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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;! u* ]1 _1 n' A
#define PH 07 M4 ?, R( P0 b0 V1 y a- r1 P) W
#define EM 1
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#define IN 2& R$ B0 ` D* P- ]: o8 s( z
int get_Top(int days, int kind);# T1 m+ m! l1 U. ]
int triple(int ptop, int etop, int itop, int cur);
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bool is_Integer(float n);0 i! H. N& J\\" I$ i. }\\" E1 j$ _$ Y/ j
int main()+ `2 g* O' C: x
{
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int ptop, etop, itop;; ]+ J4 w2 d; Y
int pdays, edays, idays, curdays;
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int count =0;
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while(cin >> pdays >> edays >> idays >> curdays) a6 c* n/ U; b. {
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;
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count++;
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ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);( c; l3 G) G: x; c% D) t
itop = get_Top(idays,IN);
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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)
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{
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case 0:
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return days%23;+ ?7 g7 y1 W4 f' G J* A& G
case 1:
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return days%28;$ ?% }/ d s: i9 J7 f
case 2:
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return days%33;
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}
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return 0;5 p' e. ]7 N8 n, B( W8 T. P
}
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int triple(int ptop, int etop, int itop, int cur)2 i' V/ d: @; h! ?: Z
{
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float x, y, z;
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int m1 = etop-ptop-5,. E( `* K+ C8 S6 e/ k5 }( h8 S
m2 = itop-etop-5;* h8 w: a# _$ V/ w
int temp = 0;* C z8 S# X+ K\\" U
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;
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x = (float)(m1+temp)/23;* L+ U2 j8 e# l' s+ H: Z
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;+ z% C# Y! u1 @3 h
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bool is_Integer(float n)
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if( (int)n - n ==0)
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return true;' l% B9 F7 b% K% c. E
else
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return false;$ I3 Q. J* m0 ~\\" w( |* ^/ X
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#include <iostream> 
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using namespace std;! u* ]1 _1 n' A

#define PH      07 M4 ?, R( P0 b0 V1 y  a- r1 P) W

#define EM      1 
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#define IN      2& R$ B0 `  D* P- ]: o8 s( z

int get_Top(int days, int kind);# T1 m+ m! l1 U. ]

int triple(int ptop, int etop, int itop, int cur); 
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bool is_Integer(float n);0 i! H. N& J\\" I$ i. }\\" E1 j$ _$ Y/ j

int main()+ `2 g* O' C: x

{ 
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 int ptop,  etop,  itop;; ]+ J4 w2 d; Y

int pdays, edays, idays, curdays; 
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 int count =0; 
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 while(cin >> pdays >> edays >> idays >> curdays)  a6 c* n/ U; b. {

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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
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   break; 
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  count++; 
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  ptop = get_Top(pdays,PH); 
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     etop = get_Top(edays,EM);( c; l3 G) G: x; c% D) t

itop = get_Top(idays,IN); 
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  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) 
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 { 
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 case 0: 
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  return days%23;+ ?7 g7 y1 W4 f' G  J* A& G

case 1: 
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  return days%28;$ ?% }/ d  s: i9 J7 f

case 2: 
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  return days%33; 
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 } 
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 return 0;5 p' e. ]7 N8 n, B( W8 T. P

} 
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int triple(int ptop, int etop, int itop, int cur)2 i' V/ d: @; h! ?: Z

{ 
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 float x, y, z; 
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 int m1   = etop-ptop-5,. E( `* K+ C8 S6 e/ k5 }( h8 S

m2   = itop-etop-5;* h8 w: a# _$ V/ w

int temp = 0;* C  z8 S# X+ K\\" U

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; 
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  x = (float)(m1+temp)/23;* L+ U2 j8 e# l' s+ H: Z

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;+ z% C# Y! u1 @3 h

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bool is_Integer(float n) 
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{- J5 b$ o! y8 c+ P

if( (int)n - n ==0) 
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  return true;' l% B9 F7 b% K% c. E

else 
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  return false;$ I3 Q. J* m0 ~\\" w( |* ^/ X

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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;
}