📄 usaco 3_1_5 contact 题解_leokan的blog.mht
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print(ans[i]);<BR> &=
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init;<BR> work;<BR>end.</STRONG></P>
<P><STRONG><BR></STRONG></P><STRONG>
<HR>
</STRONG>
<P><STRONG>USACO=B5=C4=B7=D6=CE=F6......=CE=DE=D3=EF</STRONG></P>
<P><STRONG>For this problem, we keep track of every bit sequence =
we see.=20
We could use the bit sequence itself as an index into a table of=20
frequencies, but that would not distinguish between the 2-bit =
sequence=20
"10" and the 4-bit sequence "0010". To solve this, we always add a =
1 to=20
the beginning of the number, so "10" becomes "110" and "0010" =
becomes=20
"10010". </STRONG></P>
<P><STRONG>After reading the entire bit string, we sort the =
frequency=20
table and walk through it to print out the top sequences. =
</STRONG></P><PRE><STRONG>#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#define MAXBITS 12
#define MAXSEQ (1<<(MAXBITS+1))
typedef struct Seq Seq;
struct Seq {
unsigned bits;
int count;
};
Seq seq[MAXSEQ];
/* increment the count for the n-bit sequence "bits" */
void
addseq(unsigned bits, int n)
{
bits &=3D (1<<n)-1;
bits |=3D 1<<n;
assert(seq[bits].bits =3D=3D bits);
seq[bits].count++;
}
/* print the bit sequence, decoding the 1<<n stuff */
/* recurse to print the bits most significant bit first */
void
printbits(FILE *fout, unsigned bits)
{
assert(bits >=3D 1);
if(bits =3D=3D 1) /* zero-bit sequence */
return;
printbits(fout, bits>>1);
fprintf(fout, "%d", bits&1);
}
int
seqcmp(const void *va, const void *vb)
{
Seq *a, *b;
a =3D (Seq*)va;
b =3D (Seq*)vb;
/* big counts first */
if(a->count < b->count)
return 1;
if(a->count > b->count)
return -1;
/* same count: small numbers first */
if(a->bits < b->bits)
return -1;
if(a->bits > b->bits)
return 1;
return 0;
}
void
main(void)
{
FILE *fin, *fout;
int i, a, b, n, nbit, c, j, k;
unsigned bit;
char *sep;
fin =3D fopen("contact.in", "r");
fout =3D fopen("contact.out", "w");
assert(fin !=3D NULL && fout !=3D NULL);
nbit =3D 0;
bit =3D 0;
for(i=3D0; i<=3DMAXBITS; i++)
for(j=3D0; j<(1<<i); j++)
seq[(1<<i) | j].bits =3D (1<<i) | j;
fscanf(fin, "%d %d %d", &a, &b, &n);
while((c =3D getc(fin)) !=3D EOF) {
if(c !=3D '0' && c !=3D '1')
continue;
bit <<=3D 1;
if(c =3D=3D '1')
bit |=3D 1;
if(nbit < b)
nbit++;
for(i=3Da; i<=3Dnbit; i++)
addseq(bit, i);
}
qsort(seq, MAXSEQ, sizeof(Seq), seqcmp);
/* print top n frequencies for number of bits between a and b */
j =3D 0;
for(i=3D0; i<n && j < MAXSEQ; i++) {
if(seq[j].count =3D=3D 0)
break;
c =3D seq[j].count;
fprintf(fout, "%d\n", c);
/* print all entries with frequency c */
sep =3D "";
for(k=3D0; seq[j].count =3D=3D c; j++, k++) {
fprintf(fout, sep);
printbits(fout, seq[j].bits);
if(k%6 =3D=3D 5)
sep =3D "\n";
else
sep =3D " ";
}
fprintf(fout, "\n");
}
exit(0);
}</STRONG></PRE></DIV></TD></TR></TBODY></TABLE><BR>
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