POJ-1006问题

maybe_madio

POJ-1006
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- q! _+ \" h: x) u3 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

: F5 E# V: D6 P& J2 k现将代码贴出，请帮忙看看：
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2. #include <iostream>5 h0 G; [! k; S) i* j: m
   
4. using namespace std;
   
5. 5 L) j$ ~! w8 A- [
6. #define PH      01 J2 S1 y  a( y; S
   
8. #define EM      1) Q# E& F; n8 \2 A5 z
   
10. #define IN      21 t- u* `% a- o/ z5 m
    
12. int get_Top(int days, int kind);
    
13. + r1 L) \. B6 k5 y, o0 p$ k
14. int triple(int ptop, int etop, int itop, int cur);
    
15. 4 Q' p8 O& Q. t4 a0 g
16. bool is_Integer(float n);
    
17. ( w2 q8 _\\" \% H! s5 {\\" q. x
18. int main()0 y+ h# F6 z, h! R3 H3 U
    
20. {
    
21. * d5 C* M. N! A+ a: V0 h- d
22. int ptop,  etop,  itop;\\" i. M* g! _/ ]1 H1 s1 t
    
24. int pdays, edays, idays, curdays;1 K0 |1 A. e$ `\\" ]
    
26. int count =0;* a' |% n6 U# Y9 ]* |$ B5 U5 @
    
28. 
    
29. 3 Y3 Z3 J6 P# F* c' U
30. while(cin >> pdays >> edays >> idays >> curdays)  p0 c* P! o8 C/ K\\" N
    
32. {5 m- F) f4 T9 e% L3 m, f6 \& W
    
34.   if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1): U* S# T/ i: ~( i# X( p
    
36.    break;
    
37. : p: s; B1 R9 {) P
38.   count++;0 @; x) @& D5 D0 B7 x# K' s
    
40.   ptop = get_Top(pdays,PH);
    
41. 5 n) I/ Q: O5 S3 g3 ^5 A
42.      etop = get_Top(edays,EM);
    
43. 4 O/ Q! y4 C4 S3 x! {) R
44.      itop = get_Top(idays,IN);* g0 H& ]\\" y+ f* B2 k! w; U. q
    
46.   cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";
    
47. ! O) V3 h. O0 Q
48. }& A: h) F\\" x) N  L! n* y$ N( n5 c
    
50. : [) V3 }: c+ G& Q4 I
    
52. return 0;3 K0 d/ l# F  l+ s6 E7 ]
    
54. }: B0 _7 M, b4 T  i
    
56. int get_Top(int days, int kind)
    
57. # v. a\\" Z\\" {+ o% v: _2 ]* t; D1 o
58. {
    
59. ) h+ e) _\\" P, I: f6 n4 s3 h' Y
60. switch(kind)
    
61. 4 M% O  V2 ?/ X. ]- e; ^
62. {
    
63. & X. a: r+ h# n( k1 _: M\\" i6 u
64. case 0:* z& d\\" G: w6 w1 K+ {- {* T
    
66.   return days%23;
    
67. 1 @; O* E  y8 |\\" F\\" B6 c
68. case 1:
    
69. , K- e, z4 S. D. A3 W2 I: {
70.   return days%28;
    
71. 5 ^& ~8 v2 P/ x; }' y6 z3 V
72. case 2:9 F) z9 G\\" F4 t\\" @$ e
    
74.   return days%33;1 M# |3 r1 E, ]5 f- ~* j
    
76. }# h: E, a; z% E+ e+ p( ^8 b
    
78. return 0;
    
79. 4 C. I# b  e) Y0 }
80. }
    
81. + p5 I+ C' I$ v
82. ' \7 Q% g; O' ]# z3 [9 c: C
    
84. int triple(int ptop, int etop, int itop, int cur)
    
85. ( B; b1 d& }\\" Y* @2 T
86. {
    
87. ( t0 ]\\" @; Q( i5 L. |
88. float x, y, z;
    
89. ' `, N' W  m( ]\\" y
90. int m1   = etop-ptop-5,
    
91. . T1 n# \9 k1 j2 R% W  ]+ _; x$ [
92.   m2   = itop-etop-5;
    
93. 2 M- c) }4 m( E$ m% j: r: {
94. int temp = 0;
    
95. # Y. p+ H5 C% e
96. for(z=2;  (itop+(z-1)*33) <= 21252;  z++)% z  y' m6 p$ a\\" o
    
98. {
    
99. : n: G* ]  D1 W) }; n
100.   temp = m2 + 33*z;! G' C0 b( A7 `% |
     
102.   y = (float)temp/28;; I( s' ^7 K# r7 `+ ~1 \  {\\" X, o: R
     
104.   x = (float)(m1+temp)/23;
     
105. 6 }0 v' [1 f. ~2 q3 y
106.   if(x>0 && y>0 && is_Integer(x) && is_Integer(y))
     
107.   |. }, g/ l) {) \- O
108.    return itop + (z-1)*33 -cur;
     
109. & L! j5 T9 {  f
110. }, l2 N3 [  f5 U7 E
     
112. return 0;6 P/ C$ z+ Q; Q, f# d* y
     
114. }
     
115. ' Y, x$ j2 A3 i8 Z1 F
116. $ B) m' m/ r  N\\" u7 _
     
118. bool is_Integer(float n)
     
119. . w6 z\\" v; Z  f, b' s
120. {0 ~1 g4 e/ H\\" H1 V6 v1 \8 w
     
122. if( (int)n - n ==0)\\" |1 G6 T, a\\" n$ }\\" {
     
124.   return true;\\" E3 j' f) f6 n, c
     
126. else9 Y0 A5 r8 a\\" s\\" s, k8 G, d
     
128.   return false;; B+ x$ S( ?; F
     
130. }
     
131. 5 g; d1 ]* |( z6 ~
132.   s% [+ b* m8 A& s) g  g8 d9 [
     

#include <iostream><font class=\\"jammer\\">5 h0 G; [! k; S) i* j: m</font><br /> using namespace std;<br /> <span style=\\"display:none\\">5 L) j$ ~! w8 A- [</span> #define PH      0<font class=\\"jammer\\">1 J2 S1 y  a( y; S</font><br /> #define EM      1<font class=\\"jammer\\">) Q# E& F; n8 \2 A5 z</font><br /> #define IN      2<font class=\\"jammer\\">1 t- u* `% a- o/ z5 m</font><br /> int get_Top(int days, int kind);<br /> <span style=\\"display:none\\">+ r1 L) \. B6 k5 y, o0 p$ k</span> int triple(int ptop, int etop, int itop, int cur);<br /> <span style=\\"display:none\\">4 Q' p8 O& Q. t4 a0 g</span> bool is_Integer(float n);<br /> <span style=\\"display:none\\">( w2 q8 _\\" \% H! s5 {\\" q. x</span> int main()<font class=\\"jammer\\">0 y+ h# F6 z, h! R3 H3 U</font><br /> {<br /> <span style=\\"display:none\\">* d5 C* M. N! A+ a: V0 h- d</span> int ptop,  etop,  itop;<font class=\\"jammer\\">\\" i. M* g! _/ ]1 H1 s1 t</font><br /> int pdays, edays, idays, curdays;<font class=\\"jammer\\">1 K0 |1 A. e$ `\\" ]</font><br /> int count =0;<font class=\\"jammer\\">* a' |% n6 U# Y9 ]* |$ B5 U5 @</font><br /> <br /> <span style=\\"display:none\\">3 Y3 Z3 J6 P# F* c' U</span> while(cin >> pdays >> edays >> idays >> curdays)<font class=\\"jammer\\">  p0 c* P! o8 C/ K\\" N</font><br /> {<font class=\\"jammer\\">5 m- F) f4 T9 e% L3 m, f6 \& W</font><br />   if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)<font class=\\"jammer\\">: U* S# T/ i: ~( i# X( p</font><br />    break;<br /> <span style=\\"display:none\\">: p: s; B1 R9 {) P</span>   count++;<font class=\\"jammer\\">0 @; x) @& D5 D0 B7 x# K' s</font><br />   ptop = get_Top(pdays,PH);<br /> <span style=\\"display:none\\">5 n) I/ Q: O5 S3 g3 ^5 A</span>      etop = get_Top(edays,EM);<br /> <span style=\\"display:none\\">4 O/ Q! y4 C4 S3 x! {) R</span>      itop = get_Top(idays,IN);<font class=\\"jammer\\">* g0 H& ]\\" y+ f* B2 k! w; U. q</font><br />   cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";<br /> <span style=\\"display:none\\">! O) V3 h. O0 Q</span> }<font class=\\"jammer\\">& A: h) F\\" x) N  L! n* y$ N( n5 c</font><br /> <font class=\\"jammer\\">: [) V3 }: c+ G& Q4 I</font><br /> return 0;<font class=\\"jammer\\">3 K0 d/ l# F  l+ s6 E7 ]</font><br /> }<font class=\\"jammer\\">: B0 _7 M, b4 T  i</font><br /> int get_Top(int days, int kind)<br /> <span style=\\"display:none\\"># v. a\\" Z\\" {+ o% v: _2 ]* t; D1 o</span> {<br /> <span style=\\"display:none\\">) h+ e) _\\" P, I: f6 n4 s3 h' Y</span> switch(kind)<br /> <span style=\\"display:none\\">4 M% O  V2 ?/ X. ]- e; ^</span> {<br /> <span style=\\"display:none\\">& X. a: r+ h# n( k1 _: M\\" i6 u</span> case 0:<font class=\\"jammer\\">* z& d\\" G: w6 w1 K+ {- {* T</font><br />   return days%23;<br /> <span style=\\"display:none\\">1 @; O* E  y8 |\\" F\\" B6 c</span> case 1:<br /> <span style=\\"display:none\\">, K- e, z4 S. D. A3 W2 I: {</span>   return days%28;<br /> <span style=\\"display:none\\">5 ^& ~8 v2 P/ x; }' y6 z3 V</span> case 2:<font class=\\"jammer\\">9 F) z9 G\\" F4 t\\" @$ e</font><br />   return days%33;<font class=\\"jammer\\">1 M# |3 r1 E, ]5 f- ~* j</font><br /> }<font class=\\"jammer\\"># h: E, a; z% E+ e+ p( ^8 b</font><br /> return 0;<br /> <span style=\\"display:none\\">4 C. I# b  e) Y0 }</span> }<br /> <span style=\\"display:none\\">+ p5 I+ C' I$ v</span> <font class=\\"jammer\\">' \7 Q% g; O' ]# z3 [9 c: C</font><br /> int triple(int ptop, int etop, int itop, int cur)<br /> <span style=\\"display:none\\">( B; b1 d& }\\" Y* @2 T</span> {<br /> <span style=\\"display:none\\">( t0 ]\\" @; Q( i5 L. |</span> float x, y, z;<br /> <span style=\\"display:none\\">' `, N' W  m( ]\\" y</span> int m1   = etop-ptop-5,<br /> <span style=\\"display:none\\">. T1 n# \9 k1 j2 R% W  ]+ _; x$ [</span>   m2   = itop-etop-5;<br /> <span style=\\"display:none\\">2 M- c) }4 m( E$ m% j: r: {</span> int temp = 0;<br /> <span style=\\"display:none\\"># Y. p+ H5 C% e</span> for(z=2;  (itop+(z-1)*33) <= 21252;  z++)<font class=\\"jammer\\">% z  y' m6 p$ a\\" o</font><br /> {<br /> <span style=\\"display:none\\">: n: G* ]  D1 W) }; n</span>   temp = m2 + 33*z;<font class=\\"jammer\\">! G' C0 b( A7 `% |</font><br />   y = (float)temp/28;<font class=\\"jammer\\">; I( s' ^7 K# r7 `+ ~1 \  {\\" X, o: R</font><br />   x = (float)(m1+temp)/23;<br /> <span style=\\"display:none\\">6 }0 v' [1 f. ~2 q3 y</span>   if(x>0 && y>0 && is_Integer(x) && is_Integer(y))<br /> <span style=\\"display:none\\">  |. }, g/ l) {) \- O</span>    return itop + (z-1)*33 -cur;<br /> <span style=\\"display:none\\">& L! j5 T9 {  f</span> }<font class=\\"jammer\\">, l2 N3 [  f5 U7 E</font><br /> return 0;<font class=\\"jammer\\">6 P/ C$ z+ Q; Q, f# d* y</font><br /> }<br /> <span style=\\"display:none\\">' Y, x$ j2 A3 i8 Z1 F</span> <font class=\\"jammer\\">$ B) m' m/ r  N\\" u7 _</font><br /> bool is_Integer(float n)<br /> <span style=\\"display:none\\">. w6 z\\" v; Z  f, b' s</span> {<font class=\\"jammer\\">0 ~1 g4 e/ H\\" H1 V6 v1 \8 w</font><br /> if( (int)n - n ==0)<font class=\\"jammer\\">\\" |1 G6 T, a\\" n$ }\\" {</font><br />   return true;<font class=\\"jammer\\">\\" E3 j' f) f6 n, c</font><br /> else<font class=\\"jammer\\">9 Y0 A5 r8 a\\" s\\" s, k8 G, d</font><br />   return false;<font class=\\"jammer\\">; B+ x$ S( ?; F</font><br /> }<br /> <span style=\\"display:none\\">5 g; d1 ]* |( z6 ~</span> <font class=\\"jammer\\">  s% [+ b* m8 A& s) g  g8 d9 [</font><br />

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ultra1989

用float可能会有精度问题用中国剩余定理求解
, @; ~' ?+ i% a) W参考程序：

/ ?# a6 O3 ?+ _, c& d#include<stdio.h>
) n$ a# n4 {( u, _int main(){
   int a,b,c,m,n,i,d=1;
   while(scanf("%d%d%d%d",&a,&b,&c,&m)){
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if(a==-1&&b==-1&&c==-1&&m==-1) break;

n=0;

- d( X( Q7 D, E* S8 W5 _- Ei=(5544*a+14421*b+1288*c-m+21252)%21252;
' i; A; A) y6 }) R& P) ^
if(!i)
0 T+ g! L$ I( Y& x, S$ b
i=21252;

( P7 v1 r; J3 h4 O: o. @printf("Case %d: the next triple peak occurs in %d days.\n",d++,i);
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   }
5 T( L. q8 b, p
3 c6 c' ~" f  q: a4 D' dreturn 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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& J' [* y6 q8 T: S: l9 v) 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

( i6 H6 w7 _8 Q) C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