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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>+ d2 a7 [8 J( A+ b* F( @# e
using namespace std;# S; y- n/ g' p
#define PH 0& C3 H# ]6 f: z) k% }' t
#define EM 11 e. O9 b( m {. J* \, ^% X
#define IN 2/ R1 o! S* |% p+ r; s
int get_Top(int days, int kind);7 T6 l0 L u# t, U% l6 h2 g
int triple(int ptop, int etop, int itop, int cur);1 i5 t# h3 P, `* ~/ H0 f$ Q2 I
bool is_Integer(float n);6 c% c* c$ S/ c2 S: K' T8 o& u
int main()
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{
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int ptop, etop, itop;
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int pdays, edays, idays, curdays;
* v: Y; f/ H1 `9 k
int count =0;
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while(cin >> pdays >> edays >> idays >> curdays)
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{% H1 Y/ a; ]$ k7 L5 h
if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1), w0 x9 N) l0 d\\" m
break;
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count++;( ]% I9 a7 k! }8 d1 y0 x5 G# t
ptop = get_Top(pdays,PH);1 c+ E k9 G, I0 O) U+ I
etop = get_Top(edays,EM);
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itop = get_Top(idays,IN);2 v7 ]# O+ E+ f$ ^9 }1 p6 y
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";( H0 j# q$ V/ W( N. Q6 ?, c
}
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return 0;
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int get_Top(int days, int kind)1 t; u2 F. n: J# w/ h
{
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switch(kind): }) T, u6 C! U# Y# h5 {# Y; U; o$ e
{
& W. f- e, @' C1 E
case 0:
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return days%23;
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case 1:1 e3 f& r+ X. ]
return days%28; }% J9 G) J2 ^8 H! C/ a$ v
case 2: U/ ^+ T- t0 Q
return days%33;
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return 0;1 c# ^+ l4 k3 q# H& p! r% a
}
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int triple(int ptop, int etop, int itop, int cur)% S+ i: J\\" x I. R8 f0 k' d
{+ _9 }+ e1 T\\" `
float x, y, z;
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int m1 = etop-ptop-5,
\\" g. j) H: ]' F. {
m2 = itop-etop-5;& ?; d4 ?7 C1 h- D6 s4 c$ B: P/ x
int temp = 0;7 h7 e; d% N% d8 ?% Z% |! E
for(z=2; (itop+(z-1)*33) <= 21252; z++); N# E! v) W& d
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temp = m2 + 33*z;
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y = (float)temp/28;
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x = (float)(m1+temp)/23;
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if(x>0 && y>0 && is_Integer(x) && is_Integer(y))9 e6 R( j+ n: Z0 m) m' `
return itop + (z-1)*33 -cur;; U& b# P1 a: ^8 N H! `
}# [( j. z; ^8 B
return 0;
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}
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bool is_Integer(float n)
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{
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if( (int)n - n ==0)- m1 [/ V) s! T8 b( u7 W
return true;+ b5 f3 q* {0 V\\" x\\" [! a% [
else
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return false;. {4 ~& d9 h5 G
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#include <iostream>+ d2 a7 [8 J( A+ b* F( @# e

using namespace std;# S; y- n/ g' p

#define PH      0& C3 H# ]6 f: z) k% }' t

#define EM      11 e. O9 b( m  {. J* \, ^% X

#define IN      2/ R1 o! S* |% p+ r; s

int get_Top(int days, int kind);7 T6 l0 L  u# t, U% l6 h2 g

int triple(int ptop, int etop, int itop, int cur);1 i5 t# h3 P, `* ~/ H0 f$ Q2 I

bool is_Integer(float n);6 c% c* c$ S/ c2 S: K' T8 o& u

int main() 
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{ 
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 int ptop,  etop,  itop; 
1 o& o$ l3 G- j4 H. D
 int pdays, edays, idays, curdays; 
* v: Y; f/ H1 `9 k
 int count =0; 
* v  J8 a% ]8 ~0 }- a: ^) A
 ! V& N+ Y7 i+ M$ ?! n9 j

while(cin >> pdays >> edays >> idays >> curdays) 
2 G% S2 {1 _; R/ H: n6 E3 u
 {% H1 Y/ a; ]$ k7 L5 h

if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1), w0 x9 N) l0 d\\" m

break; 
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  count++;( ]% I9 a7 k! }8 d1 y0 x5 G# t

ptop = get_Top(pdays,PH);1 c+ E  k9 G, I0 O) U+ I

etop = get_Top(edays,EM); 
  ~9 E* t9 C/ [# M* f* Z
     itop = get_Top(idays,IN);2 v7 ]# O+ E+ f$ ^9 }1 p6 y

cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";( H0 j# q$ V/ W( N. Q6 ?, c

} 
1 \' Y' v( }0 ]! X' z1 b7 Q\\" O' }9 H
 
& q. ]; a, ^, y3 S, c1 X# k
 return 0; 
6 h' O0 c8 m) |; V% S* c. @
}9 x& I* u9 T) J; p% L7 m

int get_Top(int days, int kind)1 t; u2 F. n: J# w/ h

{ 
$ B8 t! s- x# T: k
 switch(kind): }) T, u6 C! U# Y# h5 {# Y; U; o$ e

{ 
& W. f- e, @' C1 E
 case 0: 
. q7 l3 M( c9 _* S
  return days%23; 
4 F) W1 S) K! H. o5 Q2 C1 a
 case 1:1 e3 f& r+ X. ]

return days%28;  }% J9 G) J2 ^8 H! C/ a$ v

case 2:  U/ ^+ T- t0 Q

return days%33; 
; W  I& }# _- M2 w- r4 O! w
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return 0;1 c# ^+ l4 k3 q# H& p! r% a

} 
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/ i: f! Q* n9 [7 L  ~5 |
int triple(int ptop, int etop, int itop, int cur)% S+ i: J\\" x  I. R8 f0 k' d

{+ _9 }+ e1 T\\" `

float x, y, z; 
+ e( r$ P* G7 `+ Z* ^) s* l2 X6 Z) m
 int m1   = etop-ptop-5, 
\\" g. j) H: ]' F. {
  m2   = itop-etop-5;& ?; d4 ?7 C1 h- D6 s4 c$ B: P/ x

int temp = 0;7 h7 e; d% N% d8 ?% Z% |! E

for(z=2;  (itop+(z-1)*33) <= 21252;  z++); N# E! v) W& d

{* I1 `3 K' l& l1 c3 n

temp = m2 + 33*z; 
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  y = (float)temp/28; 
$ {; d7 e2 _7 W, x( ^% l
  x = (float)(m1+temp)/23; 
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  if(x>0 && y>0 && is_Integer(x) && is_Integer(y))9 e6 R( j+ n: Z0 m) m' `

return itop + (z-1)*33 -cur;; U& b# P1 a: ^8 N  H! `

}# [( j. z; ^8 B

return 0; 
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} 
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bool is_Integer(float n) 
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{ 
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 if( (int)n - n ==0)- m1 [/ V) s! T8 b( u7 W

return true;+ b5 f3 q* {0 V\\" x\\" [! a% [

else 
: {% _5 B\\" c4 i4 @* j9 X& F
  return false;. {4 ~& d9 h5 G

}3 }\\" _( U9 j! a! U

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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;
}