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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>* W; F1 i) \9 x\\" `% _
using namespace std;0 X; y3 r1 A/ b' C6 o) j
#define PH 04 x6 a, Z& j4 D/ C) {
#define EM 17 n2 o+ K+ X% y. z
#define IN 2- x8 h* d7 W9 f! p, P( F$ ~& T
int get_Top(int days, int kind);3 F3 c2 @5 Q\\" O
int triple(int ptop, int etop, int itop, int cur);
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bool is_Integer(float n);
% }; s4 b' Z7 ~6 d' E& H
int main()
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{
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int ptop, etop, itop;
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int pdays, edays, idays, curdays;) ~# Y5 \\\" n7 I/ a\\" h, o
int count =0;* r7 V2 H. c5 Z8 x3 Q
+ A1 T, \: z: L' `# `4 ~& A+ q: T: G
while(cin >> pdays >> edays >> idays >> curdays)+ L O& {) H\\" R: O/ r( e' l
{* ?\\" n+ c9 E9 w# b! |' W2 B
if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;( V! R( A$ D) h: ~3 w' z
count++;
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ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);
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itop = get_Top(idays,IN); a D. D+ ?* g: W3 L: L
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";
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}
: { L9 e$ ]: W4 i F
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return 0;
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}
+ U\\" ~5 j: e4 i, v0 r
int get_Top(int days, int kind)
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{
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switch(kind)
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{; n* L6 B% \/ C4 F
case 0:
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return days%23;1 R5 `! g; w$ q
case 1:
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return days%28;5 w0 H+ h: [1 v( I' v8 y1 ?
case 2:0 a\\" _0 X. Z; @ n7 i7 W/ o
return days%33;% F* U\\" x' `1 L
}* v h/ `* {- M
return 0;; F5 [7 C# R! \: g Z) [+ V
}* ]9 w( r: q9 H6 ]) }, T: A
# \; U+ Y) G. {6 \
int triple(int ptop, int etop, int itop, int cur)6 N! m% e8 S! ?1 f; q! p3 A! ]
{
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float x, y, z;
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int m1 = etop-ptop-5,
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m2 = itop-etop-5;
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int temp = 0;
' s8 i\\" I# C2 A( I1 G
for(z=2; (itop+(z-1)*33) <= 21252; z++)
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{
- T- G- e$ }( \
temp = m2 + 33*z;
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y = (float)temp/28;1 P* h! @0 {! O9 n' S\\" `2 U( b
x = (float)(m1+temp)/23;
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if(x>0 && y>0 && is_Integer(x) && is_Integer(y))7 K* N( F6 K+ @. E, Y% Y. o8 z' a
return itop + (z-1)*33 -cur;
% _9 g. ], @7 p. z5 f6 @* e2 K
}
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return 0;5 y$ R' h& n) V( G- ~
}\\" R: v) S# I# C. H' [
; s6 }0 }2 S, a1 T/ s
bool is_Integer(float n)3 P) Z: x4 f( P4 J3 C5 P
{+ B\\" T8 I3 t i3 v0 s0 M$ R
if( (int)n - n ==0)
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return true;/ z9 [( b+ \- a3 E
else
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return false;
# I5 v\\" R1 ~- G6 q* l7 i
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#include <iostream>* W; F1 i) \9 x\\" `% _

using namespace std;0 X; y3 r1 A/ b' C6 o) j

#define PH      04 x6 a, Z& j4 D/ C) {

#define EM      17 n2 o+ K+ X% y. z

#define IN      2- x8 h* d7 W9 f! p, P( F$ ~& T

int get_Top(int days, int kind);3 F3 c2 @5 Q\\" O

int triple(int ptop, int etop, int itop, int cur); 
8 x8 I1 u5 ^5 x& E. Y
bool is_Integer(float n); 
% }; s4 b' Z7 ~6 d' E& H
int main() 
7 b% Q$ [) g6 l7 H: j
{ 
8 }% K8 l; Z. S7 ?
 int ptop,  etop,  itop; 
+ B- |6 L- w& g- a1 Q- i3 P* l. M- @
 int pdays, edays, idays, curdays;) ~# Y5 \\\" n7 I/ a\\" h, o

int count =0;* r7 V2 H. c5 Z8 x3 Q

+ A1 T, \: z: L' `# `4 ~& A+ q: T: G
 while(cin >> pdays >> edays >> idays >> curdays)+ L  O& {) H\\" R: O/ r( e' l

{* ?\\" n+ c9 E9 w# b! |' W2 B

if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
) ]& o) @% u9 g
   break;( V! R( A$ D) h: ~3 w' z

count++; 
7 V1 e\\" u7 i- g0 G6 O6 w) n! Z2 J
  ptop = get_Top(pdays,PH); 
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     etop = get_Top(edays,EM); 
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     itop = get_Top(idays,IN);  a  D. D+ ?* g: W3 L: L

cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days."; 
! Q  n4 o: K* [& I, U3 J; O6 l4 n
 } 
: {  L9 e$ ]: W4 i  F
 ! ]* M. i& B( P# n# p2 B/ F

return 0; 
7 m6 R2 ~' N: E$ h1 Z4 L2 k* l; p+ e
} 
+ U\\" ~5 j: e4 i, v0 r
int get_Top(int days, int kind) 
. g7 n: \\\" P5 |# T  `! w$ n
{ 
8 X( f2 \! d1 e. q- h8 E8 t
 switch(kind) 
4 M; E1 U- Z- e- C
 {; n* L6 B% \/ C4 F

case 0: 
3 L7 E8 e4 A% z$ [0 U
  return days%23;1 R5 `! g; w$ q

case 1: 
' f1 v\\" G5 M8 j/ `' R
  return days%28;5 w0 H+ h: [1 v( I' v8 y1 ?

case 2:0 a\\" _0 X. Z; @  n7 i7 W/ o

return days%33;% F* U\\" x' `1 L

}* v  h/ `* {- M

return 0;; F5 [7 C# R! \: g  Z) [+ V

}* ]9 w( r: q9 H6 ]) }, T: A

# \; U+ Y) G. {6 \
int triple(int ptop, int etop, int itop, int cur)6 N! m% e8 S! ?1 f; q! p3 A! ]

{ 
5 d1 _7 [3 O9 u& H. k4 n
 float x, y, z; 
9 L' Z1 ~% q, f% C* e
 int m1   = etop-ptop-5, 
' Q( z( f5 E8 n( S% D5 b0 n
  m2   = itop-etop-5; 
\\" x2 B* J9 z& v( q
 int temp = 0; 
' s8 i\\" I# C2 A( I1 G
 for(z=2;  (itop+(z-1)*33) <= 21252;  z++) 
+ b0 V6 W- o: s( H5 I2 A! ]
 { 
- T- G- e$ }( \
  temp = m2 + 33*z; 
& z8 t3 |0 x  d5 ?\\" v0 s; x  c0 C
  y = (float)temp/28;1 P* h! @0 {! O9 n' S\\" `2 U( b

x = (float)(m1+temp)/23; 
- n+ X9 G' @# [% @  i0 j# v
  if(x>0 && y>0 && is_Integer(x) && is_Integer(y))7 K* N( F6 K+ @. E, Y% Y. o8 z' a

return itop + (z-1)*33 -cur; 
% _9 g. ], @7 p. z5 f6 @* e2 K
 } 
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 return 0;5 y$ R' h& n) V( G- ~

}\\" R: v) S# I# C. H' [

; s6 }0 }2 S, a1 T/ s

bool is_Integer(float n)3 P) Z: x4 f( P4 J3 C5 P

{+ B\\" T8 I3 t  i3 v0 s0 M$ R

if( (int)n - n ==0) 
# M. k& a# T* P& K( E5 a( p: R: `
  return true;/ z9 [( b+ \- a3 E

else 
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  return false; 
# I5 v\\" R1 ~- G6 q* l7 i
}1 j, L+ N, M/ a3 X

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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;
}