📄 usaco 4_1_1 beef mcnuggets 题解_leokan的blog.mht
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int pos; /* count position */
int last; /* last impossible count */
if ((fin =3D fopen("nuggets.in", "r")) =3D=3D NULL)
{
perror ("fopen fin");
exit(1);
}
if ((fout =3D fopen("nuggets.out", "w")) =3D=3D NULL)
{
perror ("fopen fout");
exit(1);
}
/* read in data */
fscanf (fin, "%d", &nsize);
for (lv =3D 0; lv < nsize; lv++) fscanf (fin, "%d", =
&sizes[lv]);
/* ensure gcd =3D 1 */
lv2 =3D sizes[0];
for (lv =3D 1; lv < nsize; lv++)
lv2 =3D gcd(sizes[lv], lv2);
if (lv2 !=3D 1)
{ /* gcd !=3D 1, no bound on size that cannot be purchased */
fprintf (fout, "0\n");
return 0;
}
/* we can do 0 */
cando[0] =3D 1;
lv =3D pos =3D 0;
last =3D 0;
while (pos < 2000000000)
{ /* bound as stated */
/* if last 256 were all possible, we are done */
if (pos - last > 256) break;=20
if (!cando[lv])=20
last =3D pos; /* this isn't possible, update last impossible */
else=20
{ /* this is possible */
cando[lv] =3D 0; /* mark pos+256 as impossible */
/* mark pos + size as possible for each package size */
for (lv2 =3D 0; lv2 < nsize; lv2++)
cando[(lv+sizes[lv2])%256] =3D 1;
}
/* update lv & pos */
lv =3D (++pos) % 256;=20
}
if (pos >=3D 2000000000) last =3D 0; /* shouldn't occur */
fprintf (fout, "%i\n", last);
return 0;
}
</STRONG></PRE>
<H2>Alex Schwendner's Comments</H2>
<P><STRONG>Given two relatively prime numbers N and M, the largest =
number=20
that you cannot make is NM - M - N, that is, the product minus the =
sum. We=20
do not have two numbers; however, even if we were using only two =
of them=20
the answer could not exceed 256 * 256 - 256 - 256 =3D 65024 (much =
less then=20
the 2,000,000,000 that we were guaranteed). It is therefore =
reasonable to=20
have an array of 65024 booleans and work on all of them (If =
cando[x] then=20
cando[x + sizes[p]]). If there is some number above 65024 that =
cannot be=20
made then we know that there is no bound to the largest number. =
Because=20
cando[0] is set to false, if every number can be made then the =
program=20
will output '0' automatically. This program is shorter and easier =
to code,=20
and although it is somewhat less efficient, it is easily able to =
solve the=20
problem in the time limit. </STRONG></P><PRE><STRONG>#include =
<fstream.h>
#include <string.h>
int=20
main ()
{
int n;
int sizes[10];
ifstream filein ("nuggets.in");
filein >> n;
for (int in =3D 0; in < n; ++in) {
filein >> sizes[in];
}
filein.close ();
bool cando[67000];
memset (cando, 0, 67000);
for (int loop =3D 0; loop < n; ++loop) {
cando[sizes[loop]] =3D true;
for (int loop2 =3D 0; loop2 < 66700; ++loop2) {
if (cando[loop2]) {
cando[loop2 + sizes[loop]] =3D true;
}
}
}
ofstream fileout ("nuggets.out");
for (int out =3D 66500; out >=3D 0; --out) {
if (!cando[out]) {
if (out < 66000) {
fileout << out << endl;
break;
}
else {
fileout << 0 << endl;
break;
}
}
}
fileout.close ();
return (0);
}</STRONG></PRE></DIV></TD></TR></TBODY></TABLE><BR>
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