📄 usaco 2_4_5 fractions to decimals 题解_leokan的blog.mht
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end;<BR> if=20
ans[visitp+1]<>ans[repp] then=20
=
write(ans[visitp+1]);<BR> =20
write(')');<BR> end;<BR> =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:=C0=FB=D3=C32=BA=CD5=C4=DC=CE=DE=D1=AD=
=BB=B7=D0=A1=CA=FD=B5=C4=CC=D8=D0=D4,=D4=DA=D3=E0=CA=FD=D6=D0=BD=F8=D0=D0=
=B4=A6=C0=ED</STRONG></P>
<P><STRONG>Remember long division? We know that the decimal =
expansion is=20
repeating when, after the decimal point, we see a remainder we've =
seen=20
before. The repeating part will be all the digits we've calculated =
since=20
the last time we saw that remainder. </STRONG></P>
<P><STRONG>We read in the input and print the integer part. Then =
we do=20
long division on the fractional part until we see a remainder more =
than=20
once or the remainder becomes zero. If we see a remainder more =
than once,=20
we're repeating, in which case we print the non-repeated and =
repeated part=20
appropriately. If the remainder becomes zero, we finished, in =
which case=20
we print the decimal expansion. When no digits of the decimal =
expansion=20
have been generated, the correct answer seems to be to print a =
zero.=20
</STRONG></P><PRE><STRONG>#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#define MAXDIGIT 100100
char dec[MAXDIGIT];
int lastrem[MAXDIGIT];
char buf[MAXDIGIT];
void
main(void)
{
FILE *fin, *fout;
int n, d, k, i, rem, len;
fin =3D fopen("fracdec.in", "r");
fout =3D fopen("fracdec.out", "w");
assert(fin !=3D NULL && fout !=3D NULL);
fscanf(fin, "%d %d", &n, &d);
sprintf(buf, "%d.", n/d);
/* long division keeping track of if we've seem a remainder before */
for(i=3D0; i<MAXDIGIT; i++)
lastrem[i] =3D -1;
rem =3D n % d;
for(i=3D0;; i++) {
if(rem =3D=3D 0) {
if(i =3D=3D 0)
sprintf(buf+strlen(buf), "0");
else
sprintf(buf+strlen(buf), "%s", dec);
break;
}
if(lastrem[rem] !=3D -1) {
k =3D lastrem[rem];
sprintf(buf+strlen(buf), "%.*s(%s)", k, dec, dec+k);
break;
}
lastrem[rem] =3D i;
n =3D rem * 10;
dec[i] =3D n/d + '0';
rem =3D n%d;
}
/* print buf 76 chars per line */
len =3D strlen(buf);
for(i=3D0; i<len; i+=3D76) {
fprintf(fout, "%.76s\n", buf+i);
}
exit(0);
}
</STRONG></PRE>
<P><STRONG>Here's a another, more elegant solution from Anatoly =
Preygel.=20
</STRONG></P>
<P><STRONG>Compute the number of digits before the repeat starts, =
and then=20
you don't even have to store the digits or remainders, making the =
program=20
use much less memory and go faster. We know that powers of 2 and 5 =
are the=20
only numbers which do not result in a repeat, so to find the =
number of=20
digits before the repeat, we just find the maximum of the =
differences=20
between the powers of 2 and 5 in the denominator and numerator =
(see code=20
snippet). Then we just use the first remainder, and output each =
digit as=20
we calculate it: </STRONG></P><PRE><STRONG>#include =
<iostream.h>
#include <fstream.h>
#include <math.h>
ofstream out("fracdec.out");
int colcount=3D0;
int numBeforeRepeat(int n, int d) {
int c2=3D0, c5=3D0;
if (n =3D=3D 0) return 1;
while (d%2=3D=3D0) { d/=3D2; c2++; }
while (d%5=3D=3D0) { d/=3D5; c5++; }
while (n%2=3D=3D0) { n/=3D2; c2--; } /* can go negative */
while (n%5=3D=3D0) { n/=3D5; c5--; } /* can go negative */
if (c2>c5)
if (c2>0) return c2;
else return 0;
else
if (c5>0) return c5;
else return 0;
}
void print (char c) {
if (colcount=3D=3D76) {
out<<endl;
colcount=3D0;
}
out<<c;
colcount++;
}
void print (int n) {
if (n>=3D10) print (n/10);
print ((char)('0'+(n%10)));
}
int main() {
int n, d;
ifstream in("fracdec.in");
in>>n>>d;
in.close();
print (n/d);
print ('.');
n=3Dn%d;
int m=3DnumBeforeRepeat(n,d);
for(int i=3D0;i<m;i++) {
n*=3D10;
print (n/d);
n%=3Dd;
}
int r=3Dn;
if(r!=3D0) {
print ('(');
do {
n*=3D10;
print (n/d);
n%=3Dd;
} while (n!=3Dr);
print (')');
}
out<<endl;
out.close();
exit (0);
}
</STRONG></PRE></DIV></TD></TR></TBODY></TABLE><BR>
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