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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 08 j# ]4 ]! U! l9 o. N* B
#define EM 1
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#define IN 21 K( c& e+ \; `2 G2 d, R
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);
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int main()
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{
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int ptop, etop, itop;4 h6 m$ |7 \% d- \# T; n* r
int pdays, edays, idays, curdays;
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int count =0;
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while(cin >> pdays >> edays >> idays >> curdays)* ]% w) l8 v. B/ t* t
{
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)+ d' ?/ _! {\\" ]4 n
break;; E5 O/ ?# i4 l+ A3 z- m5 }
count++;9 \2 J\\" L+ r/ Q% E
ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);4 M% ] m# D0 r
itop = get_Top(idays,IN);$ }) T& E1 U% G0 u; y. b2 f* A
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): K$ p+ y8 g\\" y2 ?) R
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switch(kind)
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{
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case 0:0 M- a- `. P\\" \! C
return days%23;
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case 1:8 t1 U/ z9 I. s$ \6 [3 g9 v# c
return days%28;
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case 2:6 U( o% L s! z- V
return days%33;
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}
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return 0;
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int triple(int ptop, int etop, int itop, int cur)
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float x, y, z;1 x0 ]7 e! N. W% Y
int m1 = etop-ptop-5,
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m2 = itop-etop-5;5 L* D {& K N* R6 b' e
int temp = 0;/ }% ~- l- w% X& U
for(z=2; (itop+(z-1)*33) <= 21252; z++)
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temp = m2 + 33*z; F8 S+ q& B) g% t; A
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;2 z. \( o% Z' x+ `+ s; I
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return 0;
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bool is_Integer(float n)4 z0 B% _5 ^! Z8 E1 y
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if( (int)n - n ==0). A7 b9 q* b7 ^! P }% v
return true;# Q( V6 M( ^: W, L( Z6 n: G7 d( _
else
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return false;( K, ~( _; ^6 p Z9 y8 Z$ U' V
}
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#include <iostream> 
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using namespace std; 
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#define PH      08 j# ]4 ]! U! l9 o. N* B

#define EM      1 
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#define IN      21 K( c& e+ \; `2 G2 d, R

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); 
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int main() 
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{ 
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 int ptop,  etop,  itop;4 h6 m$ |7 \% d- \# T; n* r

int pdays, edays, idays, curdays; 
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 int count =0; 
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 while(cin >> pdays >> edays >> idays >> curdays)* ]% w) l8 v. B/ t* t

{ 
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  if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)+ d' ?/ _! {\\" ]4 n

break;; E5 O/ ?# i4 l+ A3 z- m5 }

count++;9 \2 J\\" L+ r/ Q% E

ptop = get_Top(pdays,PH); 
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     etop = get_Top(edays,EM);4 M% ]  m# D0 r

itop = get_Top(idays,IN);$ }) T& E1 U% G0 u; y. b2 f* A

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): K$ p+ y8 g\\" y2 ?) R

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switch(kind) 
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 { 
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 case 0:0 M- a- `. P\\" \! C

return days%23; 
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 case 1:8 t1 U/ z9 I. s$ \6 [3 g9 v# c

return days%28; 
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 case 2:6 U( o% L  s! z- V

return days%33; 
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 } 
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 return 0; 
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int triple(int ptop, int etop, int itop, int cur) 
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float x, y, z;1 x0 ]7 e! N. W% Y

int m1   = etop-ptop-5, 
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  m2   = itop-etop-5;5 L* D  {& K  N* R6 b' e

int temp = 0;/ }% ~- l- w% X& U

for(z=2;  (itop+(z-1)*33) <= 21252;  z++) 
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temp = m2 + 33*z;  F8 S+ q& B) g% t; A

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;2 z. \( o% Z' x+ `+ s; I

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return 0; 
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bool is_Integer(float n)4 z0 B% _5 ^! Z8 E1 y

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if( (int)n - n ==0). A7 b9 q* b7 ^! P  }% v

return true;# Q( V6 M( ^: W, L( Z6 n: G7 d( _

else 
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  return false;( K, ~( _; ^6 p  Z9 y8 Z$ U' V

} 
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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;
}