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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>
% f1 i, @5 r0 h7 t- H
using namespace std;
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#define PH 0
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#define EM 1, b% d% E5 e0 B+ Y8 b2 ?
#define IN 2
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int get_Top(int days, int kind);
( b: \; ^+ |; n& y
int triple(int ptop, int etop, int itop, int cur);) _+ l% i- q; |: l& V- r
bool is_Integer(float n);* R+ |. Q3 I$ W( [
int main()
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{
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int ptop, etop, itop;- j- E+ }6 }# A. M- j
int pdays, edays, idays, curdays;. i: ^' d# A4 [7 d
int count =0;. j! z& q$ I: _7 J0 F) V4 K
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while(cin >> pdays >> edays >> idays >> curdays) J% h5 U6 S B. k9 r+ e/ T
{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)0 Z6 D' p( H' y3 X7 D9 y% c4 P
break;
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count++;2 E, \3 s( w% ~$ v4 h9 S9 @2 A
ptop = get_Top(pdays,PH);2 O3 t2 f& Q0 B6 N5 w8 _) j0 U
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.";6 m; C/ f! ^- S- w) k
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return 0;9 o! E' ^) Y! c2 L+ Q: P
}
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int get_Top(int days, int kind)
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{
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switch(kind) |' ` Z/ V+ @% h9 r' t3 ]
{\\" ]! Z\\" F5 v- ^- e4 x% E) j1 k
case 0:
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return days%23;3 P) y; {6 g' [4 S! Z\\" X
case 1:- M- f3 k* q x\\" n7 ~
return days%28;, P0 E2 o\\" y$ T: C0 I
case 2:
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return days%33;5 {# f/ q4 G. z\\" s
}
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return 0;9 _( [# x8 l5 z2 P% O; Q& w- E
}
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int triple(int ptop, int etop, int itop, int cur)
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{
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float x, y, z;1 h2 d\\" ^ L; c9 w( F
int m1 = etop-ptop-5,: N% t3 R2 ]9 {+ @% c
m2 = itop-etop-5;
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int temp = 0;3 O |/ \8 E; {5 h2 i4 n
for(z=2; (itop+(z-1)*33) <= 21252; z++); r8 V. c0 T2 f6 f$ P, N
{$ K9 p1 H- C0 l; x$ ?
temp = m2 + 33*z;1 d3 O2 I+ n$ F
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;
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}
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return 0;( B5 j& E\\" G+ d4 k
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bool is_Integer(float n)
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if( (int)n - n ==0)
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return true;) v S- q) z$ o- C8 [& g. Z
else% N5 M/ J- S2 y: K9 p! ~
return false;4 H+ W2 J1 D$ e/ E7 M
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#include <iostream> 
% f1 i, @5 r0 h7 t- H
using namespace std; 
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#define PH      0 
2 z) x' J. ?0 z, R$ B& Y
#define EM      1, b% d% E5 e0 B+ Y8 b2 ?

#define IN      2 
4 x$ M! t: W( s\\" n9 R
int get_Top(int days, int kind); 
( b: \; ^+ |; n& y
int triple(int ptop, int etop, int itop, int cur);) _+ l% i- q; |: l& V- r

bool is_Integer(float n);* R+ |. Q3 I$ W( [

int main() 
2 l* S  }  o( ^, [
{ 
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 int ptop,  etop,  itop;- j- E+ }6 }# A. M- j

int pdays, edays, idays, curdays;. i: ^' d# A4 [7 d

int count =0;. j! z& q$ I: _7 J0 F) V4 K

6 V3 @6 j& c3 I. G( z, m; ~6 `

while(cin >> pdays >> edays >> idays >> curdays)  J% h5 U6 S  B. k9 r+ e/ T

{ 
$ Q: ]- U, ]4 V7 w) u& {& D& O2 a9 K& Y
  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)0 Z6 D' p( H' y3 X7 D9 y% c4 P

break; 
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  count++;2 E, \3 s( w% ~$ v4 h9 S9 @2 A

ptop = get_Top(pdays,PH);2 O3 t2 f& Q0 B6 N5 w8 _) j0 U

etop = get_Top(edays,EM); 
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     itop = get_Top(idays,IN); 
  s: A. A8 T2 `  c0 ^3 o
  cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";6 m; C/ f! ^- S- w) k

}& M% Y2 t' g  p6 s: s

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 return 0;9 o! E' ^) Y! c2 L+ Q: P

} 
\\" x) s: z7 q  b% W, t2 W& o
int get_Top(int days, int kind) 
4 ]8 P. ]# C1 n* X1 F% A
{ 
; u4 z4 t5 d) y2 Z8 w) B1 C
 switch(kind)  |' `  Z/ V+ @% h9 r' t3 ]

{\\" ]! Z\\" F5 v- ^- e4 x% E) j1 k

case 0: 
1 G$ ]$ U3 \4 \1 I$ h6 x; O
  return days%23;3 P) y; {6 g' [4 S! Z\\" X

case 1:- M- f3 k* q  x\\" n7 ~

return days%28;, P0 E2 o\\" y$ T: C0 I

case 2: 
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  return days%33;5 {# f/ q4 G. z\\" s

} 
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 return 0;9 _( [# x8 l5 z2 P% O; Q& w- E

} 
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& Q5 N8 N/ c* u, j. y5 W8 q
int triple(int ptop, int etop, int itop, int cur) 
& L9 o; l# S/ A3 \/ ]/ k
{ 
& [. j, X\\" v* ^1 O# K. a
 float x, y, z;1 h2 d\\" ^  L; c9 w( F

int m1   = etop-ptop-5,: N% t3 R2 ]9 {+ @% c

m2   = itop-etop-5; 
4 P  ]5 ?) p9 y  ?9 y- C7 x
 int temp = 0;3 O  |/ \8 E; {5 h2 i4 n

for(z=2;  (itop+(z-1)*33) <= 21252;  z++); r8 V. c0 T2 f6 f$ P, N

{$ K9 p1 H- C0 l; x$ ?

temp = m2 + 33*z;1 d3 O2 I+ n$ F

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; 
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 } 
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 return 0;( B5 j& E\\" G+ d4 k

}) q* f) q$ k- @9 k

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bool is_Integer(float n) 
% c7 v& \2 H# _
{' l! S  l- \5 I6 D; v

if( (int)n - n ==0) 
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  return true;) v  S- q) z$ o- C8 [& g. Z

else% N5 M/ J- S2 y: K9 p! ~

return false;4 H+ W2 J1 D$ e/ E7 M

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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;
}