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POJ-1006问题

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maybe_madio

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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>* {3 m- k* f1 M! W' C
using namespace std;! w( _0 k$ O- o\\" N5 `- b
#define PH 0
\\" j. }) O4 L! ^! T
#define EM 1
6 }# a: {) }9 w2 Q4 R& g# `
#define IN 2: f) N0 N+ p$ S% l& ^7 z# }$ k
int get_Top(int days, int kind);9 ?' G4 Q8 L; O\\" d: e( b' n
int triple(int ptop, int etop, int itop, int cur);
% U\\" T9 @7 t. f7 m+ w
bool is_Integer(float n);5 u1 w' Q) G# D$ s% u
int main()
+ ^' A6 D0 D: c H
{1 e0 d( V1 G1 Q
int ptop, etop, itop;
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int pdays, edays, idays, curdays;$ Z y* L8 A6 |- a8 h, J% b
int count =0;' O+ u( p3 q# \8 j/ P+ v7 q
. b# M: a7 K1 `, H# I) v
while(cin >> pdays >> edays >> idays >> curdays)$ w' s( c1 l C4 U/ {; Y( b8 x w7 V
{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)7 P$ @4 F! S& b' o
break;1 M% d8 C5 d; E2 K, f e5 Y
count++;
) Y: E5 s3 c+ } M2 ?& n. J
ptop = get_Top(pdays,PH);4 w% O, E) w: K% u- |
etop = get_Top(edays,EM);# x- Z0 N' @+ D$ Y5 A
itop = get_Top(idays,IN);
t; I- p n/ d+ I# m$ n
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";2 O, v& j1 m1 s* n\\" j
}
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return 0;4 O7 Z& G) w$ F) K4 B( U U
}) R0 |; f% g, I9 X' ~
int get_Top(int days, int kind)
; x p# m# ]1 f' ]
{
) Q! Z5 i3 {% M- y& `: d4 v
switch(kind): L8 ^5 x; x& M- y0 \8 { c* ]
{1 Z8 f: v$ M/ R/ G# m' g+ {
case 0:0 W3 a7 n# J7 z. ?( T9 n* X# ]4 h
return days%23;; d6 ?2 s: f S7 N, P. t% S; D
case 1:% k3 k- l% `2 B# b3 ^5 Q' j6 ]5 Q( e! a N
return days%28;
& m7 @2 u4 ~: ? Z g% P
case 2:
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return days%33;% T& d+ i7 ^& u6 s4 N% w( W5 d
}
# Z: N! P5 m2 }$ K
return 0;
4 @3 k# h# X4 v' |$ H) d( U
}
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int triple(int ptop, int etop, int itop, int cur)8 l3 A( r% k( h' C
{
% F! U. F. k2 z% _/ ~! M7 I+ {
float x, y, z;* ]8 o _ U! u' v5 a+ c
int m1 = etop-ptop-5,
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m2 = itop-etop-5;
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int temp = 0;
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for(z=2; (itop+(z-1)*33) <= 21252; z++)
\\" F# Y# Z: \/ W4 ?- @. K9 h
{
% c# m+ L9 a5 p* g9 I) r
temp = m2 + 33*z;
: o! C( Q4 N: Q% i) a
y = (float)temp/28;
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x = (float)(m1+temp)/23;
. c( V1 h0 |9 u+ t$ y
if(x>0 && y>0 && is_Integer(x) && is_Integer(y))2 B3 \* Y/ O' E\\" H8 g! M; ~- a% u* r
return itop + (z-1)*33 -cur;0 |3 z9 V$ x: v0 |8 [2 ^
}
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return 0;
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}
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bool is_Integer(float n)
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{$ x\\" n6 L: C) E; Q: E$ V$ ?
if( (int)n - n ==0)# W2 H. s8 n$ \7 p, G( r
return true;6 e2 i/ D9 B% E. G
else
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return false;
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}$ ?3 d7 U! w& E$ j; Q
, M\\" ]* e/ t- ~; Z* I

#include <iostream>* {3 m- k* f1 M! W' C

using namespace std;! w( _0 k$ O- o\\" N5 `- b

#define PH      0 
\\" j. }) O4 L! ^! T
#define EM      1 
6 }# a: {) }9 w2 Q4 R& g# `
#define IN      2: f) N0 N+ p$ S% l& ^7 z# }$ k

int get_Top(int days, int kind);9 ?' G4 Q8 L; O\\" d: e( b' n

int triple(int ptop, int etop, int itop, int cur); 
% U\\" T9 @7 t. f7 m+ w
bool is_Integer(float n);5 u1 w' Q) G# D$ s% u

int main() 
+ ^' A6 D0 D: c  H
{1 e0 d( V1 G1 Q

int ptop,  etop,  itop; 
4 N) A& k! \2 f3 g4 R/ A. m
 int pdays, edays, idays, curdays;$ Z  y* L8 A6 |- a8 h, J% b

int count =0;' O+ u( p3 q# \8 j/ P+ v7 q

. b# M: a7 K1 `, H# I) v
 while(cin >> pdays >> edays >> idays >> curdays)$ w' s( c1 l  C4 U/ {; Y( b8 x  w7 V

{ 
2 j* p1 r1 d) K5 t1 M
  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)7 P$ @4 F! S& b' o

break;1 M% d8 C5 d; E2 K, f  e5 Y

count++; 
) Y: E5 s3 c+ }  M2 ?& n. J
  ptop = get_Top(pdays,PH);4 w% O, E) w: K% u- |

etop = get_Top(edays,EM);# x- Z0 N' @+ D$ Y5 A

itop = get_Top(idays,IN); 
  t; I- p  n/ d+ I# m$ n
  cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";2 O, v& j1 m1 s* n\\" j

} 
! a4 `! T* F8 a$ F\\" C1 r& }
 
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 return 0;4 O7 Z& G) w$ F) K4 B( U  U

}) R0 |; f% g, I9 X' ~

int get_Top(int days, int kind) 
; x  p# m# ]1 f' ]
{ 
) Q! Z5 i3 {% M- y& `: d4 v
 switch(kind): L8 ^5 x; x& M- y0 \8 {  c* ]

{1 Z8 f: v$ M/ R/ G# m' g+ {

case 0:0 W3 a7 n# J7 z. ?( T9 n* X# ]4 h

return days%23;; d6 ?2 s: f  S7 N, P. t% S; D

case 1:% k3 k- l% `2 B# b3 ^5 Q' j6 ]5 Q( e! a  N

return days%28; 
& m7 @2 u4 ~: ?  Z  g% P
 case 2: 
8 ?4 w. I' K) P: ?# f# S* q! X0 j
  return days%33;% T& d+ i7 ^& u6 s4 N% w( W5 d

} 
# Z: N! P5 m2 }$ K
 return 0; 
4 @3 k# h# X4 v' |$ H) d( U
} 
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1 ?- T% x+ M  E# p+ n  q7 X

int triple(int ptop, int etop, int itop, int cur)8 l3 A( r% k( h' C

{ 
% F! U. F. k2 z% _/ ~! M7 I+ {
 float x, y, z;* ]8 o  _  U! u' v5 a+ c

int m1   = etop-ptop-5, 
$ N$ b5 g+ `# j
  m2   = itop-etop-5; 
& p5 G- i: I5 ^, w% O9 z0 a! ~7 ^
 int temp = 0; 
7 f\\" @  ~1 J4 S- q
 for(z=2;  (itop+(z-1)*33) <= 21252;  z++) 
\\" F# Y# Z: \/ W4 ?- @. K9 h
 { 
% c# m+ L9 a5 p* g9 I) r
  temp = m2 + 33*z; 
: o! C( Q4 N: Q% i) a
  y = (float)temp/28; 
3 {1 Y5 `& d/ o( c- y6 Z, U
  x = (float)(m1+temp)/23; 
. c( V1 h0 |9 u+ t$ y
  if(x>0 && y>0 && is_Integer(x) && is_Integer(y))2 B3 \* Y/ O' E\\" H8 g! M; ~- a% u* r

return itop + (z-1)*33 -cur;0 |3 z9 V$ x: v0 |8 [2 ^

} 
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 return 0; 
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} 
% h& S, `  m) o+ x: S8 K) X
 
/ Q4 W; t) F8 y# x% O* k9 E
bool is_Integer(float n) 
& }' M: V8 e+ a( `
{$ x\\" n6 L: C) E; Q: E$ V$ ?

if( (int)n - n ==0)# W2 H. s8 n$ \7 p, G( r

return true;6 e2 i/ D9 B% E. G

else 
% c\\" \/ T6 `( e' |' P- ~
  return false; 
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}$ ?3 d7 U! w& E$ j; Q

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ultra1989

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	开心 2015-2-25 23:29

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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;
}