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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;; ^8 Z: r9 E4 ^, G. o
#define PH 0
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#define EM 13 J1 m ]+ y' C' V9 c
#define IN 2
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int get_Top(int days, int kind); ~5 G$ \. D& G; s& s
int triple(int ptop, int etop, int itop, int cur);( G1 V# E- ^, H
bool is_Integer(float n);
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int main(), l5 N3 A& @: l% r& }' |
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int ptop, etop, itop;' `/ h6 x0 E2 e( {( }$ s& |& x
int pdays, edays, idays, curdays;+ b7 N a% `$ l& c\\" ?1 p
int count =0;
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while(cin >> pdays >> edays >> idays >> curdays)' r' O\\" y1 K+ n, M& R
{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;/ n; ]# Y6 g! W- Y4 U
count++;
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ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);% w5 |* |/ }6 f# c
itop = get_Top(idays,IN);1 O: M4 [7 y$ |9 D
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";2 z3 T! K: E7 K( y+ a) u5 S
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return 0;; O5 {; Q* B/ S. d% N1 ~; l1 M( T
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int get_Top(int days, int kind)$ M: ~7 i* D. ?, T
{
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switch(kind)1 @8 X5 H6 P& o5 S: z$ s1 Z
{
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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;( B8 N\\" |8 M- c& r+ c\\" W5 I4 N
case 2:
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return days%33;
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}
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return 0;3 G* G\\" }( @& e- W! d- z8 G. P6 e
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int triple(int ptop, int etop, int itop, int cur)! b% @1 F z+ A- N6 @) k$ U+ x @6 Q
{
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float x, y, z;1 s8 u- j9 A0 Z3 A
int m1 = etop-ptop-5,, ], N7 A( B; P) D\\" e! F/ }! L Z
m2 = itop-etop-5;
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int temp = 0;
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for(z=2; (itop+(z-1)*33) <= 21252; z++) \, s! R- z5 K. f. m
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temp = m2 + 33*z;
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y = (float)temp/28;/ \& z7 e+ {7 C% E4 e: A/ M\\" M
x = (float)(m1+temp)/23;2 R0 {7 ?( M e6 K3 N
if(x>0 && y>0 && is_Integer(x) && is_Integer(y))
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return itop + (z-1)*33 -cur;9 @9 f1 R\\" U) F$ a
}+ d1 W d\\" g3 o5 Z( R: e! a
return 0;
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bool is_Integer(float n)
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{
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if( (int)n - n ==0)
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return true;& g& j* |, C* A4 K
else1 k p\\" m1 O2 c( f
return false;3 ?6 T# W: ]\\" E! ]9 \& _
}
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#include <iostream> 
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using namespace std;; ^8 Z: r9 E4 ^, G. o

#define PH      0 
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#define EM      13 J1 m  ]+ y' C' V9 c

#define IN      2 
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int get_Top(int days, int kind);  ~5 G$ \. D& G; s& s

int triple(int ptop, int etop, int itop, int cur);( G1 V# E- ^, H

bool is_Integer(float n); 
; D8 P8 S# u2 g3 Q8 t( {+ V
int main(), l5 N3 A& @: l% r& }' |

{! s# |9 R0 F\\" N1 z9 W* v

int ptop,  etop,  itop;' `/ h6 x0 E2 e( {( }$ s& |& x

int pdays, edays, idays, curdays;+ b7 N  a% `$ l& c\\" ?1 p

int count =0; 
. b5 C! D) S; P/ K- h( o, P
 5 |+ f1 X8 J: E

while(cin >> pdays >> edays >> idays >> curdays)' r' O\\" y1 K+ n, M& R

{ 
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  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
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   break;/ n; ]# Y6 g! W- Y4 U

count++; 
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  ptop = get_Top(pdays,PH); 
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     etop = get_Top(edays,EM);% w5 |* |/ }6 f# c

itop = get_Top(idays,IN);1 O: M4 [7 y$ |9 D

cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";2 z3 T! K: E7 K( y+ a) u5 S

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 return 0;; O5 {; Q* B/ S. d% N1 ~; l1 M( T

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int get_Top(int days, int kind)$ M: ~7 i* D. ?, T

{ 
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 switch(kind)1 @8 X5 H6 P& o5 S: z$ s1 Z

{ 
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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;( B8 N\\" |8 M- c& r+ c\\" W5 I4 N

case 2: 
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  return days%33; 
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 } 
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 return 0;3 G* G\\" }( @& e- W! d- z8 G. P6 e

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int triple(int ptop, int etop, int itop, int cur)! b% @1 F  z+ A- N6 @) k$ U+ x  @6 Q

{ 
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 float x, y, z;1 s8 u- j9 A0 Z3 A

int m1   = etop-ptop-5,, ], N7 A( B; P) D\\" e! F/ }! L  Z

m2   = itop-etop-5; 
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 int temp = 0; 
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 for(z=2;  (itop+(z-1)*33) <= 21252;  z++)  \, s! R- z5 K. f. m

{) }& N- E: i' Q- x  r

temp = m2 + 33*z; 
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  y = (float)temp/28;/ \& z7 e+ {7 C% E4 e: A/ M\\" M

x = (float)(m1+temp)/23;2 R0 {7 ?( M  e6 K3 N

if(x>0 && y>0 && is_Integer(x) && is_Integer(y)) 
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   return itop + (z-1)*33 -cur;9 @9 f1 R\\" U) F$ a

}+ d1 W  d\\" g3 o5 Z( R: e! a

return 0; 
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}3 e  S8 r2 s  ^7 M2 u3 v8 h

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bool is_Integer(float n) 
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{ 
& a\\" J$ l' P8 M8 J9 G8 O4 B8 m& D8 b
 if( (int)n - n ==0) 
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  return true;& g& j* |, C* A4 K

else1 k  p\\" m1 O2 c( f

return false;3 ?6 T# W: ]\\" E! ]9 \& _

} 
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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;
}