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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! [: s2 J. P5 f% V& [; p- {5 Q
#define EM 1; y0 D! L\\" j) z1 N' T
#define IN 2
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int get_Top(int days, int kind);
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int triple(int ptop, int etop, int itop, int cur);
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bool is_Integer(float n);/ j) A; Y* W5 i% A6 p
int main()8 X1 h- j% y N/ @! d+ s! _+ d\\" p
{
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int ptop, etop, itop;* N# r. U4 r4 f1 Y. R( e. f) R
int pdays, edays, idays, curdays;
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int count =0;$ D- I& W( ] q; c: W* b& C! b
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while(cin >> pdays >> edays >> idays >> curdays)2 o& B% a$ A2 I
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)* B1 _) K' z/ V: O: w w6 m
break;
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count++;5 @# [4 A3 \5 P E% |, R
ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);( Q, }% g+ h4 t0 ^
itop = get_Top(idays,IN);
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cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";% C* V7 }: J$ k- {
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return 0;8 a' O$ D g+ ~5 x1 D
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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;$ a% D' X5 ]& K* E1 [$ m
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return days%28; s% j/ B+ d$ ]: `. q! |
case 2:
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return days%33;7 p ?7 l( ]' i0 i U/ @\\" l
}; @. w# w3 J6 M/ R! ]4 h
return 0;
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int triple(int ptop, int etop, int itop, int cur)
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float x, y, z;- a3 Q7 x6 K% p# b6 n/ I
int m1 = etop-ptop-5,
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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;
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y = (float)temp/28;9 o4 M) o, {/ c2 u2 Y5 m$ h2 _, J
x = (float)(m1+temp)/23;
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if(x>0 && y>0 && is_Integer(x) && is_Integer(y))- F S* E8 ^; N$ e2 V
return itop + (z-1)*33 -cur;
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}
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return 0;5 D! n- r& R: z* D- p) S
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bool is_Integer(float n)
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if( (int)n - n ==0)1 \# {, n4 ]$ Y: R7 o0 T
return true;
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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! [: s2 J. P5 f% V& [; p- {5 Q

#define EM      1; y0 D! L\\" j) z1 N' T

#define IN      2 
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int get_Top(int days, int kind); 
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int triple(int ptop, int etop, int itop, int cur); 
% i2 s; q8 O1 z3 i
bool is_Integer(float n);/ j) A; Y* W5 i% A6 p

int main()8 X1 h- j% y  N/ @! d+ s! _+ d\\" p

{ 
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 int ptop,  etop,  itop;* N# r. U4 r4 f1 Y. R( e. f) R

int pdays, edays, idays, curdays; 
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 int count =0;$ D- I& W( ]  q; c: W* b& C! b

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while(cin >> pdays >> edays >> idays >> curdays)2 o& B% a$ A2 I

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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)* B1 _) K' z/ V: O: w  w6 m

break; 
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  count++;5 @# [4 A3 \5 P  E% |, R

ptop = get_Top(pdays,PH); 
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     etop = get_Top(edays,EM);( Q, }% g+ h4 t0 ^

itop = get_Top(idays,IN); 
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  cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";% C* V7 }: J$ k- {

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return 0;8 a' O$ D  g+ ~5 x1 D

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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;$ a% D' X5 ]& K* E1 [$ m

case 1:+ t9 g. L7 k4 a# ^& @- h% G$ d. Z! A

return days%28;  s% j/ B+ d$ ]: `. q! |

case 2: 
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  return days%33;7 p  ?7 l( ]' i0 i  U/ @\\" l

}; @. w# w3 J6 M/ R! ]4 h

return 0; 
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int triple(int ptop, int etop, int itop, int cur) 
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float x, y, z;- a3 Q7 x6 K% p# b6 n/ I

int m1   = etop-ptop-5, 
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  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; 
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  y = (float)temp/28;9 o4 M) o, {/ c2 u2 Y5 m$ h2 _, J

x = (float)(m1+temp)/23; 
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  if(x>0 && y>0 && is_Integer(x) && is_Integer(y))- F  S* E8 ^; N$ e2 V

return itop + (z-1)*33 -cur; 
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 } 
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 return 0;5 D! n- r& R: z* D- p) S

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bool is_Integer(float n) 
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if( (int)n - n ==0)1 \# {, n4 ]$ Y: R7 o0 T

return true; 
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return false; 
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} 
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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;
}