POJ 1840 Eqs
解题思路：hash+枚举
0 0 0 0 0 输出 1e10，因此需要64位整数
某一系数为0，做一点优化
NULL
#include
<
iostream
>
using
namespace
std;
#define
PRIME 100003
#define
MAXN 1000001
#define
Cube(n) ((n)*(n)*(n))
int
p[MAXN],first[PRIME],next[MAXN];
int
main()
{
int
i,j,a,b,c,h,t,coef[
5
],iter
=
1000000
,zero
=
0
;
long
long
ans
=
0
;
for
(i
=
0
;i
<
PRIME;i
++
)first[i]
=-
1
;
for
(i
=
0
;i
<
5
;i
++
)scanf(
"
%d
"
,
&
coef[i]);
for
(i
=
2
;i
>=
0
;i
--
)
if
(
!
coef[i]){iter
/=
100
;zero
++
;}
else
if
(zero){coef[i
+
zero]
=
coef[i];coef[i]
=
0
;}
for
(j
=
i
=
0
;i
<
iter;i
++
,j
=
i)
{
if
(
!
coef[
0
])a
=
0
;
else
{a
=
j
/
10000
-
50
;
if
(a
>=
0
)a
++
;j
%=
10000
;}
if
(
!
coef[
1
])b
=
0
;
else
{b
=
j
/
100
-
50
;
if
(b
>=
0
)b
++
;j
%=
100
;}
if
(
!
coef[
2
])c
=
0
;
else
{c
=
j
-
50
;
if
(c
>=
0
)c
++
;}
h
=
coef[
0
]
*
Cube(a)
+
coef[
1
]
*
Cube(b)
+
coef[
2
]
*
Cube(c);
p[i]
=
h;
h
%=
PRIME;
if
(h
<
0
)h
+=
PRIME;
next[i]
=
first[h];first[h]
=
i;
}
for
(iter
=
1e4,i
=
3
;i
<
5
;i
++
)
if
(
!
coef[i])iter
/=
100
,zero
++
;
if
(
!
coef[
4
]){coef[
4
]
=
coef[
3
];coef[
3
]
=
0
;}
for
(j
=
i
=
0
;i
<
iter;i
++
,j
=
i)
{
if
(
!
coef[
3
])a
=
0
;
else
{a
=
j
/
100
-
50
;
if
(a
>=
0
)a
++
;}
if
(
!
coef[
4
])b
=
0
;
else
{b
=
j
%
100
-
50
;
if
(b
>=
0
)b
++
;}
c
=-
coef[
3
]
*
Cube(a)
-
coef[
4
]
*
Cube(b);
h
=
c
%
PRIME;
if
(h
<
0
)h
+=
PRIME;
for
(t
=
first[h];t
!=-
1
;t
=
next[t])
if
(p[t]
==
c)ans
++
;
}
while
(zero
--
)ans
*=
100
;
printf(
"
%lld\n
"
,ans);
return
0
;
}