POJ-1006问题

maybe_madio

POJ-1006
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在本地VC6.0下测试，得出正确结果， 为什么提交后总是Wrong Answer
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) O& A1 {# H" T* Z, q/ _6 h* J现将代码贴出，请帮忙看看：

2. #include <iostream>
   
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4. using namespace std;
   
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6. #define PH      09 _+ p; K+ F$ o  ~) N, o
   
8. #define EM      1
   
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10. #define IN      2
    
11. ' ~4 S, d9 l. B9 l1 `
12. int get_Top(int days, int kind);
    
13. : t0 `9 c( q  F( ]
14. int triple(int ptop, int etop, int itop, int cur);; w, C3 P; ^, ?4 A% I; y
    
16. bool is_Integer(float n);8 f% e; S) C* I/ w- t1 q\\" U1 _9 `
    
18. int main()
    
19. 4 ^& p1 x3 l. U9 ?
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22. int ptop,  etop,  itop;& c: q9 x9 x2 `/ ?( v3 U\\" G/ \* w
    
24. int pdays, edays, idays, curdays;& u4 q8 ?( t# f) p1 T, h. E
    
26. int count =0;
    
27. $ V% j4 W. w; J: b! o1 y5 z% s1 r# i9 q
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30. while(cin >> pdays >> edays >> idays >> curdays). V# D. z! V* Q2 E/ c
    
32. {\\" z% x* e+ n4 |% Y/ D/ _) B2 n
    
34.   if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)4 q  m7 G' O) J( Z! }
    
36.    break;
    
37. # ]; {9 x6 f6 v' D3 s
38.   count++;2 {5 P; Q1 j' A\\" {( R4 b
    
40.   ptop = get_Top(pdays,PH);$ U' q+ k7 ?. `: U
    
42.      etop = get_Top(edays,EM);1 ]. x) G) m4 b3 c+ M
    
44.      itop = get_Top(idays,IN);
    
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46.   cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";
    
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50. 
    
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52. return 0;' P8 S8 t) J4 n2 p
    
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56. int get_Top(int days, int kind)
    
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60. switch(kind)
    
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62. {' }. W8 g* ]1 C# s
    
64. case 0:! F, T' M0 |) \5 R0 R2 I+ m
    
66.   return days%23;+ E\\" `& r8 k/ D0 m1 T, j
    
68. case 1:% ]5 X4 Y2 o+ L8 s\\" ]
    
70.   return days%28;
    
71. ! l' H# y. ~7 P0 v7 c8 t, l; L7 S
72. case 2:
    
73. % }/ q/ ^3 p% H
74.   return days%33;
    
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78. return 0;
    
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82. 
    
83. 5 j- U/ r\\" ~8 q; W
84. int triple(int ptop, int etop, int itop, int cur)5 g3 T. M3 A3 c
    
86. {
    
87. , Q8 r- }4 x: x
88. float x, y, z;
    
89. + u+ ]4 w, g% C& }. r+ F- f4 h
90. int m1   = etop-ptop-5,
    
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92.   m2   = itop-etop-5;; f0 V- u: Z7 D
    
94. int temp = 0;
    
95.   e3 I- h9 N$ i! S
96. for(z=2;  (itop+(z-1)*33) <= 21252;  z++)
    
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100.   temp = m2 + 33*z;
     
101. $ l  x8 J\\" F5 x- L7 d1 B! Q4 o  |
102.   y = (float)temp/28;2 t8 u, k! b, ^
     
104.   x = (float)(m1+temp)/23;
     
105. 0 Y7 N- [' @& ~2 q
106.   if(x>0 && y>0 && is_Integer(x) && is_Integer(y))
     
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108.    return itop + (z-1)*33 -cur;7 d\\" D% X. y! {* G
     
110. }
     
111. \\" D0 X( v$ h  z% W\\" J/ R7 ]
112. return 0;
     
113.   h& g* q, {  a
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118. bool is_Integer(float n)
     
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120. {/ h3 G5 L0 G4 p\\" U
     
122. if( (int)n - n ==0)
     
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124.   return true;
     
125. + u: E# P- I/ k! c$ v
126. else
     
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128.   return false;
     
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130. }4 j: I% \& @7 q* X+ ]8 z
     
132. 
     
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#include <iostream><br /> <span style=\\"display:none\\">/ x4 n1 l$ v' E# z$ q6 D# z+ N# _</span> using namespace std;<br /> <span style=\\"display:none\\">- W$ R. o* b3 ^& p+ ^\\" n5 e) y, D</span> #define PH      0<font class=\\"jammer\\">9 _+ p; K+ F$ o  ~) N, o</font><br /> #define EM      1<br /> <span style=\\"display:none\\">1 l2 z( c, R6 S7 m7 \</span> #define IN      2<br /> <span style=\\"display:none\\">' ~4 S, d9 l. B9 l1 `</span> int get_Top(int days, int kind);<br /> <span style=\\"display:none\\">: t0 `9 c( q  F( ]</span> int triple(int ptop, int etop, int itop, int cur);<font class=\\"jammer\\">; w, C3 P; ^, ?4 A% I; y</font><br /> bool is_Integer(float n);<font class=\\"jammer\\">8 f% e; S) C* I/ w- t1 q\\" U1 _9 `</font><br /> int main()<br /> <span style=\\"display:none\\">4 ^& p1 x3 l. U9 ?</span> {<font class=\\"jammer\\"># L2 N/ _( Y# f! E# J8 I) G. D6 T</font><br /> int ptop,  etop,  itop;<font class=\\"jammer\\">& c: q9 x9 x2 `/ ?( v3 U\\" G/ \* w</font><br /> int pdays, edays, idays, curdays;<font class=\\"jammer\\">& u4 q8 ?( t# f) p1 T, h. E</font><br /> int count =0;<br /> <span style=\\"display:none\\">$ V% j4 W. w; J: b! o1 y5 z% s1 r# i9 q</span> <font class=\\"jammer\\">+ V9 R3 q$ }7 d1 {\\" M</font><br /> while(cin >> pdays >> edays >> idays >> curdays)<font class=\\"jammer\\">. V# D. z! V* Q2 E/ c</font><br /> {<font class=\\"jammer\\">\\" z% x* e+ n4 |% Y/ D/ _) B2 n</font><br />   if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)<font class=\\"jammer\\">4 q  m7 G' O) J( Z! }</font><br />    break;<br /> <span style=\\"display:none\\"># ]; {9 x6 f6 v' D3 s</span>   count++;<font class=\\"jammer\\">2 {5 P; Q1 j' A\\" {( R4 b</font><br />   ptop = get_Top(pdays,PH);<font class=\\"jammer\\">$ U' q+ k7 ?. `: U</font><br />      etop = get_Top(edays,EM);<font class=\\"jammer\\">1 ]. x) G) m4 b3 c+ M</font><br />      itop = get_Top(idays,IN);<br /> <span style=\\"display:none\\">6 T5 p# B, u  A3 ~6 y/ \- ^</span>   cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";<br /> <span style=\\"display:none\\">$ e5 }' [: d' c9 g</span> }<font class=\\"jammer\\">0 Y& j, l  v% y' b3 k</font><br /> <br /> <span style=\\"display:none\\">3 Z/ ^& l- E; f* {</span> return 0;<font class=\\"jammer\\">' P8 S8 t) J4 n2 p</font><br /> }<font class=\\"jammer\\">! X3 O* v\\" p2 d# d1 T</font><br /> int get_Top(int days, int kind)<br /> <span style=\\"display:none\\">* n3 r( [- r* u</span> {<font class=\\"jammer\\">; F& `: f; k! M3 `' L2 f% U</font><br /> switch(kind)<br /> <span style=\\"display:none\\">' ^* {/ {  B. \5 W) H* w</span> {<font class=\\"jammer\\">' }. W8 g* ]1 C# s</font><br /> case 0:<font class=\\"jammer\\">! F, T' M0 |) \5 R0 R2 I+ m</font><br />   return days%23;<font class=\\"jammer\\">+ E\\" `& r8 k/ D0 m1 T, j</font><br /> case 1:<font class=\\"jammer\\">% ]5 X4 Y2 o+ L8 s\\" ]</font><br />   return days%28;<br /> <span style=\\"display:none\\">! l' H# y. ~7 P0 v7 c8 t, l; L7 S</span> case 2:<br /> <span style=\\"display:none\\">% }/ q/ ^3 p% H</span>   return days%33;<br /> <span style=\\"display:none\\">8 z9 j9 e$ [7 r: x/ K0 N' R</span> }<font class=\\"jammer\\">/ U7 j\\" j8 l: l: q4 Z+ y5 i, z3 x</font><br /> return 0;<br /> <span style=\\"display:none\\">' z1 Q2 E, O/ q& d! U' Z- j7 F</span> }<font class=\\"jammer\\">3 ^( Z- n( }6 ~3 G8 [  S5 z6 J0 y</font><br /> <br /> <span style=\\"display:none\\">5 j- U/ r\\" ~8 q; W</span> int triple(int ptop, int etop, int itop, int cur)<font class=\\"jammer\\">5 g3 T. M3 A3 c</font><br /> {<br /> <span style=\\"display:none\\">, Q8 r- }4 x: x</span> float x, y, z;<br /> <span style=\\"display:none\\">+ u+ ]4 w, g% C& }. r+ F- f4 h</span> int m1   = etop-ptop-5,<br /> <span style=\\"display:none\\">/ f3 w/ O1 U& o7 @) Q+ b</span>   m2   = itop-etop-5;<font class=\\"jammer\\">; f0 V- u: Z7 D</font><br /> int temp = 0;<br /> <span style=\\"display:none\\">  e3 I- h9 N$ i! S</span> for(z=2;  (itop+(z-1)*33) <= 21252;  z++)<br /> <span style=\\"display:none\\">2 j& g! A3 \( m% N( c</span> {<font class=\\"jammer\\">7 ^( [& [* n: o, Z8 q0 C</font><br />   temp = m2 + 33*z;<br /> <span style=\\"display:none\\">$ l  x8 J\\" F5 x- L7 d1 B! Q4 o  |</span>   y = (float)temp/28;<font class=\\"jammer\\">2 t8 u, k! b, ^</font><br />   x = (float)(m1+temp)/23;<br /> <span style=\\"display:none\\">0 Y7 N- [' @& ~2 q</span>   if(x>0 && y>0 && is_Integer(x) && is_Integer(y))<br /> <span style=\\"display:none\\">8 K2 n# C& T1 x+ O</span>    return itop + (z-1)*33 -cur;<font class=\\"jammer\\">7 d\\" D% X. y! {* G</font><br /> }<br /> <span style=\\"display:none\\">\\" D0 X( v$ h  z% W\\" J/ R7 ]</span> return 0;<br /> <span style=\\"display:none\\">  h& g* q, {  a</span> }<font class=\\"jammer\\">: T. o: f( X4 D; H4 G</font><br /> <font class=\\"jammer\\">- x1 U1 r$ v% P  o6 l1 F</font><br /> bool is_Integer(float n)<br /> <span style=\\"display:none\\">; B- [2 V  ?' L5 C& z5 |8 A1 s* C</span> {<font class=\\"jammer\\">/ h3 G5 L0 G4 p\\" U</font><br /> if( (int)n - n ==0)<br /> <span style=\\"display:none\\">) ?\\" a2 V3 \5 r! W! K</span>   return true;<br /> <span style=\\"display:none\\">+ u: E# P- I/ k! c$ v</span> else<br /> <span style=\\"display:none\\">* B3 i! S4 s9 r; e  I  E</span>   return false;<br /> <span style=\\"display:none\\"># I8 o/ T( u- M1 Z3 y$ c</span> }<font class=\\"jammer\\">4 j: I% \& @7 q* X+ ]8 z</font><br /> <br /> <span style=\\"display:none\\">7 `( ^/ ^# Z\\" [3 D7 t* o</span>

ultra1989

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用float可能会有精度问题用中国剩余定理求解
参考程序：
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* M, v! o  l! Y6 J$ N#include<stdio.h>
! A2 K# S0 J4 @2 j3 F. o2 Lint main(){
   int a,b,c,m,n,i,d=1;
, V' Y) F. D, T/ t   while(scanf("%d%d%d%d",&a,&b,&c,&m)){
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if(a==-1&&b==-1&&c==-1&&m==-1) break;
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n=0;
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( l" ]# p% {7 u7 Ti=(5544*a+14421*b+1288*c-m+21252)%21252;

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i=21252;

; |; N$ I7 _- g6 q) [. Cprintf("Case %d: the next triple peak occurs in %d days.\n",d++,i);

   }
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- g+ w8 |/ w5 h% y( Hreturn 0;
}
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qbist

小草远在天涯

好复杂啊，我是初级的。。。路过。。。

maybe_madio

回复 ultra1989 的帖子
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5 L+ Y% P) w( S% M    确实这道题使用剩余定理解效率太高了！ 哈哈！ 学习了！  我还没有看过数论方面的书！ 以后肯定要看看
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7 n1 |; r% L# ^4 N4 o两个程序一对比，一下就体现出数学的强大力量！

39133120

好复杂啊，得仔细

鸿光满面

好复杂啊