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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;\\" a6 B; N! `+ d2 v. B; c4 M\\" J, ]/ z
#define PH 0
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#define EM 1; v' N5 M9 C) h1 q% w; p! K! W% y
#define IN 2
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int get_Top(int days, int kind);& \: q& U2 c$ ~. ?: U6 ?+ I
int triple(int ptop, int etop, int itop, int cur);
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bool is_Integer(float n);3 g8 n) z% {9 a+ |. |$ n
int main()) Z' \1 E: k) B, L9 s3 M9 \
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int ptop, etop, itop;
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int pdays, edays, idays, curdays;
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int count =0;4 ?4 _& p+ V2 [( E, w! A G; n# l% V
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while(cin >> pdays >> edays >> idays >> curdays)6 |7 {6 i8 i3 F$ J
{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;9 {. Q+ | B% p' t1 \! b
count++;
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ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);
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itop = get_Top(idays,IN);) u7 t1 B8 T% B\\" R/ \+ G
cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";
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}
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return 0;) [* ]1 Z- C) c% V
}
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int get_Top(int days, int kind)
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switch(kind)
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case 0:# D' @+ n\\" \1 W [* t! c
return days%23;7 `2 y! _! H- Z
case 1:* X$ q\\" J [& u- G
return days%28;
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case 2:
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return days%33;) F; ~& D0 k2 d4 Y4 } P: r
}
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return 0;4 A5 G9 R: O1 f/ V; S* J
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int triple(int ptop, int etop, int itop, int cur)6 Y- f) n8 i\\" v6 Y! X9 ?
{% p; ~6 b4 |6 ]2 k3 G# T4 p
float x, y, z;
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int m1 = etop-ptop-5,- ]0 {9 V, ^% A1 X4 l! Z
m2 = itop-etop-5;4 `\\" N: t# D1 j, D\\" [7 _1 O: s
int temp = 0;- h8 ~; {0 T, F8 M8 @
for(z=2; (itop+(z-1)*33) <= 21252; z++)
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{
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temp = m2 + 33*z;2 n% i+ G9 V* Q, g
y = (float)temp/28;2 G+ c: U- q) s\\" m: L\\" d
x = (float)(m1+temp)/23;0 D& C7 y) w5 ~8 }9 x1 o
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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return 0;
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}
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bool is_Integer(float n)7 b8 {7 Y+ l5 o% l' R
{
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if( (int)n - n ==0)
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return true;
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else5 y) ^0 Q. h! h# n P2 {9 Y
return false;
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#include <iostream> 
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using namespace std;\\" a6 B; N! `+ d2 v. B; c4 M\\" J, ]/ z

#define PH      0 
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#define EM      1; v' N5 M9 C) h1 q% w; p! K! W% y

#define IN      2 
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int get_Top(int days, int kind);& \: q& U2 c$ ~. ?: U6 ?+ I

int triple(int ptop, int etop, int itop, int cur); 
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bool is_Integer(float n);3 g8 n) z% {9 a+ |. |$ n

int main()) Z' \1 E: k) B, L9 s3 M9 \

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int ptop,  etop,  itop; 
& |8 F6 W\\" W\\" `5 \1 T$ I( c' _
 int pdays, edays, idays, curdays; 
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 int count =0;4 ?4 _& p+ V2 [( E, w! A  G; n# l% V

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while(cin >> pdays >> edays >> idays >> curdays)6 |7 {6 i8 i3 F$ J

{ 
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  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1) 
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   break;9 {. Q+ |  B% p' t1 \! b

count++; 
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  ptop = get_Top(pdays,PH); 
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     etop = get_Top(edays,EM); 
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     itop = get_Top(idays,IN);) u7 t1 B8 T% B\\" R/ \+ G

cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days."; 
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 } 
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 return 0;) [* ]1 Z- C) c% V

} 
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int get_Top(int days, int kind) 
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switch(kind) 
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case 0:# D' @+ n\\" \1 W  [* t! c

return days%23;7 `2 y! _! H- Z

case 1:* X$ q\\" J  [& u- G

return days%28; 
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 case 2: 
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  return days%33;) F; ~& D0 k2 d4 Y4 }  P: r

} 
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 return 0;4 A5 G9 R: O1 f/ V; S* J

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int triple(int ptop, int etop, int itop, int cur)6 Y- f) n8 i\\" v6 Y! X9 ?

{% p; ~6 b4 |6 ]2 k3 G# T4 p

float x, y, z; 
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 int m1   = etop-ptop-5,- ]0 {9 V, ^% A1 X4 l! Z

m2   = itop-etop-5;4 `\\" N: t# D1 j, D\\" [7 _1 O: s

int temp = 0;- h8 ~; {0 T, F8 M8 @

for(z=2;  (itop+(z-1)*33) <= 21252;  z++) 
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 { 
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  temp = m2 + 33*z;2 n% i+ G9 V* Q, g

y = (float)temp/28;2 G+ c: U- q) s\\" m: L\\" d

x = (float)(m1+temp)/23;0 D& C7 y) w5 ~8 }9 x1 o

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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return 0; 
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} 
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bool is_Integer(float n)7 b8 {7 Y+ l5 o% l' R

{ 
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 if( (int)n - n ==0) 
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  return true; 
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 else5 y) ^0 Q. h! h# n  P2 {9 Y

return false; 
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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;
}