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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>
# @0 R* p; K0 X L$ `* ^
using namespace std;/ X& a. P3 N9 p* x2 B- g& A- ?
#define PH 0+ [+ F' J! B% L# M* A3 ^
#define EM 12 c4 ] L* a, l# ~* z1 a+ c- Z
#define IN 2
8 }4 Y8 c! m; M9 L h. e$ P
int get_Top(int days, int kind);4 w+ ^3 _( S1 I7 e
int triple(int ptop, int etop, int itop, int cur);. l& ^4 E% R3 d
bool is_Integer(float n);
. q9 ^/ B+ U* ~: U. I7 M! d% L& ^
int main()
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{- X2 g$ x* b \6 U
int ptop, etop, itop;( q5 Y8 [0 M+ }$ G
int pdays, edays, idays, curdays;8 t1 Z+ ?- z/ U& D# v
int count =0;- Y9 q9 Z6 Y9 R
/ g3 R- K* e+ m9 g' U* g2 W `
while(cin >> pdays >> edays >> idays >> curdays)
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{# b m! ?% A& I0 @0 L
if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
R- f/ l1 n% V1 A: Y2 }8 A! p
break;
& C2 d9 V\\" d9 }4 U% N
count++;% x1 \+ s' d3 j
ptop = get_Top(pdays,PH);* P! S, u; h# y, E7 b. T- ?
etop = get_Top(edays,EM);5 ?/ s+ J4 d* c% P7 M! J! W! ~
itop = get_Top(idays,IN);
. `, c+ C, Q' |( l4 e, |! k
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";+ B& z& Z* L3 N% F, C/ n+ U Z
}6 E4 b- ~6 Z\\" h
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return 0;
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}& |+ c: J/ G' d/ N- T% j\\" S
int get_Top(int days, int kind)8 _\\" g; z; f# _; k
{; [* M( r9 H3 _
switch(kind): U. n* M; g+ @$ Q& A
{6 D) ]6 }0 t1 l' p1 z0 r5 L1 K
case 0:4 v4 k# D: v* c4 @2 W
return days%23;: w0 n. o+ N! v\\" D
case 1:! c7 \7 ^6 U! n3 ~5 R2 w) o
return days%28;! M- ]( M# U( Q1 L* v4 c* R
case 2:+ P, V\\" f9 B& r
return days%33;- i, X/ P- e0 ]* S0 M1 V
}
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return 0;: {3 K) H* a- w- b
}
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2 ?2 W: N+ X1 Z& U3 D
int triple(int ptop, int etop, int itop, int cur)
$ P8 c\\" X' |: l7 x$ G7 |
{
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float x, y, z;
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int m1 = etop-ptop-5, J9 J' ^7 r) `1 S. S, O
m2 = itop-etop-5;% W+ i$ q8 r# l- O3 l\\" d% x
int temp = 0;0 u# T. F$ ?4 Z\\" M
for(z=2; (itop+(z-1)*33) <= 21252; z++)4 u# d, X1 a8 P
{
8 K6 @! h/ U5 v# R0 Z* a
temp = m2 + 33*z;& ^6 ?% ]7 I# n7 T/ x
y = (float)temp/28;7 Z% k' {; z5 B! a( B
x = (float)(m1+temp)/23;1 a/ D; ~: ~0 M: p0 ^+ G2 d
if(x>0 && y>0 && is_Integer(x) && is_Integer(y))
^% b* }1 @. }; G
return itop + (z-1)*33 -cur;
$ o3 ?9 l7 ~+ N1 R; \) M
}7 @- j+ K; s5 n
return 0;; ~' n2 D0 h! a4 p9 {\\" p
}
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bool is_Integer(float n)
+ c( U/ V3 X; `) J! |
{' p\\" w/ b# Y5 Q
if( (int)n - n ==0)7 G) W2 Z) @, [5 z\\" D: [
return true;
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else
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return false;
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}
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#include <iostream> 
# @0 R* p; K0 X  L$ `* ^
using namespace std;/ X& a. P3 N9 p* x2 B- g& A- ?

#define PH      0+ [+ F' J! B% L# M* A3 ^

#define EM      12 c4 ]  L* a, l# ~* z1 a+ c- Z

#define IN      2 
8 }4 Y8 c! m; M9 L  h. e$ P
int get_Top(int days, int kind);4 w+ ^3 _( S1 I7 e

int triple(int ptop, int etop, int itop, int cur);. l& ^4 E% R3 d

bool is_Integer(float n); 
. q9 ^/ B+ U* ~: U. I7 M! d% L& ^
int main() 
4 o\\" n$ q' y& I\\" d' ?; K- N
{- X2 g$ x* b  \6 U

int ptop,  etop,  itop;( q5 Y8 [0 M+ }$ G

int pdays, edays, idays, curdays;8 t1 Z+ ?- z/ U& D# v

int count =0;- Y9 q9 Z6 Y9 R

/ g3 R- K* e+ m9 g' U* g2 W  `

while(cin >> pdays >> edays >> idays >> curdays) 
5 i# g7 k: z& U+ \: |
 {# b  m! ?% A& I0 @0 L

if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
  R- f/ l1 n% V1 A: Y2 }8 A! p
   break; 
& C2 d9 V\\" d9 }4 U% N
  count++;% x1 \+ s' d3 j

ptop = get_Top(pdays,PH);* P! S, u; h# y, E7 b. T- ?

etop = get_Top(edays,EM);5 ?/ s+ J4 d* c% P7 M! J! W! ~

itop = get_Top(idays,IN); 
. `, c+ C, Q' |( l4 e, |! k
  cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";+ B& z& Z* L3 N% F, C/ n+ U  Z

}6 E4 b- ~6 Z\\" h

4 w# }2 [, u. `! d

return 0; 
, I6 l$ {. k2 k! y( Y3 \
}& |+ c: J/ G' d/ N- T% j\\" S

int get_Top(int days, int kind)8 _\\" g; z; f# _; k

{; [* M( r9 H3 _

switch(kind): U. n* M; g+ @$ Q& A

{6 D) ]6 }0 t1 l' p1 z0 r5 L1 K

case 0:4 v4 k# D: v* c4 @2 W

return days%23;: w0 n. o+ N! v\\" D

case 1:! c7 \7 ^6 U! n3 ~5 R2 w) o

return days%28;! M- ]( M# U( Q1 L* v4 c* R

case 2:+ P, V\\" f9 B& r

return days%33;- i, X/ P- e0 ]* S0 M1 V

} 
6 ~- }9 V8 N: ?/ R$ f) }
 return 0;: {3 K) H* a- w- b

} 
6 z; ]: l# c# \( x% Z
 
2 ?2 W: N+ X1 Z& U3 D
int triple(int ptop, int etop, int itop, int cur) 
$ P8 c\\" X' |: l7 x$ G7 |
{ 
7 B! ^( D: E* f\\" q
 float x, y, z; 
6 q' ^, E7 q; W0 ?3 C
 int m1   = etop-ptop-5,  J9 J' ^7 r) `1 S. S, O

m2   = itop-etop-5;% W+ i$ q8 r# l- O3 l\\" d% x

int temp = 0;0 u# T. F$ ?4 Z\\" M

for(z=2;  (itop+(z-1)*33) <= 21252;  z++)4 u# d, X1 a8 P

{ 
8 K6 @! h/ U5 v# R0 Z* a
  temp = m2 + 33*z;& ^6 ?% ]7 I# n7 T/ x

y = (float)temp/28;7 Z% k' {; z5 B! a( B

x = (float)(m1+temp)/23;1 a/ D; ~: ~0 M: p0 ^+ G2 d

if(x>0 && y>0 && is_Integer(x) && is_Integer(y)) 
  ^% b* }1 @. }; G
   return itop + (z-1)*33 -cur; 
$ o3 ?9 l7 ~+ N1 R; \) M
 }7 @- j+ K; s5 n

return 0;; ~' n2 D0 h! a4 p9 {\\" p

} 
1 c) z4 A7 q\\" {4 p
! c% M' K! n( V6 V/ b

bool is_Integer(float n) 
+ c( U/ V3 X; `) J! |
{' p\\" w/ b# Y5 Q

if( (int)n - n ==0)7 G) W2 Z) @, [5 z\\" D: [

return true; 
- X. N3 `/ m9 i3 W  T
 else 
0 o2 x\\" H- g+ Z, g4 T
  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;
}