><STRONG>{<BR>TASK:hamming<BR>LANG:PASCAL<BR>}<BR>program=20
hamming;<BR>var<BR> =
n,b,d:integer;<BR> =20
ans:array[0..10000] of integer;<BR>procedure=20
init;<BR>begin<BR> =20
assign(input,'hamming.in');reset(input);<BR> =20
readln(n,b,d);<BR> =
close(input);<BR>end;<BR>function=20
check(x:longint):boolean;<BR>var<BR> =20
i:integer;<BR> =20
pos:longint;<BR>begin<BR> for i:=3D1 to ans[0]=20
do<BR> =20
=
begin<BR> &nbs=
p;=20
pos:=3Dx xor=20
=
ans[i];<BR> &n=
bsp;=20
pos :=3D (pos and $55555555) + ((pos shr 1) and=20
=
$55555555);<BR> &nbs=
p; =20
pos :=3D (pos and $33333333) + ((pos shr 2) and=20
=
$33333333);<BR> &nbs=
p; =20
pos :=3D (pos and $0F0F0F0F) + ((pos shr 4) and=20
=
$0F0F0F0F);<BR> &nbs=
p; =20
pos :=3D (pos and $00FF00FF) + ((pos shr 8) and=20
=
$00FF00FF);<BR> &nbs=
p; =20
pos :=3D (pos and $0000FFFF) + ((pos shr 16) and=20
=
$0000FFFF);<BR> &nbs=
p; =20
if pos<d then=20
exit(false);<BR> =20
end;<BR> exit(true);<BR>end;<BR>procedure=20
work;<BR>var<BR> =20
num:longint;<BR>begin<BR> =20
ans[0]:=3D1;<BR> =
ans[1]:=3D0;<BR> =20
num:=3D0;<BR> for num:=3D1 to 1 shl b=20
do<BR> =20
=
begin<BR> &nbs=
p;=20
if check(num)=20
=
then<BR>  =
; =20
=
begin<BR> &nbs=
p; =20
=
inc(ans[0]);<BR> &nb=
sp; =20
=
ans[ans[0]]:=3Dnum;<BR> &n=
bsp; =20
=
end;<BR>  =
;=20
if ans[0]=3Dn then =
break;<BR> =20
end;<BR> =20
=
assign(output,'hamming.out');rewrite(output);<BR> for=20
num:=3D1 to ans[0] =
do<BR> =20
=
begin<BR> &nbs=
p;=20
if (num-1)mod 10=3D0 then=20
=
write(ans[num])<BR> =
=20
else write('=20
=
',ans[num]);<BR> &nb=
sp; =20
if num mod 10=3D0 then=20
writeln;<BR> =20
end;<BR> if num mod 10<>0 then=20
writeln;<BR> =20
close(output);<BR>end;<BR>begin<BR> =20
init;<BR> work;<BR>end.</STRONG></P>
<P></P><STRONG>
<HR>
</STRONG>
=
<P><STRONG>USACO=B5=C4=B7=D6=CE=F6=A3=AC=D3=C3=B5=DD=B9=E9=C7=F3=BD=E2</S=
TRONG></P>
<P><STRONG>There are only a few tools we need to solve this =
problem. First=20
of all, we can use basic techniques to find the Nth bit of a =
number M:=20
counting the least significant bit as bit 0, the next bit as bit =
1, etc.,=20
we can find the value of that bit through the expression =
</STRONG></P><PRE><STRONG> int Nth_bit =3D (1 << N) & =
M;
</STRONG></PRE>
<P><STRONG>In other words, we are shifting the number 1 left by N =
spaces,=20
and then performing a binary AND on the resulting number with our =
original=20
number, which will leave either a 1 in the Nth bit or a 0. So the =
first=20
thing we have to do is find out the distance between each pair of =
numbers=20
within the set of all numbers with B bits (0..2<SUP>B</SUP>-1).=20
</STRONG></P>
<P><STRONG>We also know that 0 can be one of the numbers in the =
set,=20
because if the minimum number in the set were N instead of 0, we =
could=20
subtract N from each number in the set and they would still be the =
same=20
distance apart. The limits on the problem are low enough that we =
can do a=20
DFS, and as soon as we hit a solution we can output it and exit.=20
</STRONG></P><PRE><STRONG>#include <stdio.h>
#include <stdlib.h>
#include <iostream.h>
#define MAX (1 << 8 + 1)
#define NMAX 65
#define BMAX 10
#define DMAX 10
int nums[NMAX], dist[MAX][MAX];
int N, B, D, maxval;
FILE *in, *out;
void findgroups(int cur, int start) {
int a, b, canuse;
char ch;
if (cur =3D=3D N) {
for (a =3D 0; a < cur; a++) {
if (a % 10)
fprintf(out, " ");
fprintf(out, "%d", nums[a]);
if (a % 10 =3D=3D 9 || a =3D=3D cur-1)
fprintf(out, "\n");
}
exit(0);
}
for (a =3D start; a < maxval; a++) {
canuse =3D 1;
for (b =3D 0; b < cur; b++)
if (dist[nums[b]][a] < D) {
canuse =3D 0;
break;
}
if (canuse) {
nums[cur] =3D a;
findgroups(cur+1, a+1);
}
}
}
int main() {
in =3D fopen("hamming.in", "r");
out =3D fopen("hamming.out", "w");
int a, b, c;
fscanf(in, "%d%d%d", &N, &B, &D);
maxval =3D (1 << B);
for (a =3D 0; a < maxval; a++)
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