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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;
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#define PH 03 P' W7 @) U6 ?9 h- {- l! s, Q% G- y. a
#define EM 1) t5 Q8 o5 h' k
#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);
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int main()
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{
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int ptop, etop, itop;# y ?4 `; A, k& R% C
int pdays, edays, idays, curdays;
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int count =0;, h2 h' X( n* r0 w( P# z7 j
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while(cin >> pdays >> edays >> idays >> curdays); ~4 }. ]3 E( K S0 C$ W
{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;
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count++;
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ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);' @7 \& g# ?, }/ N
itop = get_Top(idays,IN);
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cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";
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}\\" x2 Z( N/ S2 a J* S8 T
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return 0;
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}
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int get_Top(int days, int kind)\\" E4 k$ v/ ^7 ^2 F( G$ B
{
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switch(kind)
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case 0:
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return days%23;
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case 1:
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return days%28;
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case 2:
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return days%33;& V! b; y8 `4 n/ c1 r. Y6 \
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return 0;
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int triple(int ptop, int etop, int itop, int cur)) V. Z g! \/ e* r, L5 t7 F7 l
{
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float x, y, z;) U7 z: X, D# e' c5 B
int m1 = etop-ptop-5,2 W& b2 L2 I% a3 ~8 n
m2 = itop-etop-5;
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int temp = 0;\\" x$ ?5 n4 H* v7 T. T6 [ O0 k. _
for(z=2; (itop+(z-1)*33) <= 21252; z++)) n$ W( M5 ^' J8 i+ D* B- g/ N
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temp = m2 + 33*z;- Z7 D! E! R/ J9 D* i
y = (float)temp/28;+ u- d/ Z) [1 ]3 I
x = (float)(m1+temp)/23;/ [( c. g8 W4 x9 q
if(x>0 && y>0 && is_Integer(x) && is_Integer(y))' j: h* E3 q. C
return itop + (z-1)*33 -cur;( w* [+ g4 {# ^) N, C
}
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return 0;( K- U. W& f# W) f( |) Y# I d\\" p. E
}
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bool is_Integer(float n)# w. Q, l. ]% C$ ^% p1 S3 a% H\\" z\\" i
{
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if( (int)n - n ==0)
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return true;
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else m. W# J' y, h6 j: @
return false;& H2 ~- ~0 r( ]3 h- N7 N
}
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#include <iostream> 
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using namespace std; 
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#define PH      03 P' W7 @) U6 ?9 h- {- l! s, Q% G- y. a

#define EM      1) t5 Q8 o5 h' k

#define IN      2 
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int get_Top(int days, int kind); 
; c+ C, B, M  g\\" G
int triple(int ptop, int etop, int itop, int cur); 
' b/ S; x& f0 A8 J9 g
bool is_Integer(float n); 
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int main() 
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{ 
- w\\" @% k+ n0 m+ S/ x2 n
 int ptop,  etop,  itop;# y  ?4 `; A, k& R% C

int pdays, edays, idays, curdays; 
2 ]\\" A2 ~6 a: }  F
 int count =0;, h2 h' X( n* r0 w( P# z7 j

' R1 n5 {4 v- U! T* U; Z  i
 while(cin >> pdays >> edays >> idays >> curdays); ~4 }. ]3 E( K  S0 C$ W

{ 
* p5 m6 u! Z\\" I7 x5 |8 }
  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
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   break; 
( K* |- w  m2 ]1 ?
  count++; 
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  ptop = get_Top(pdays,PH); 
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     etop = get_Top(edays,EM);' @7 \& g# ?, }/ N

itop = get_Top(idays,IN); 
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  cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days."; 
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 return 0; 
0 ^* @# r1 g* A5 L* d
} 
% m3 \3 Y9 i7 `% c7 c
int get_Top(int days, int kind)\\" E4 k$ v/ ^7 ^2 F( G$ B

{ 
, n5 ~! S! N4 @
 switch(kind) 
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 {6 c! r% j; [4 [  s\\" ?' z

case 0: 
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  return days%23; 
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 case 1: 
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  return days%28; 
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 case 2: 
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  return days%33;& V! b; y8 `4 n/ c1 r. Y6 \

}! a* D3 X: ^6 U5 p\\" y

return 0; 
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int triple(int ptop, int etop, int itop, int cur)) V. Z  g! \/ e* r, L5 t7 F7 l

{ 
; a+ R  Y: D5 A
 float x, y, z;) U7 z: X, D# e' c5 B

int m1   = etop-ptop-5,2 W& b2 L2 I% a3 ~8 n

m2   = itop-etop-5; 
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 int temp = 0;\\" x$ ?5 n4 H* v7 T. T6 [  O0 k. _

for(z=2;  (itop+(z-1)*33) <= 21252;  z++)) n$ W( M5 ^' J8 i+ D* B- g/ N

{8 ?/ r. o\\" c) l* _8 e

temp = m2 + 33*z;- Z7 D! E! R/ J9 D* i

y = (float)temp/28;+ u- d/ Z) [1 ]3 I

x = (float)(m1+temp)/23;/ [( c. g8 W4 x9 q

if(x>0 && y>0 && is_Integer(x) && is_Integer(y))' j: h* E3 q. C

return itop + (z-1)*33 -cur;( w* [+ g4 {# ^) N, C

} 
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 return 0;( K- U. W& f# W) f( |) Y# I  d\\" p. E

} 
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bool is_Integer(float n)# w. Q, l. ]% C$ ^% p1 S3 a% H\\" z\\" i

{ 
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 if( (int)n - n ==0) 
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  return true; 
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 else  m. W# J' y, h6 j: @

return false;& H2 ~- ~0 r( ]3 h- N7 N

} 
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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;
}