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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/ Y\\" a, ?, H7 a! w4 A: s
#define EM 1
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#define IN 2
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int get_Top(int days, int kind);; S3 Y: `7 _$ q2 n
int triple(int ptop, int etop, int itop, int cur);. g0 M; \& G0 [* K6 E0 O1 y0 ?' i\\" ?
bool is_Integer(float n);
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int main()
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int ptop, etop, itop;& v& s x6 ?' [- ]\\" k( W\\" k! F% O& D
int pdays, edays, idays, curdays; y* r6 `- u$ Q: H: P
int count =0;
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while(cin >> pdays >> edays >> idays >> curdays); \7 o/ g. D6 k9 T9 k( [
{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;* \8 u/ c$ z/ ]9 k
count++;$ ?' ^- C\\" l @+ K' {/ q
ptop = get_Top(pdays,PH);' ^. q7 D0 D% @1 Z. t
etop = get_Top(edays,EM);4 D6 x9 o/ h/ E6 d2 O9 p
itop = get_Top(idays,IN);* G( l) S$ k4 ~; ]. j$ x
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";
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return 0;
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int get_Top(int days, int kind)
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{
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switch(kind)* s7 P2 E' b( y1 x( x v
{
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case 0:& i7 z$ g5 C) U; D* s1 ]2 J& ~
return days%23;
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case 1:
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return days%28;
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case 2:
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return days%33;. t# W2 @& w! q\\" @. A' y+ u/ t$ S i
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return 0;
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}
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int triple(int ptop, int etop, int itop, int cur)3 n0 s3 x. Q8 L- R; p
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float x, y, z;
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int m1 = etop-ptop-5,. i3 N) Q1 U7 ]4 I) _
m2 = itop-etop-5;! X5 F4 ?% @& K$ W, |# z
int temp = 0;
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for(z=2; (itop+(z-1)*33) <= 21252; z++)' U! g' y- ^% D7 l' X. Y( z
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temp = m2 + 33*z;
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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;$ Q! G/ G: o3 H; L4 }4 D
}
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bool is_Integer(float n)- \\\" J# B/ K: e+ X5 E
{
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if( (int)n - n ==0)
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return true;
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else
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return false;
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}
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#include <iostream> 
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using namespace std; 
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#define PH      0/ Y\\" a, ?, H7 a! w4 A: s

#define EM      1 
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#define IN      2 
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int get_Top(int days, int kind);; S3 Y: `7 _$ q2 n

int triple(int ptop, int etop, int itop, int cur);. g0 M; \& G0 [* K6 E0 O1 y0 ?' i\\" ?

bool is_Integer(float n); 
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int main() 
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int ptop,  etop,  itop;& v& s  x6 ?' [- ]\\" k( W\\" k! F% O& D

int pdays, edays, idays, curdays;  y* r6 `- u$ Q: H: P

int count =0; 
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 \\" r\\" h! Q: W$ G( l. Z( T

while(cin >> pdays >> edays >> idays >> curdays); \7 o/ g. D6 k9 T9 k( [

{ 
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  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
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   break;* \8 u/ c$ z/ ]9 k

count++;$ ?' ^- C\\" l  @+ K' {/ q

ptop = get_Top(pdays,PH);' ^. q7 D0 D% @1 Z. t

etop = get_Top(edays,EM);4 D6 x9 o/ h/ E6 d2 O9 p

itop = get_Top(idays,IN);* G( l) S$ k4 ~; ]. j$ x

cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days."; 
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return 0; 
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int get_Top(int days, int kind) 
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{ 
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 switch(kind)* s7 P2 E' b( y1 x( x  v

{ 
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 case 0:& i7 z$ g5 C) U; D* s1 ]2 J& ~

return days%23; 
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 case 1: 
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  return days%28; 
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 case 2: 
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  return days%33;. t# W2 @& w! q\\" @. A' y+ u/ t$ S  i

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return 0; 
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} 
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int triple(int ptop, int etop, int itop, int cur)3 n0 s3 x. Q8 L- R; p

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float x, y, z; 
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 int m1   = etop-ptop-5,. i3 N) Q1 U7 ]4 I) _

m2   = itop-etop-5;! X5 F4 ?% @& K$ W, |# z

int temp = 0; 
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 for(z=2;  (itop+(z-1)*33) <= 21252;  z++)' U! g' y- ^% D7 l' X. Y( z

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temp = m2 + 33*z; 
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  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;$ Q! G/ G: o3 H; L4 }4 D

} 
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bool is_Integer(float n)- \\\" J# B/ K: e+ X5 E

{ 
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 if( (int)n - n ==0) 
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  return true; 
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 else 
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  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;
}