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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;
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#define PH 0
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#define EM 1
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#define IN 2
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int get_Top(int days, int kind);6 i+ \3 T! |8 [3 v. ^9 @
int triple(int ptop, int etop, int itop, int cur);
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bool is_Integer(float n);
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int main()0 w+ z! x2 R2 h( {7 k
{
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int ptop, etop, itop;' v7 _! |0 [. W- s _. y9 n
int pdays, edays, idays, curdays;
% a: i. K, m. X0 q# ]. ?) l
int count =0;4 a( V: T5 a8 o9 g5 m* G
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while(cin >> pdays >> edays >> idays >> curdays). ]. G( w/ b\\" o
{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;$ ~ A; U1 a* B
count++;
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ptop = get_Top(pdays,PH);7 F/ x* P6 K) c# L9 t$ V
etop = get_Top(edays,EM);
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itop = get_Top(idays,IN);9 K% V+ W h. v% a* m6 U\\" U& I
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";1 n& U9 D0 l( I6 n3 M6 d9 n% e
}
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return 0;\\" v! m1 |& L. V8 t
}
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int get_Top(int days, int kind)
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{
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switch(kind)
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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;
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case 2:$ S7 e1 H' e2 ]
return days%33;5 B# m) S3 Y( W- ]( G* u\\" {& z- |* l
}
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return 0;
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}\\" q/ C3 ~( C' `5 g\\" Q
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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-5,6 W4 Z+ I0 Z, W- ~* I
m2 = itop-etop-5;, ~8 N% A# c0 e! _6 Z$ E
int temp = 0;+ _$ T+ w8 D; M; q
for(z=2; (itop+(z-1)*33) <= 21252; z++)5 V2 j; Y8 W1 ], B( L3 V
{
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temp = m2 + 33*z;
y = (float)temp/28;. N' N3 g+ I+ X) {
x = (float)(m1+temp)/23;
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if(x>0 && y>0 && is_Integer(x) && is_Integer(y))4 Y: s- ^' f# l% `+ Q% j% l f; r
return itop + (z-1)*33 -cur;( l3 w3 l8 R4 d& T
}
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return 0;
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bool is_Integer(float n)
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{
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if( (int)n - n ==0)% p! E% `& z$ o1 @; _2 S5 r
return true;4 O2 X) i8 U7 S
else
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return false;+ }\\" w2 n ]% j$ S! g
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#include <iostream> 
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using namespace std; 
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#define PH      0 
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#define EM      1 
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#define IN      2 
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int get_Top(int days, int kind);6 i+ \3 T! |8 [3 v. ^9 @

int triple(int ptop, int etop, int itop, int cur); 
\\" x% A  s4 t$ y: b  y\\" Z) w+ j
bool is_Integer(float n); 
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int main()0 w+ z! x2 R2 h( {7 k

{ 
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 int ptop,  etop,  itop;' v7 _! |0 [. W- s  _. y9 n

int pdays, edays, idays, curdays; 
% a: i. K, m. X0 q# ]. ?) l
 int count =0;4 a( V: T5 a8 o9 g5 m* G

& h, r, x) _) l9 ?# D# ^

while(cin >> pdays >> edays >> idays >> curdays). ]. G( w/ b\\" o

{ 
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  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
\\" C% d* s0 g% T  r* g. u
   break;$ ~  A; U1 a* B

count++; 
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  ptop = get_Top(pdays,PH);7 F/ x* P6 K) c# L9 t$ V

etop = get_Top(edays,EM); 
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     itop = get_Top(idays,IN);9 K% V+ W  h. v% a* m6 U\\" U& I

cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";1 n& U9 D0 l( I6 n3 M6 d9 n% e

} 
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 return 0;\\" v! m1 |& L. V8 t

} 
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int get_Top(int days, int kind) 
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{ 
( z6 @0 Y* D) V0 c
 switch(kind) 
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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; 
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 case 2:$ S7 e1 H' e2 ]

return days%33;5 B# m) S3 Y( W- ]( G* u\\" {& z- |* l

} 
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 return 0; 
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}\\" q/ C3 ~( C' `5 g\\" Q

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int triple(int ptop, int etop, int itop, int cur) 
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{3 N9 o. E: y2 T% g

float x, y, z; 
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 int m1   = etop-ptop-5,6 W4 Z+ I0 Z, W- ~* I

m2   = itop-etop-5;, ~8 N% A# c0 e! _6 Z$ E

int temp = 0;+ _$ T+ w8 D; M; q

for(z=2;  (itop+(z-1)*33) <= 21252;  z++)5 V2 j; Y8 W1 ], B( L3 V

{ 
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  temp = m2 + 33*z; 
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  y = (float)temp/28;. N' N3 g+ I+ X) {

x = (float)(m1+temp)/23; 
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  if(x>0 && y>0 && is_Integer(x) && is_Integer(y))4 Y: s- ^' f# l% `+ Q% j% l  f; r

return itop + (z-1)*33 -cur;( l3 w3 l8 R4 d& T

} 
$ F- d! i, M# w# B  q
 return 0; 
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}0 t! t& w. i) s

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bool is_Integer(float n) 
3 m* X% o\\" z, V# U- V* o
{ 
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 if( (int)n - n ==0)% p! E% `& z$ o1 @; _2 S5 r

return true;4 O2 X) i8 U7 S

else 
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  return false;+ }\\" w2 n  ]% j$ S! g

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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;
}