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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 02 ]- X/ R$ h2 B W5 `2 v
#define EM 1! o: J w6 I2 n7 h' C
#define IN 2
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int get_Top(int days, int kind);; v# P, k4 b9 @- e7 N3 g
int triple(int ptop, int etop, int itop, int cur);
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bool is_Integer(float n);: l- d0 r% w, Y& k
int main()$ {, f+ H7 }4 a Z+ W8 H
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int ptop, etop, itop;
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int pdays, edays, idays, curdays;
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int count =0;6 M6 C8 g, @9 Y& |% S1 I
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while(cin >> pdays >> edays >> idays >> curdays)# A( H0 v5 h- \. _2 D
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if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
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break;
count++;
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ptop = get_Top(pdays,PH);
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etop = get_Top(edays,EM);5 [1 c F3 J4 ]9 t3 q
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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}
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return 0;/ t/ Q, t\\" _( i' h1 w9 n
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int get_Top(int days, int kind)' E( @9 D) h& m8 M+ a' R
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switch(kind)
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{
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case 0:
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return days%23;
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case 1:
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return days%28;
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case 2:- q. l* X# Z2 Q
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;
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int m1 = etop-ptop-5,
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m2 = itop-etop-5;& r* h3 y+ F/ g7 ^9 L% ]! g% E2 J
int temp = 0;
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for(z=2; (itop+(z-1)*33) <= 21252; z++)& V5 S7 c( v( ~( m: n7 f
{
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temp = m2 + 33*z;, S7 U, _9 I2 }7 \
y = (float)temp/28;2 X8 }\\" Z6 E) U+ e# u9 }4 R
x = (float)(m1+temp)/23;
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if(x>0 && y>0 && is_Integer(x) && is_Integer(y))$ l$ u9 Y\\" {6 {7 n- y+ O# k- q
return itop + (z-1)*33 -cur;
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return 0;( I3 x4 |7 i2 ^9 ?3 [4 S2 e% v# W
}
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bool is_Integer(float n)/ @2 D$ W$ |9 h& I* k: |
{
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if( (int)n - n ==0)
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return true;
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else- d4 l1 z+ n/ I8 S/ f
return false;
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#include <iostream> 
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using namespace std; 
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#define PH      02 ]- X/ R$ h2 B  W5 `2 v

#define EM      1! o: J  w6 I2 n7 h' C

#define IN      2 
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int get_Top(int days, int kind);; v# P, k4 b9 @- e7 N3 g

int triple(int ptop, int etop, int itop, int cur); 
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bool is_Integer(float n);: l- d0 r% w, Y& k

int main()$ {, f+ H7 }4 a  Z+ W8 H

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int ptop,  etop,  itop; 
* [$ u: F- h$ L$ I( P
 int pdays, edays, idays, curdays; 
& w$ t! n% I3 p
 int count =0;6 M6 C8 g, @9 Y& |% S1 I

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 while(cin >> pdays >> edays >> idays >> curdays)# A( H0 v5 h- \. _2 D

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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);5 [1 c  F3 J4 ]9 t3 q

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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 } 
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 return 0;/ t/ Q, t\\" _( i' h1 w9 n

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int get_Top(int days, int kind)' E( @9 D) h& m8 M+ a' R

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switch(kind) 
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 { 
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 case 0: 
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  return days%23; 
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 case 1: 
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  return days%28; 
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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; 
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 int m1   = etop-ptop-5, 
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  m2   = itop-etop-5;& r* h3 y+ F/ g7 ^9 L% ]! g% E2 J

int temp = 0; 
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 for(z=2;  (itop+(z-1)*33) <= 21252;  z++)& V5 S7 c( v( ~( m: n7 f

{ 
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  temp = m2 + 33*z;, S7 U, _9 I2 }7 \

y = (float)temp/28;2 X8 }\\" Z6 E) U+ e# u9 }4 R

x = (float)(m1+temp)/23; 
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  if(x>0 && y>0 && is_Integer(x) && is_Integer(y))$ l$ u9 Y\\" {6 {7 n- y+ O# k- q

return itop + (z-1)*33 -cur; 
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return 0;( I3 x4 |7 i2 ^9 ?3 [4 S2 e% v# W

} 
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bool is_Integer(float n)/ @2 D$ W$ |9 h& I* k: |

{ 
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 if( (int)n - n ==0) 
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  return true; 
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 else- d4 l1 z+ n/ I8 S/ f

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;
}