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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>4 Y2 p\\" k, ?% [' O/ t5 H; [
using namespace std;
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#define PH 00 s5 l4 N. K5 P, V
#define EM 1
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#define IN 2. h8 K* u* T& T\\" {
int get_Top(int days, int kind);( m# p' O* O- \+ F9 O
int triple(int ptop, int etop, int itop, int cur);
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bool is_Integer(float n);\\" ?7 Y k1 \4 K8 V, q& A
int main()7 Q; t$ z, f! q% \' l
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int ptop, etop, itop;
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int pdays, edays, idays, curdays;/ N) e+ q' o* [
int count =0;1 K q _, |; Y, W. B! X
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while(cin >> pdays >> edays >> idays >> curdays)
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1); d' U# Z: S! Y) x6 X \/ Y! A# P7 {
break;
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count++;0 c\\" W _' B8 e$ n
ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);
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itop = get_Top(idays,IN);; K1 {$ u- M: x/ p
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";/ j9 n2 O! E& Z
}
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return 0;0 U2 _2 s' @9 \3 P
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int get_Top(int days, int kind)
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switch(kind)& a0 C* t2 E3 w2 T$ _: V5 e
{: A c0 O2 `* S
case 0:
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return days%23;
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case 1:) a6 e/ A2 f$ P& O: V& h' [5 L
return days%28;
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case 2:7 s6 K: f4 f! ~0 O$ d
return days%33;\\" }! w9 \) i8 t; E7 d' T. O
}
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return 0; r% H# J\\" A, U, ]/ s
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int triple(int ptop, int etop, int itop, int cur)
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float x, y, z;
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int m1 = etop-ptop-5,6 T, \/ G. M, D
m2 = itop-etop-5;
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int temp = 0;
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for(z=2; (itop+(z-1)*33) <= 21252; z++)
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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))7 Z4 u$ t; _: n% B\\" W' r
return itop + (z-1)*33 -cur;) X( c6 r2 v: i, Z* }0 `5 T6 q3 K
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return 0;$ G( l/ Q+ S% N' K) o6 F+ P
}
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bool is_Integer(float n)
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if( (int)n - n ==0)
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return true;- }/ ~: R- W6 h
else* Q. \' j3 ?: I1 T: x' b
return false;9 Q: r4 J! j0 c\\" B6 a
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#include <iostream>4 Y2 p\\" k, ?% [' O/ t5 H; [

using namespace std; 
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#define PH      00 s5 l4 N. K5 P, V

#define EM      1 
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#define IN      2. h8 K* u* T& T\\" {

int get_Top(int days, int kind);( m# p' O* O- \+ F9 O

int triple(int ptop, int etop, int itop, int cur); 
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bool is_Integer(float n);\\" ?7 Y  k1 \4 K8 V, q& A

int main()7 Q; t$ z, f! q% \' l

{. ]; f8 e! ?& B% M% g/ D5 @

int ptop,  etop,  itop; 
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 int pdays, edays, idays, curdays;/ N) e+ q' o* [

int count =0;1 K  q  _, |; Y, W. B! X

6 V! C% o3 \# s\\" z\\" c
 while(cin >> pdays >> edays >> idays >> curdays) 
5 f5 B5 [( a, [
 {; k, h& W\\" p* g) f

if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1); d' U# Z: S! Y) x6 X  \/ Y! A# P7 {

break; 
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  count++;0 c\\" W  _' B8 e$ n

ptop = get_Top(pdays,PH); 
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     etop = get_Top(edays,EM); 
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     itop = get_Top(idays,IN);; K1 {$ u- M: x/ p

cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";/ j9 n2 O! E& Z

} 
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return 0;0 U2 _2 s' @9 \3 P

}+ }$ h0 }9 u8 w) u0 A- t; h- ]

int get_Top(int days, int kind) 
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{; O! x* o  C6 @9 }# ]! e

switch(kind)& a0 C* t2 E3 w2 T$ _: V5 e

{: A  c0 O2 `* S

case 0: 
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  return days%23; 
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 case 1:) a6 e/ A2 f$ P& O: V& h' [5 L

return days%28; 
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 case 2:7 s6 K: f4 f! ~0 O$ d

return days%33;\\" }! w9 \) i8 t; E7 d' T. O

} 
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 return 0;  r% H# J\\" A, U, ]/ s

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int triple(int ptop, int etop, int itop, int cur) 
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float x, y, z; 
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 int m1   = etop-ptop-5,6 T, \/ G. M, D

m2   = itop-etop-5; 
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 int temp = 0; 
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 for(z=2;  (itop+(z-1)*33) <= 21252;  z++) 
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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))7 Z4 u$ t; _: n% B\\" W' r

return itop + (z-1)*33 -cur;) X( c6 r2 v: i, Z* }0 `5 T6 q3 K

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return 0;$ G( l/ Q+ S% N' K) o6 F+ P

} 
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\\" m4 s+ V; ?) R' E/ |; O
bool is_Integer(float n) 
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{+ @. C6 c4 j- C- ]+ A( l

if( (int)n - n ==0) 
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  return true;- }/ ~: R- W6 h

else* Q. \' j3 ?: I1 T: x' b

return false;9 Q: r4 J! j0 c\\" B6 a

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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;
}