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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>: n* _9 i6 C\\" D* D
using namespace std;
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#define PH 0
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#define EM 1$ ~0 }7 y$ c' H1 R
#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);1 l\\" Y# Z+ p5 y1 M+ U. R5 [6 N
int main()
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{
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int ptop, etop, itop;
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int pdays, edays, idays, curdays;
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int count =0;5 t7 w. f8 y5 `1 M
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while(cin >> pdays >> edays >> idays >> curdays)
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{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;7 ^; @5 M+ _- `% X\\" I1 D+ c1 B( s# Y) z
count++;
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ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);! B2 G# U1 C( @8 I F1 w
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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}
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return 0;/ C X7 N$ v+ A* `+ ^( W, {6 u, H. H
}+ `3 R# W0 Z5 E\\" C; g C% N
int get_Top(int days, int kind)
0 ?( I& x\\" R3 ^( {) ]
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switch(kind)& _7 ~9 n$ E3 O
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case 0:
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return days%23;; `; f, X6 i. h9 o- R\\" J
case 1:
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return days%28;
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case 2:0 y9 @7 U5 M4 I/ I5 t: }
return days%33;# p3 b8 G3 B/ \- a
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return 0;
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int triple(int ptop, int etop, int itop, int cur)
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float x, y, z;\\" r1 ]- Q: p% @$ X8 a* a
int m1 = etop-ptop-5,& }* d& i: {/ ~/ o3 o7 r
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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{
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temp = m2 + 33*z;. k! I- A: @' H: N) R, ~+ F0 z7 a8 n
y = (float)temp/28;/ Y0 o0 c, T8 M$ v# p
x = (float)(m1+temp)/23;) C/ v) ^& d3 p. g
if(x>0 && y>0 && is_Integer(x) && is_Integer(y))
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return itop + (z-1)*33 -cur;
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}
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return 0;- y\\" |& x, l ^$ M1 c( m v
}
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bool is_Integer(float n). F9 U; g$ @& i* p
{
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if( (int)n - n ==0)
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return true;
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else
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return false;+ a: j, x5 z2 L+ g\\" b
}
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#include <iostream>: n* _9 i6 C\\" D* D

using namespace std; 
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#define PH      0 
) S. [0 {7 t, k. a
#define EM      1$ ~0 }7 y$ c' H1 R

#define IN      2 
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int get_Top(int days, int kind); 
( {# X. v  [. L8 m9 P! y3 e5 B
int triple(int ptop, int etop, int itop, int cur); 
  z' p; V. B5 d( d8 j% d& [
bool is_Integer(float n);1 l\\" Y# Z+ p5 y1 M+ U. R5 [6 N

int main() 
+ p8 E( }( I2 O% p1 G* H9 N1 e
{ 
, w) m4 |* h! c, Q+ {' N5 F
 int ptop,  etop,  itop; 
. z* |! ]% o$ L# E4 W* ]% @
 int pdays, edays, idays, curdays; 
3 l: ^; W7 T8 ~# E+ B# c! V
 int count =0;5 t7 w. f8 y5 `1 M

7 X, G3 J\\" b4 V9 w4 c\\" n
 while(cin >> pdays >> edays >> idays >> curdays) 
  D' z! a- q: ^/ w9 \( b3 U
 { 
9 F: M2 }& Y+ q
  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
3 q) q! ^5 v' m* V
   break;7 ^; @5 M+ _- `% X\\" I1 D+ c1 B( s# Y) z

count++; 
\\" d0 I( r4 s0 d- m' m
  ptop = get_Top(pdays,PH); 
. @! ]: c3 S; Q( N0 y
     etop = get_Top(edays,EM);! B2 G# U1 C( @8 I  F1 w

itop = get_Top(idays,IN); 
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  cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days."; 
* f7 p0 I0 l$ Z4 ]  w
 } 
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return 0;/ C  X7 N$ v+ A* `+ ^( W, {6 u, H. H

}+ `3 R# W0 Z5 E\\" C; g  C% N

int get_Top(int days, int kind) 
0 ?( I& x\\" R3 ^( {) ]
{- y# O$ u+ _9 H8 b

switch(kind)& _7 ~9 n$ E3 O

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case 0: 
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  return days%23;; `; f, X6 i. h9 o- R\\" J

case 1: 
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  return days%28; 
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 case 2:0 y9 @7 U5 M4 I/ I5 t: }

return days%33;# p3 b8 G3 B/ \- a

}- v& G8 r! t/ g/ c6 W/ u

return 0; 
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}' d\\" G# F! @$ o2 r7 R$ w

4 ]\\" g7 M4 n% a1 _! Q& A. a# o
int triple(int ptop, int etop, int itop, int cur) 
3 _8 S  T) Y- e# m# |4 Y/ i1 x5 |
{: S  M- c* I) o( \3 c3 q

float x, y, z;\\" r1 ]- Q: p% @$ X8 a* a

int m1   = etop-ptop-5,& }* d& i: {/ ~/ o3 o7 r

m2   = itop-etop-5; 
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 int temp = 0; 
! `! \: [5 U* ^
 for(z=2;  (itop+(z-1)*33) <= 21252;  z++) 
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 { 
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  temp = m2 + 33*z;. k! I- A: @' H: N) R, ~+ F0 z7 a8 n

y = (float)temp/28;/ Y0 o0 c, T8 M$ v# p

x = (float)(m1+temp)/23;) C/ v) ^& d3 p. g

if(x>0 && y>0 && is_Integer(x) && is_Integer(y)) 
# g; V! `/ \# X/ W  k' e% n
   return itop + (z-1)*33 -cur; 
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 } 
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 return 0;- y\\" |& x, l  ^$ M1 c( m  v

} 
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bool is_Integer(float n). F9 U; g$ @& i* p

{ 
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 if( (int)n - n ==0) 
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  return true; 
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 else 
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  return false;+ a: j, x5 z2 L+ g\\" b

} 
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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;
}