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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>
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using namespace std;
% C: a- L# @1 [
#define PH 0$ H5 o% E. K, }3 l( D
#define EM 1
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#define IN 2
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int get_Top(int days, int kind);
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int triple(int ptop, int etop, int itop, int cur);
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bool is_Integer(float n);0 u* Q$ y/ Q6 d* D* d/ |9 q: A
int main()
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int ptop, etop, itop;+ Z' {; d7 E0 u' N# U
int pdays, edays, idays, curdays;
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int count =0;: A, ?% S! o: `4 }7 p7 D\\" m
% ~3 t1 ~0 \$ T* g2 p: K D\\" Z
while(cin >> pdays >> edays >> idays >> curdays)% G' z$ b\\" ^* D+ _5 p, B8 l\\" Z
{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1): H: T3 d6 A }. f
break;' p( D8 |! u( q- o- S7 v
count++;: P7 A4 K* c6 @' x) X/ M- l, I
ptop = get_Top(pdays,PH);! Q7 u' }2 t- v
etop = get_Top(edays,EM);
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itop = get_Top(idays,IN);
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cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";7 @+ p, o9 B6 m! A- W
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return 0;
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int get_Top(int days, int kind)
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switch(kind)! D4 r. `0 F3 W) e6 T: J6 L4 c5 a$ D2 |
{
\\" Q\\" L+ i- r3 r# x1 Y/ n
case 0: ~/ T\\" A+ \3 h: H x S# u# k
return days%23; K! i2 K' J) w
case 1:
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return days%28;\\" x; c! w: \\\" y G! W2 d
case 2:' U+ b6 {# M$ \# u: ?* D% U2 X
return days%33;7 n( L ?5 A2 y: S\\" }
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return 0;0 s. V' F2 X' N# J+ y- m6 o1 L
}' D$ S- Q- X: p5 w: e
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int triple(int ptop, int etop, int itop, int cur) w V' N4 _3 `7 Q( n! e
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float x, y, z; c @, x3 f r1 x: n) {
int m1 = etop-ptop-5,7 ]\\" k% [+ }2 z) }+ }2 e+ G& b* \
m2 = itop-etop-5;. `3 E5 y. q4 @) y
int temp = 0;, L0 i( j0 v' a! ^8 e
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))
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return itop + (z-1)*33 -cur;$ q0 p; d! I1 C
}
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return 0;
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}
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bool is_Integer(float n)' }8 i2 h1 m+ b- ?+ O G
{; A1 [' g4 Z& C. Q- s$ d
if( (int)n - n ==0)* o, w4 \/ f# a4 @% `. ?4 B& p' x9 m
return true;! y/ Z\\" a) f, _
else+ |9 ?: J4 a) p: I4 e; X) j
return false;
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}
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#include <iostream> 
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using namespace std; 
% C: a- L# @1 [
#define PH      0$ H5 o% E. K, }3 l( D

#define EM      1 
' r+ B  y! j- s# \# ^4 _& k! ^1 j( O
#define IN      2 
; c\\" R; Q) g9 V& x+ r8 S
int get_Top(int days, int kind); 
- J9 p' ^- `+ {8 \) d: t% d
int triple(int ptop, int etop, int itop, int cur); 
+ M+ E3 |, h% _8 ^$ V# h
bool is_Integer(float n);0 u* Q$ y/ Q6 d* D* d/ |9 q: A

int main() 
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{8 S1 g4 n' P2 `7 Z$ s2 [4 ^5 ^

int ptop,  etop,  itop;+ Z' {; d7 E0 u' N# U

int pdays, edays, idays, curdays; 
* g3 M- l& C3 @\\" T; Z
 int count =0;: A, ?% S! o: `4 }7 p7 D\\" m

% ~3 t1 ~0 \$ T* g2 p: K  D\\" Z
 while(cin >> pdays >> edays >> idays >> curdays)% G' z$ b\\" ^* D+ _5 p, B8 l\\" Z

{ 
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  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1): H: T3 d6 A  }. f

break;' p( D8 |! u( q- o- S7 v

count++;: P7 A4 K* c6 @' x) X/ M- l, I

ptop = get_Top(pdays,PH);! Q7 u' }2 t- v

etop = get_Top(edays,EM); 
% w& [/ }% B$ I- J6 N
     itop = get_Top(idays,IN); 
  S& z* O( b. [: N& @5 s
  cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";7 @+ p, o9 B6 m! A- W

}  m9 h1 W4 w  A; e# X

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return 0; 
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}  l5 a1 d1 `$ ?( l& u: W2 ]% |

int get_Top(int days, int kind) 
( n% o7 x' L; g; K! i
{, K. j/ [0 s2 z$ Y6 A

switch(kind)! D4 r. `0 F3 W) e6 T: J6 L4 c5 a$ D2 |

{ 
\\" Q\\" L+ i- r3 r# x1 Y/ n
 case 0:  ~/ T\\" A+ \3 h: H  x  S# u# k

return days%23;  K! i2 K' J) w

case 1: 
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  return days%28;\\" x; c! w: \\\" y  G! W2 d

case 2:' U+ b6 {# M$ \# u: ?* D% U2 X

return days%33;7 n( L  ?5 A2 y: S\\" }

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return 0;0 s. V' F2 X' N# J+ y- m6 o1 L

}' D$ S- Q- X: p5 w: e

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int triple(int ptop, int etop, int itop, int cur)  w  V' N4 _3 `7 Q( n! e

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float x, y, z;  c  @, x3 f  r1 x: n) {

int m1   = etop-ptop-5,7 ]\\" k% [+ }2 z) }+ }2 e+ G& b* \

m2   = itop-etop-5;. `3 E5 y. q4 @) y

int temp = 0;, L0 i( j0 v' a! ^8 e

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; 
  }: D3 C/ r! n* h0 F& K# S0 F: h
  x = (float)(m1+temp)/23; 
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  if(x>0 && y>0 && is_Integer(x) && is_Integer(y)) 
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   return itop + (z-1)*33 -cur;$ q0 p; d! I1 C

} 
% B2 C# S* J- R, ~
 return 0; 
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} 
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bool is_Integer(float n)' }8 i2 h1 m+ b- ?+ O  G

{; A1 [' g4 Z& C. Q- s$ d

if( (int)n - n ==0)* o, w4 \/ f# a4 @% `. ?4 B& p' x9 m

return true;! y/ Z\\" a) f, _

else+ |9 ?: J4 a) p: I4 e; X) j

return false; 
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} 
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zan

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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;
}