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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>/ A0 A- F3 Y% W, [5 H
using namespace std;
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#define PH 0
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#define EM 1
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#define IN 2- c# A1 K* f Q, H0 \+ {* q. i
int get_Top(int days, int kind);
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int triple(int ptop, int etop, int itop, int cur);, T5 N5 B' p: t+ ^6 a- l$ O. p\\" L8 o
bool is_Integer(float n);
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int main()
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{
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int ptop, etop, itop;5 B7 a4 {\\" M* C$ n
int pdays, edays, idays, curdays;& ^+ w\\" n; j5 S/ c6 d
int count =0;
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while(cin >> pdays >> edays >> idays >> curdays)
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{# q* l, k! M* U
if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;. t5 ^. n( t I: D; D( x# N( N0 Q
count++;
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ptop = get_Top(pdays,PH);
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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.";+ M2 j5 B4 \$ g6 v9 T) m# C
}
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return 0;+ k, ?( T2 G; V, U4 s! U ~
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int get_Top(int days, int kind)
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{
\\" a# X- } M0 Q0 ?. P; A
switch(kind)
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{
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case 0:
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return days%23;+ x9 t2 C5 P1 `, P1 Y
case 1:, o9 G/ ^8 D% i\\" v/ g. [0 E9 A
return days%28;
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case 2:; u8 ~$ ^# L* [2 L* T. l, o- u
return days%33;4 A7 ~: F# i, |# E9 |5 E% Z0 O# F3 _; u
}\\" K0 Q0 W$ w! ] ?
return 0;
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}
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int triple(int ptop, int etop, int itop, int cur)
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float x, y, z;
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int m1 = etop-ptop-5,% D9 B/ w& \0 _! y7 v
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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temp = m2 + 33*z;, |; T0 O. d6 J, y8 Z
y = (float)temp/28;* z, ?! x+ E\\" G9 z! l# R
x = (float)(m1+temp)/23;. U5 C: R# `' d! r
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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return 0;
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}
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bool is_Integer(float n) m+ Q# @4 W% c1 |1 C. L. l\\" B
{\\" P0 p1 M' g# D6 r w- g
if( (int)n - n ==0)- G$ X) j- B\\" H+ v0 j
return true;8 U( Q V7 N% p* ?6 z+ n( e
else2 t w3 h) `+ a* S( ?: v
return false;* j* N' {$ D\\" w\\" X8 U( I& R
}
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#include <iostream>/ A0 A- F3 Y% W, [5 H

using namespace std; 
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#define PH      0 
; G% O. @$ O( V/ y' e
#define EM      1 
& X! [# L% K, {' k
#define IN      2- c# A1 K* f  Q, H0 \+ {* q. i

int get_Top(int days, int kind); 
' h) J8 t! Z8 F6 f% J- ?, g
int triple(int ptop, int etop, int itop, int cur);, T5 N5 B' p: t+ ^6 a- l$ O. p\\" L8 o

bool is_Integer(float n); 
5 A8 Y5 f  o. [7 n0 `* q* Z3 E1 N
int main() 
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{ 
4 a& {: ^& }* N: e! k! p$ A: o
 int ptop,  etop,  itop;5 B7 a4 {\\" M* C$ n

int pdays, edays, idays, curdays;& ^+ w\\" n; j5 S/ c6 d

int count =0; 
/ @: r2 U2 P' O1 R! z+ s7 w
 + y. f! S0 M7 t4 W

while(cin >> pdays >> edays >> idays >> curdays) 
: S4 I* H# k6 ~1 g1 n
 {# q* l, k! M* U

if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
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   break;. t5 ^. n( t  I: D; D( x# N( N0 Q

count++; 
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  ptop = get_Top(pdays,PH); 
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     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.";+ M2 j5 B4 \$ g6 v9 T) m# C

} 
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 return 0;+ k, ?( T2 G; V, U4 s! U  ~

}/ j/ l4 J3 \& v

int get_Top(int days, int kind) 
1 \! i9 p) T0 d8 Y
{ 
\\" a# X- }  M0 Q0 ?. P; A
 switch(kind) 
* \: A9 O7 t3 y3 ^, }
 { 
# F4 `. W: r  K1 x8 X7 {; D& l
 case 0: 
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  return days%23;+ x9 t2 C5 P1 `, P1 Y

case 1:, o9 G/ ^8 D% i\\" v/ g. [0 E9 A

return days%28; 
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 case 2:; u8 ~$ ^# L* [2 L* T. l, o- u

return days%33;4 A7 ~: F# i, |# E9 |5 E% Z0 O# F3 _; u

}\\" K0 Q0 W$ w! ]  ?

return 0; 
% u) S& J9 y3 h4 Z: e- _: o7 c
} 
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int triple(int ptop, int etop, int itop, int cur) 
0 ]; F# d4 y( F$ V
{; C- }9 d* y% E/ K) T8 l+ V5 o

float x, y, z; 
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 int m1   = etop-ptop-5,% D9 B/ w& \0 _! y7 v

m2   = itop-etop-5; 
  M( ^( z+ t8 q/ I) G% e4 x\\" k9 K
 int temp = 0; 
' `2 r: B1 o8 g! C' \3 m1 @
 for(z=2;  (itop+(z-1)*33) <= 21252;  z++) 
; D2 s* h' t' M5 p( O
 {, b+ [* z) x2 a5 L7 Z

temp = m2 + 33*z;, |; T0 O. d6 J, y8 Z

y = (float)temp/28;* z, ?! x+ E\\" G9 z! l# R

x = (float)(m1+temp)/23;. U5 C: R# `' d! r

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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return 0; 
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} 
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bool is_Integer(float n)  m+ Q# @4 W% c1 |1 C. L. l\\" B

{\\" P0 p1 M' g# D6 r  w- g

if( (int)n - n ==0)- G$ X) j- B\\" H+ v0 j

return true;8 U( Q  V7 N% p* ?6 z+ n( e

else2 t  w3 h) `+ a* S( ?: v

return false;* j* N' {$ D\\" w\\" X8 U( I& R

} 
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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;
}