查看: 4865|回复: 6

POJ-1006问题

[复制链接]

字体大小: 正常放大

maybe_madio

2 主题	3 听众	16 积分

升级 11.58%

该用户从未签到

电梯直达

1^#

发表于 2010-10-26 23:30 |只看该作者 |倒序浏览

|招呼Ta 关注Ta

POJ-1006

在本地VC6.0下测试，得出正确结果，为什么提交后总是Wrong Answer

现将代码贴出，请帮忙看看：

#include <iostream>
* H8 i' F0 Y! D8 j+ ` C) J
using namespace std;0 D7 Y$ ]- ?* O3 O: f
#define PH 0+ w U3 ?: K( z. P' T7 I
#define EM 14 n% a; i' u/ o* T, q0 t( a7 p) F
#define IN 2
# e- [5 f# e; s6 { |
int get_Top(int days, int kind);; D) m7 }7 ?0 [\\" ?. z1 f% N# z- K
int triple(int ptop, int etop, int itop, int cur);1 F3 o# A$ E/ _$ Y8 Q( i\\" b6 q
bool is_Integer(float n);
# r9 `8 q1 t\\" ^9 q% \
int main()
{3 S5 t# c7 f- p$ g7 l0 ?: h) S8 e8 w& }, _
int ptop, etop, itop;* z) ]+ P$ q# T- F
int pdays, edays, idays, curdays;' z6 A* G/ a0 h3 C: t
int count =0;) ]0 Y\\" U+ a8 m9 ~# }% z
( W, N$ v5 f/ g/ D7 H; l
while(cin >> pdays >> edays >> idays >> curdays)
4 u7 c& `/ e% j
{
! S N& U( t7 W* D6 U6 T% x
if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)6 g- t2 w3 y) [\\" d* Q/ E
break;
9 ^8 o4 d. Q) [5 N) d
count++;\\" j0 Q- b- Q# i4 [; f+ J
ptop = get_Top(pdays,PH);1 I$ @+ H) @2 T0 u* q
etop = get_Top(edays,EM);
: S( g8 ~2 c! M3 _
itop = get_Top(idays,IN);
7 P7 W/ F8 V! |# C, w& w* ~# U
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";, j8 u+ I2 K- y u
}/ O+ f\\" u8 }\\" q9 x0 I. W
; R0 f2 e& m, p; j2 G/ ]4 f0 m
return 0;
! e: @/ Y1 ^/ Y- J
}
# h4 j: ~/ B3 y
int get_Top(int days, int kind)- @4 x! v' _- i
{( g* w4 _' o9 L- D, d5 g8 A' G, {. f
switch(kind)
% c ~. ^\\" J! o' N1 c
{
( i9 G9 R' `5 Z* b
case 0:! a1 h3 {( v, ?
return days%23;
, W4 e8 T5 Y: {2 K\\" E- V5 w1 \
case 1:; ?; R* \6 N$ O0 x\\" M% F
return days%28;
. L1 z( d& `2 u
case 2:
, h C) C! {$ X; I3 Z* C
return days%33;
; T$ x$ o3 b+ p9 D5 P l1 _+ e* A
}: T; U7 o: V9 }( V# |2 A
return 0;+ ?$ f: B/ J5 ]4 @5 E
}/ j3 P: u# ~/ N( f I
9 i+ o! k5 M5 S8 H& c
int triple(int ptop, int etop, int itop, int cur)
' J+ {7 ~, Q4 r e7 D3 e' e
{
7 `& E1 }+ y2 y' c- |9 p
float x, y, z;6 y+ t! f4 V0 A/ D& L2 C
int m1 = etop-ptop-5,
. c! r4 y# z! C% D7 s$ M ]+ _
m2 = itop-etop-5;: _# ?3 h2 U\\" n& _) r
int temp = 0; [& p) O& P {2 n) `/ T
for(z=2; (itop+(z-1)*33) <= 21252; z++)
\\" ^7 M* {& v, n) \( v
{
; b* y0 |, h \6 c9 ~7 z) V
temp = m2 + 33*z;
! V* T7 d; [9 e6 H
y = (float)temp/28;
* ? `7 M6 T) I3 C
x = (float)(m1+temp)/23;
if(x>0 && y>0 && is_Integer(x) && is_Integer(y))
( D) D3 @, T* p
return itop + (z-1)*33 -cur;0 L# _5 X\\" ?* }5 g O& L& q# `$ }
}
1 g/ s8 N- J# o& X
return 0;
; M+ j0 T/ s9 H
}3 M) w' |- f4 V! R- n: c
2 _1 y* G% f) x0 a
bool is_Integer(float n)
0 _/ p, N, ^, R' {1 s
{
9 _3 b( B: i1 i0 M7 c/ D
if( (int)n - n ==0)
9 f& M2 P! |& C7 J \
return true;
6 n+ \( }! o0 g# ^/ o) p
else
, q1 {) B# K0 \& X I
return false;
; `) D$ K1 c: D& m6 Y% `+ n5 }0 X
}
/ k/ @5 J\\" e( U6 M% P' J6 v: \
2 Z9 Z* {2 k7 w8 X- z0 X7 g' g& p

#include <iostream> 
* H8 i' F0 Y! D8 j+ `  C) J
using namespace std;0 D7 Y$ ]- ?* O3 O: f

#define PH      0+ w  U3 ?: K( z. P' T7 I

#define EM      14 n% a; i' u/ o* T, q0 t( a7 p) F

#define IN      2 
# e- [5 f# e; s6 {  |
int get_Top(int days, int kind);; D) m7 }7 ?0 [\\" ?. z1 f% N# z- K

int triple(int ptop, int etop, int itop, int cur);1 F3 o# A$ E/ _$ Y8 Q( i\\" b6 q

bool is_Integer(float n); 
# r9 `8 q1 t\\" ^9 q% \
int main() 
1 _/ u& ~  [8 `4 u  ~9 o+ d2 }' ~9 A
{3 S5 t# c7 f- p$ g7 l0 ?: h) S8 e8 w& }, _

int ptop,  etop,  itop;* z) ]+ P$ q# T- F

int pdays, edays, idays, curdays;' z6 A* G/ a0 h3 C: t

int count =0;) ]0 Y\\" U+ a8 m9 ~# }% z

( W, N$ v5 f/ g/ D7 H; l

while(cin >> pdays >> edays >> idays >> curdays) 
4 u7 c& `/ e% j
 { 
! S  N& U( t7 W* D6 U6 T% x
  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)6 g- t2 w3 y) [\\" d* Q/ E

break; 
9 ^8 o4 d. Q) [5 N) d
  count++;\\" j0 Q- b- Q# i4 [; f+ J

ptop = get_Top(pdays,PH);1 I$ @+ H) @2 T0 u* q

etop = get_Top(edays,EM); 
: S( g8 ~2 c! M3 _
     itop = get_Top(idays,IN); 
7 P7 W/ F8 V! |# C, w& w* ~# U
  cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";, j8 u+ I2 K- y  u

}/ O+ f\\" u8 }\\" q9 x0 I. W

; R0 f2 e& m, p; j2 G/ ]4 f0 m

return 0; 
! e: @/ Y1 ^/ Y- J
} 
# h4 j: ~/ B3 y
int get_Top(int days, int kind)- @4 x! v' _- i

{( g* w4 _' o9 L- D, d5 g8 A' G, {. f

switch(kind) 
% c  ~. ^\\" J! o' N1 c
 { 
( i9 G9 R' `5 Z* b
 case 0:! a1 h3 {( v, ?

return days%23; 
, W4 e8 T5 Y: {2 K\\" E- V5 w1 \
 case 1:; ?; R* \6 N$ O0 x\\" M% F

return days%28; 
. L1 z( d& `2 u
 case 2: 
, h  C) C! {$ X; I3 Z* C
  return days%33; 
; T$ x$ o3 b+ p9 D5 P  l1 _+ e* A
 }: T; U7 o: V9 }( V# |2 A

return 0;+ ?$ f: B/ J5 ]4 @5 E

}/ j3 P: u# ~/ N( f  I

9 i+ o! k5 M5 S8 H& c

int triple(int ptop, int etop, int itop, int cur) 
' J+ {7 ~, Q4 r  e7 D3 e' e
{ 
7 `& E1 }+ y2 y' c- |9 p
 float x, y, z;6 y+ t! f4 V0 A/ D& L2 C

int m1   = etop-ptop-5, 
. c! r4 y# z! C% D7 s$ M  ]+ _
  m2   = itop-etop-5;: _# ?3 h2 U\\" n& _) r

int temp = 0;  [& p) O& P  {2 n) `/ T

for(z=2;  (itop+(z-1)*33) <= 21252;  z++) 
\\" ^7 M* {& v, n) \( v
 { 
; b* y0 |, h  \6 c9 ~7 z) V
  temp = m2 + 33*z; 
! V* T7 d; [9 e6 H
  y = (float)temp/28; 
* ?  `7 M6 T) I3 C
  x = (float)(m1+temp)/23; 
\\" [- }- y1 A( Y& v* `( g\\" u: Y) G9 F' f
  if(x>0 && y>0 && is_Integer(x) && is_Integer(y)) 
( D) D3 @, T* p
   return itop + (z-1)*33 -cur;0 L# _5 X\\" ?* }5 g  O& L& q# `$ }

} 
1 g/ s8 N- J# o& X
 return 0; 
; M+ j0 T/ s9 H
}3 M) w' |- f4 V! R- n: c

2 _1 y* G% f) x0 a

bool is_Integer(float n) 
0 _/ p, N, ^, R' {1 s
{ 
9 _3 b( B: i1 i0 M7 c/ D
 if( (int)n - n ==0) 
9 f& M2 P! |& C7 J  \
  return true; 
6 n+ \( }! o0 g# ^/ o) p
 else 
, q1 {) B# K0 \& X  I
  return false; 
; `) D$ K1 c: D& m6 Y% `+ n5 }0 X
} 
/ k/ @5 J\\" e( U6 M% P' J6 v: \
 
2 Z9 Z* {2 k7 w8 X- z0 X7 g' g& p

zan

转播0 淘帖0 分享0 收藏0 支持0 反对0 微信

使用道具举报

ultra1989

20 主题	65 听众	2071 积分

ACMer

升级 2.37%

TA的每日心情

	开心 2015-2-25 23:29

签到天数: 240 天

[LV.8]以坛为家I

群组: Linux推广

群组: Matlab讨论组

群组: 数学建模

2^#

发表于 2010-10-27 08:45 |只看该作者 |招呼Ta 关注Ta

本帖最后由 ultra1989 于 2010-10-27 08:54 编辑

用float可能会有精度问题用中国剩余定理求解
参考程序：

#include<stdio.h>
int main(){
int a,b,c,m,n,i,d=1;
while(scanf("%d%d%d%d",&a,&b,&c,&m)){

if(a==-1&&b==-1&&c==-1&&m==-1) break;

n=0;

i=(5544*a+14421*b+1288*c-m+21252)%21252;

if(!i)

i=21252;

printf("Case %d: the next triple peak occurs in %d days.\n",d++,i);

}

return 0;
}