POJ-1006问题

maybe_madio

POJ-1006
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, Q1 c2 Q: r" o" X& F5 Z/ f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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) f7 F0 @" d& w* C3 n; H# u# V现将代码贴出，请帮忙看看：
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2. #include <iostream>
   
3. 6 a: o) N' G5 Z' p  S5 N
4. using namespace std;\\" Z3 Y* f- T# a7 R2 `# R+ t& p
   
6. #define PH      02 V$ H; u1 ^\\" q2 n
   
8. #define EM      1
   
9.   ^6 N4 Z' e0 z
10. #define IN      2
    
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12. int get_Top(int days, int kind);* d' Y& }, D% M9 M, P. q\\" z# B
    
14. int triple(int ptop, int etop, int itop, int cur);. U* ?4 ~1 j8 x
    
16. bool is_Integer(float n);
    
17. \\" n. C3 W& x) {; P/ _1 J
18. int main()( g) ^) x\\" Y( Y6 ?1 r( e' W4 y
    
20. {  G8 T; E- L4 H6 X; k* y
    
22. int ptop,  etop,  itop;7 |9 e2 ~2 d: J5 h* I
    
24. int pdays, edays, idays, curdays;
    
25. 9 h3 s* D# Y# h3 N# o# h\\" I
26. int count =0;4 q% B+ Z\\" c* v
    
28. 
    
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30. while(cin >> pdays >> edays >> idays >> curdays)$ P% v/ z) l* R( c% [
    
32. {
    
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34.   if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
    
35. 1 ~, j  g/ v0 [8 f+ H& L\\" T1 ?
36.    break;7 M! A& y* c  Z. [% H
    
38.   count++;
    
39.   R6 U7 A9 A/ a
40.   ptop = get_Top(pdays,PH);
    
41. \\" N$ ?& h, \( I( M6 i
42.      etop = get_Top(edays,EM);; c* e8 X5 L7 Y0 L; ?
    
44.      itop = get_Top(idays,IN);
    
45. 7 c2 |. O9 ]3 t8 u+ |! e9 y: C
46.   cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";  X/ W8 G: N8 y! d. T' c
    
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50. 
    
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52. return 0;
    
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56. int get_Top(int days, int kind)
    
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60. switch(kind)
    
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64. case 0:
    
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66.   return days%23;$ v  h+ R5 ]( [. R! Q% j
    
68. case 1:, g\\" b) @( b5 r( H\\" f
    
70.   return days%28;3 ]0 Y& D% F* |' S, N& b9 l
    
72. case 2:
    
73. $ _3 z& Y/ Y3 l
74.   return days%33;: |% N% j\\" F5 ?6 M( b8 S/ T
    
76. }
    
77. + j) V& S8 S; O8 q. a
78. return 0;
    
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84. int triple(int ptop, int etop, int itop, int cur)
    
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86. {: L8 }, d$ w# Q% u7 w5 V
    
88. float x, y, z;
    
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90. int m1   = etop-ptop-5,
    
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92.   m2   = itop-etop-5;
    
93. ; s, Z4 m8 r; h
94. int temp = 0;
    
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96. for(z=2;  (itop+(z-1)*33) <= 21252;  z++)\\" U* w8 p$ s% a, w  e4 {
    
98. {; I7 Y4 {6 e& B* o$ m
    
100.   temp = m2 + 33*z;8 G4 e( Z5 r$ M0 v) [\\" g3 k
     
102.   y = (float)temp/28;
     
103. : u7 V5 s: H3 C+ I& A# a+ e% P
104.   x = (float)(m1+temp)/23;# ~/ l  V/ w3 y$ S2 Z9 O. B: S( m  H+ U
     
106.   if(x>0 && y>0 && is_Integer(x) && is_Integer(y))
     
107. . T7 Z3 ?' E' V; B
108.    return itop + (z-1)*33 -cur;
     
109. 8 G/ H7 x\\" w5 y. Q3 E. |/ q
110. }
     
111. ; \; f% i* G2 X% A\\" x
112. return 0;
     
113. ) ^: t1 l; H; w$ Y3 ^- W# J( B# l
114. }
     
115. * l: |9 Y' r' U  F0 \, \$ N0 b\\" M
116. 
     
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118. bool is_Integer(float n)
     
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120. {
     
121. 1 |  D& k, v! G/ B* J. C
122. if( (int)n - n ==0)
     
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124.   return true;
     
125. 9 q6 @* j' b: t( U2 i! M/ z
126. else: B& g  h* U4 P
     
128.   return false;; c7 p( [. v9 O& N
     
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132. 
     
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#include <iostream><br /> <span style=\\"display:none\\">6 a: o) N' G5 Z' p  S5 N</span> using namespace std;<font class=\\"jammer\\">\\" Z3 Y* f- T# a7 R2 `# R+ t& p</font><br /> #define PH      0<font class=\\"jammer\\">2 V$ H; u1 ^\\" q2 n</font><br /> #define EM      1<br /> <span style=\\"display:none\\">  ^6 N4 Z' e0 z</span> #define IN      2<br /> <span style=\\"display:none\\">: R7 |\\" e! U\\" i, r- `. g</span> int get_Top(int days, int kind);<font class=\\"jammer\\">* d' Y& }, D% M9 M, P. q\\" z# B</font><br /> int triple(int ptop, int etop, int itop, int cur);<font class=\\"jammer\\">. U* ?4 ~1 j8 x</font><br /> bool is_Integer(float n);<br /> <span style=\\"display:none\\">\\" n. C3 W& x) {; P/ _1 J</span> int main()<font class=\\"jammer\\">( g) ^) x\\" Y( Y6 ?1 r( e' W4 y</font><br /> {<font class=\\"jammer\\">  G8 T; E- L4 H6 X; k* y</font><br /> int ptop,  etop,  itop;<font class=\\"jammer\\">7 |9 e2 ~2 d: J5 h* I</font><br /> int pdays, edays, idays, curdays;<br /> <span style=\\"display:none\\">9 h3 s* D# Y# h3 N# o# h\\" I</span> int count =0;<font class=\\"jammer\\">4 q% B+ Z\\" c* v</font><br /> <br /> <span style=\\"display:none\\">+ r\\" @- A0 q- i) c- a</span> while(cin >> pdays >> edays >> idays >> curdays)<font class=\\"jammer\\">$ P% v/ z) l* R( c% [</font><br /> {<br /> <span style=\\"display:none\\">! ?& O) f' b' @\\" b  [/ z\\" T</span>   if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)<br /> <span style=\\"display:none\\">1 ~, j  g/ v0 [8 f+ H& L\\" T1 ?</span>    break;<font class=\\"jammer\\">7 M! A& y* c  Z. [% H</font><br />   count++;<br /> <span style=\\"display:none\\">  R6 U7 A9 A/ a</span>   ptop = get_Top(pdays,PH);<br /> <span style=\\"display:none\\">\\" N$ ?& h, \( I( M6 i</span>      etop = get_Top(edays,EM);<font class=\\"jammer\\">; c* e8 X5 L7 Y0 L; ?</font><br />      itop = get_Top(idays,IN);<br /> <span style=\\"display:none\\">7 c2 |. O9 ]3 t8 u+ |! e9 y: C</span>   cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";<font class=\\"jammer\\">  X/ W8 G: N8 y! d. T' c</font><br /> }<font class=\\"jammer\\">9 p\\" B$ e# `; A! U% o3 i</font><br /> <br /> <span style=\\"display:none\\">$ X2 S5 m2 e1 S+ q</span> return 0;<br /> <span style=\\"display:none\\">: `% f/ X8 U; P: |7 m</span> }<font class=\\"jammer\\">! v  ]# G, F( N; d</font><br /> int get_Top(int days, int kind)<br /> <span style=\\"display:none\\">9 A8 u, |. K6 m1 G. e. Z, e# b4 Q</span> {<font class=\\"jammer\\">+ v' ]' P$ B7 n' |</font><br /> switch(kind)<br /> <span style=\\"display:none\\">: @. }9 F- m! S1 H) D</span> {<font class=\\"jammer\\">, D9 d& |4 g, n. O6 o' o/ g</font><br /> case 0:<br /> <span style=\\"display:none\\">/ s* @! E& h0 o+ Z5 o: F( ?</span>   return days%23;<font class=\\"jammer\\">$ v  h+ R5 ]( [. R! Q% j</font><br /> case 1:<font class=\\"jammer\\">, g\\" b) @( b5 r( H\\" f</font><br />   return days%28;<font class=\\"jammer\\">3 ]0 Y& D% F* |' S, N& b9 l</font><br /> case 2:<br /> <span style=\\"display:none\\">$ _3 z& Y/ Y3 l</span>   return days%33;<font class=\\"jammer\\">: |% N% j\\" F5 ?6 M( b8 S/ T</font><br /> }<br /> <span style=\\"display:none\\">+ j) V& S8 S; O8 q. a</span> return 0;<br /> <span style=\\"display:none\\">4 Y* H$ C+ w5 O1 [3 R) [9 ~</span> }<font class=\\"jammer\\">8 p* y5 l. z( R. R7 @  C' W</font><br /> <font class=\\"jammer\\">9 q' c/ J5 `/ r- j. m) k! q</font><br /> int triple(int ptop, int etop, int itop, int cur)<br /> <span style=\\"display:none\\">' d0 m* _& ^0 Y( m2 y</span> {<font class=\\"jammer\\">: L8 }, d$ w# Q% u7 w5 V</font><br /> float x, y, z;<br /> <span style=\\"display:none\\">' g7 Z, R1 q; ~! n7 w7 @; m</span> int m1   = etop-ptop-5,<br /> <span style=\\"display:none\\">: u  m7 R( t3 ~, I! A, ~' `( `6 B</span>   m2   = itop-etop-5;<br /> <span style=\\"display:none\\">; s, Z4 m8 r; h</span> int temp = 0;<br /> <span style=\\"display:none\\">: C* f2 I. A8 y: E</span> for(z=2;  (itop+(z-1)*33) <= 21252;  z++)<font class=\\"jammer\\">\\" U* w8 p$ s% a, w  e4 {</font><br /> {<font class=\\"jammer\\">; I7 Y4 {6 e& B* o$ m</font><br />   temp = m2 + 33*z;<font class=\\"jammer\\">8 G4 e( Z5 r$ M0 v) [\\" g3 k</font><br />   y = (float)temp/28;<br /> <span style=\\"display:none\\">: u7 V5 s: H3 C+ I& A# a+ e% P</span>   x = (float)(m1+temp)/23;<font class=\\"jammer\\"># ~/ l  V/ w3 y$ S2 Z9 O. B: S( m  H+ U</font><br />   if(x>0 && y>0 && is_Integer(x) && is_Integer(y))<br /> <span style=\\"display:none\\">. T7 Z3 ?' E' V; B</span>    return itop + (z-1)*33 -cur;<br /> <span style=\\"display:none\\">8 G/ H7 x\\" w5 y. Q3 E. |/ q</span> }<br /> <span style=\\"display:none\\">; \; f% i* G2 X% A\\" x</span> return 0;<br /> <span style=\\"display:none\\">) ^: t1 l; H; w$ Y3 ^- W# J( B# l</span> }<br /> <span style=\\"display:none\\">* l: |9 Y' r' U  F0 \, \$ N0 b\\" M</span> <br /> <span style=\\"display:none\\">0 e! i2 g: s5 F$ E</span> bool is_Integer(float n)<br /> <span style=\\"display:none\\">$ z& I; R. E) A/ g  l2 W2 O2 \- S</span> {<br /> <span style=\\"display:none\\">1 |  D& k, v! G/ B* J. C</span> if( (int)n - n ==0)<br /> <span style=\\"display:none\\">& A4 n0 P/ |3 d2 D& s</span>   return true;<br /> <span style=\\"display:none\\">9 q6 @* j' b: t( U2 i! M/ z</span> else<font class=\\"jammer\\">: B& g  h* U4 P</font><br />   return false;<font class=\\"jammer\\">; c7 p( [. v9 O& N</font><br /> }<font class=\\"jammer\\">4 B$ U( y  N3 P( K4 q# ]* U' z</font><br /> <br /> <span style=\\"display:none\\">( n7 n$ H, e& o- l1 d</span>

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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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#include<stdio.h>
int main(){
   int a,b,c,m,n,i,d=1;
( x; v5 Z% k! i0 w   while(scanf("%d%d%d%d",&a,&b,&c,&m)){

+ I3 w& V4 R! T6 @+ Q9 n& N+ gif(a==-1&&b==-1&&c==-1&&m==-1) break;
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2 @1 X% z/ A3 n% T1 a; Y: Kn=0;
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( a5 }  E6 C% a' V' N, G2 w& `& {i=(5544*a+14421*b+1288*c-m+21252)%21252;
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if(!i)
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i=21252;
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$ N: b6 k/ z# G: b0 T8 Aprintf("Case %d: the next triple peak occurs in %d days.\n",d++,i);
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& C' _5 l1 b+ Q' Kreturn 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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- e) i8 E. k2 w+ W4 R% b! @( u    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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2 ~7 P% K  ^7 e6 A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