查看: 4813|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>
/ p- ?# A: h6 B+ L2 l d9 b
using namespace std;+ ]) G! W8 q! C$ b/ I4 I% l0 {
#define PH 01 `\\" q. Y3 L\\" K, p
#define EM 1
& v* h* |6 j! R3 `\\" ]1 [; @. B
#define IN 2
& r# w, @: `& C$ O5 W; n l
int get_Top(int days, int kind);' q L* P M! K0 t/ }
int triple(int ptop, int etop, int itop, int cur);& s+ u4 ]\\" Y\\" C1 f9 U1 i
bool is_Integer(float n);% {( y2 i' S) n1 g! i( }\\" g; m
int main()
4 T- h* _. P$ N( S
{
5 V( D7 D+ |. z0 R) D% Z5 o
int ptop, etop, itop; a. V7 e- F. z7 d' m' C) U) X7 g6 N
int pdays, edays, idays, curdays;
Z# m& S5 x: c4 F* y
int count =0;
2 _2 R. Q$ q+ x( ~- }. ~. b3 c
! ~. r2 C; G; X' S9 `
while(cin >> pdays >> edays >> idays >> curdays)
: N3 j' O* ]3 D; R, G; S
{( h* j( v0 E4 ~. r
if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
; ]. T4 N, E& S: G* J
break;8 o+ G8 q2 s; J/ } X
count++;0 q+ @& T+ e- u5 z8 E( c; ^
ptop = get_Top(pdays,PH);
9 i: u, d/ g/ a8 h\\" D9 v
etop = get_Top(edays,EM);
3 E# Q( ?, ~% i\\" `\\" h2 I
itop = get_Top(idays,IN);8 ], `! w4 Y! X9 J7 `( I; H; T
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";
; j. p: B! D F' l- Y6 R
}8 r3 T2 r/ I2 g6 b8 h2 I
4 u3 W. k7 r\\" O0 H& v; [- I% s
return 0;
- G\\" k$ o' A l
}; n/ Y5 q1 _5 d( [0 p' x
int get_Top(int days, int kind)
\\" c4 H% F* x& y8 C\\" S
{
# Z3 e6 v3 }2 l/ G6 h1 c7 q
switch(kind)
- \$ _: G8 W2 c! K1 k( g
{
' u4 v5 @% q$ W6 g5 J
case 0:5 d% o* }, x2 M7 f; u* T
return days%23;
\\" k! Y0 |- h2 {, l, m9 Z, P
case 1:5 B: e& F! N* D1 z$ U$ I3 K
return days%28;$ Z, f+ b0 g) } L- `
case 2:' C\\" d- j6 F% q, M
return days%33;6 F( o1 o0 L* Q+ H! w! q1 Q
}0 G( X7 S- }1 l, m# z# e6 q
return 0;
\\" l1 i0 ]6 L8 o3 u$ r
}, S z# _ w7 @. e
6 `( b* M\\" h/ L. y7 n4 B+ i
int triple(int ptop, int etop, int itop, int cur); I6 [9 Y$ P# I6 n$ b7 W! [5 r
{
' O, N( t. Y. r6 ~. K
float x, y, z;
& L\\" S; `+ S2 r\\" Q% p
int m1 = etop-ptop-5,/ M9 ], x/ m5 S+ Y4 z
m2 = itop-etop-5;# C# o% `: G/ t Y\\" E
int temp = 0;5 A9 H0 L4 F\\" D- \: u9 n
for(z=2; (itop+(z-1)*33) <= 21252; z++)
) E* I) s9 q4 @+ l
{* P4 s- R4 g+ u' U4 x\\" W- W5 D
temp = m2 + 33*z;7 u6 W/ c- z% s9 s\\" h# L
y = (float)temp/28;7 X# V5 A( ~3 C( h' M
x = (float)(m1+temp)/23;$ K9 [4 T, P9 z( P# [- T$ L
if(x>0 && y>0 && is_Integer(x) && is_Integer(y))
% {7 y# c, `3 s4 d
return itop + (z-1)*33 -cur;
% g1 O* S; c$ p
}9 M: f7 d& j; ?& r' Y& q1 O: t
return 0;
$ o- I3 h4 I) l\\" [/ }\\" X
}
\\" S* [% Y' t( _4 _7 ^
7 }, H6 s: Q5 ]! t\\" |: N7 g9 y9 [
bool is_Integer(float n)
1 [# O+ U4 N) y4 K
{
% e\\" T3 ?) x C\\" {! \: K8 b
if( (int)n - n ==0)9 G M4 o) A% b, e' l2 [
return true;. ]( `* [( ~) [- f! u% d
else. F% P! @1 N. V9 z. @
return false;
6 r4 w. q7 I. d6 S
}
4 v0 j, y. y; h) n7 J
6 C2 g2 i9 o. Z2 C) E& M

#include <iostream> 
/ p- ?# A: h6 B+ L2 l  d9 b
using namespace std;+ ]) G! W8 q! C$ b/ I4 I% l0 {

#define PH      01 `\\" q. Y3 L\\" K, p

#define EM      1 
& v* h* |6 j! R3 `\\" ]1 [; @. B
#define IN      2 
& r# w, @: `& C$ O5 W; n  l
int get_Top(int days, int kind);' q  L* P  M! K0 t/ }

int triple(int ptop, int etop, int itop, int cur);& s+ u4 ]\\" Y\\" C1 f9 U1 i

bool is_Integer(float n);% {( y2 i' S) n1 g! i( }\\" g; m

int main() 
4 T- h* _. P$ N( S
{ 
5 V( D7 D+ |. z0 R) D% Z5 o
 int ptop,  etop,  itop;  a. V7 e- F. z7 d' m' C) U) X7 g6 N

int pdays, edays, idays, curdays; 
  Z# m& S5 x: c4 F* y
 int count =0; 
2 _2 R. Q$ q+ x( ~- }. ~. b3 c
 
! ~. r2 C; G; X' S9 `
 while(cin >> pdays >> edays >> idays >> curdays) 
: N3 j' O* ]3 D; R, G; S
 {( h* j( v0 E4 ~. r

if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
; ]. T4 N, E& S: G* J
   break;8 o+ G8 q2 s; J/ }  X

count++;0 q+ @& T+ e- u5 z8 E( c; ^

ptop = get_Top(pdays,PH); 
9 i: u, d/ g/ a8 h\\" D9 v
     etop = get_Top(edays,EM); 
3 E# Q( ?, ~% i\\" `\\" h2 I
     itop = get_Top(idays,IN);8 ], `! w4 Y! X9 J7 `( I; H; T

cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days."; 
; j. p: B! D  F' l- Y6 R
 }8 r3 T2 r/ I2 g6 b8 h2 I

4 u3 W. k7 r\\" O0 H& v; [- I% s

return 0; 
- G\\" k$ o' A  l
}; n/ Y5 q1 _5 d( [0 p' x

int get_Top(int days, int kind) 
\\" c4 H% F* x& y8 C\\" S
{ 
# Z3 e6 v3 }2 l/ G6 h1 c7 q
 switch(kind) 
- \$ _: G8 W2 c! K1 k( g
 { 
' u4 v5 @% q$ W6 g5 J
 case 0:5 d% o* }, x2 M7 f; u* T

return days%23; 
\\" k! Y0 |- h2 {, l, m9 Z, P
 case 1:5 B: e& F! N* D1 z$ U$ I3 K

return days%28;$ Z, f+ b0 g) }  L- `

case 2:' C\\" d- j6 F% q, M

return days%33;6 F( o1 o0 L* Q+ H! w! q1 Q

}0 G( X7 S- }1 l, m# z# e6 q

return 0; 
\\" l1 i0 ]6 L8 o3 u$ r
}, S  z# _  w7 @. e

6 `( b* M\\" h/ L. y7 n4 B+ i
int triple(int ptop, int etop, int itop, int cur); I6 [9 Y$ P# I6 n$ b7 W! [5 r

{ 
' O, N( t. Y. r6 ~. K
 float x, y, z; 
& L\\" S; `+ S2 r\\" Q% p
 int m1   = etop-ptop-5,/ M9 ], x/ m5 S+ Y4 z

m2   = itop-etop-5;# C# o% `: G/ t  Y\\" E

int temp = 0;5 A9 H0 L4 F\\" D- \: u9 n

for(z=2;  (itop+(z-1)*33) <= 21252;  z++) 
) E* I) s9 q4 @+ l
 {* P4 s- R4 g+ u' U4 x\\" W- W5 D

temp = m2 + 33*z;7 u6 W/ c- z% s9 s\\" h# L

y = (float)temp/28;7 X# V5 A( ~3 C( h' M

x = (float)(m1+temp)/23;$ K9 [4 T, P9 z( P# [- T$ L

if(x>0 && y>0 && is_Integer(x) && is_Integer(y)) 
% {7 y# c, `3 s4 d
   return itop + (z-1)*33 -cur; 
% g1 O* S; c$ p
 }9 M: f7 d& j; ?& r' Y& q1 O: t

return 0; 
$ o- I3 h4 I) l\\" [/ }\\" X
} 
\\" S* [% Y' t( _4 _7 ^
7 }, H6 s: Q5 ]! t\\" |: N7 g9 y9 [

bool is_Integer(float n) 
1 [# O+ U4 N) y4 K
{ 
% e\\" T3 ?) x  C\\" {! \: K8 b
 if( (int)n - n ==0)9 G  M4 o) A% b, e' l2 [

return true;. ]( `* [( ~) [- f! u% d

else. F% P! @1 N. V9 z. @

return false; 
6 r4 w. q7 I. d6 S
} 
4 v0 j, y. y; h) n7 J
 
6 C2 g2 i9 o. Z2 C) E& M

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;
}