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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 00 ]' A6 w9 T) w( W
#define EM 1* _4 k! P8 u; f9 S& w
#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);4 M3 H9 Y7 b( J& y) C$ }! ~3 M
bool is_Integer(float n);
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int main()' w\\" k1 u2 ^; Q; c. I; M+ ~
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int ptop, etop, itop;- w ^! ~* i4 j9 E @3 U+ u7 e
int pdays, edays, idays, curdays;
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int count =0;
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while(cin >> pdays >> edays >> idays >> curdays)
# O: b' y% E3 c: P
{1 ?# `0 m1 j% ^
if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1), e7 Y W: ~4 n
break;
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count++;- d2 c/ S. H3 d& {, N# Y3 d
ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);7 l5 \$ 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.";7 x( p# Q% c3 g# a' Y# S5 [! V2 {
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return 0;
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int get_Top(int days, int kind)9 n: _! h$ B+ m' \0 ~+ n/ P
{/ Z2 T1 j/ x& w& W/ V' d
switch(kind)& z$ i) \) K5 d) b$ f
{% P; V9 y+ Y, ?$ q+ f; T1 H
case 0: o# \6 h$ X# B0 ? ?9 D5 k
return days%23;
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case 1:
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return days%28;
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case 2:0 j- v/ n0 o( O+ F; z& f9 Q; J# \
return days%33;\\" w% J9 l- f# a+ E+ E
}
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return 0;3 H% a8 I& ~3 N& Z2 L
}
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int triple(int ptop, int etop, int itop, int cur)% |: i* k: x( t* o4 \, z- ^6 }\\" l$ T8 ^
{' Y% p3 x+ a0 H: P' ]0 }
float x, y, z;/ H' J) w0 d2 }0 S6 Q+ A& }
int m1 = etop-ptop-5,
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m2 = itop-etop-5;. N' |$ I+ B- b
int temp = 0;
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for(z=2; (itop+(z-1)*33) <= 21252; z++)1 w! `# l( _$ u3 [
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temp = m2 + 33*z;
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y = (float)temp/28;) J; F) f! b8 ]+ M
x = (float)(m1+temp)/23;
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if(x>0 && y>0 && is_Integer(x) && is_Integer(y))' z/ [\\" l- t2 p
return itop + (z-1)*33 -cur;
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}
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return 0;
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bool is_Integer(float n)3 f a$ ]4 s4 o$ t; ~2 Z8 X$ A
{
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if( (int)n - n ==0)
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return true;
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else! m `8 a6 q# n8 {3 p) S) ]
return false;
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}
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#include <iostream> 
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using namespace std; 
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#define PH      00 ]' A6 w9 T) w( W

#define EM      1* _4 k! P8 u; f9 S& w

#define IN      2 
. I) B* R( R6 y. U
int get_Top(int days, int kind); 
4 U' ~) p5 T\\" F/ N5 z/ A
int triple(int ptop, int etop, int itop, int cur);4 M3 H9 Y7 b( J& y) C$ }! ~3 M

bool is_Integer(float n); 
3 S\\" ^6 a1 N5 o  [. ^
int main()' w\\" k1 u2 ^; Q; c. I; M+ ~

{- t- U% \* n  k: [. O6 Q+ h

int ptop,  etop,  itop;- w  ^! ~* i4 j9 E  @3 U+ u7 e

int pdays, edays, idays, curdays; 
. f& z# G  @3 k
 int count =0; 
. S! z! H$ c$ s2 V6 ]; j
   ^1 m2 F- B7 g+ b

while(cin >> pdays >> edays >> idays >> curdays) 
# O: b' y% E3 c: P
 {1 ?# `0 m1 j% ^

if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1), e7 Y  W: ~4 n

break; 
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  count++;- d2 c/ S. H3 d& {, N# Y3 d

ptop = get_Top(pdays,PH); 
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     etop = get_Top(edays,EM);7 l5 \$ 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.";7 x( p# Q% c3 g# a' Y# S5 [! V2 {

}% ^4 g+ Y- [3 r9 n  p% Y! b$ W; S

\\" O; A9 `* b! p

return 0; 
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}& x/ B, a* A! C4 A! l

int get_Top(int days, int kind)9 n: _! h$ B+ m' \0 ~+ n/ P

{/ Z2 T1 j/ x& w& W/ V' d

switch(kind)& z$ i) \) K5 d) b$ f

{% P; V9 y+ Y, ?$ q+ f; T1 H

case 0:  o# \6 h$ X# B0 ?  ?9 D5 k

return days%23; 
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 case 1: 
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  return days%28; 
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 case 2:0 j- v/ n0 o( O+ F; z& f9 Q; J# \

return days%33;\\" w% J9 l- f# a+ E+ E

} 
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 return 0;3 H% a8 I& ~3 N& Z2 L

} 
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int triple(int ptop, int etop, int itop, int cur)% |: i* k: x( t* o4 \, z- ^6 }\\" l$ T8 ^

{' Y% p3 x+ a0 H: P' ]0 }

float x, y, z;/ H' J) w0 d2 }0 S6 Q+ A& }

int m1   = etop-ptop-5, 
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  m2   = itop-etop-5;. N' |$ I+ B- b

int temp = 0; 
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 for(z=2;  (itop+(z-1)*33) <= 21252;  z++)1 w! `# l( _$ u3 [

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temp = m2 + 33*z; 
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  y = (float)temp/28;) J; F) f! b8 ]+ M

x = (float)(m1+temp)/23; 
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  if(x>0 && y>0 && is_Integer(x) && is_Integer(y))' z/ [\\" l- t2 p

return itop + (z-1)*33 -cur; 
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 } 
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 return 0; 
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}6 z# V, d' Y! }: @; N( Q9 Q

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bool is_Integer(float n)3 f  a$ ]4 s4 o$ t; ~2 Z8 X$ A

{ 
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 if( (int)n - n ==0) 
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  return true; 
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 else! m  `8 a6 q# n8 {3 p) S) ]

return false; 
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} 
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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;
}