POJ-1006问题

maybe_madio

POJ-1006
. l- i6 i& U4 K1 l' H, r& z/ G9 N
9 B9 U' D- e# l& ~0 c在本地VC6.0下测试，得出正确结果， 为什么提交后总是Wrong Answer
  m$ E! a3 ?$ b: P7 g  e) T
) v  w& N" d4 x6 Q9 y现将代码贴出，请帮忙看看：

2. #include <iostream>4 i% Z) K1 D\\" y3 f. A1 Z$ Y1 R+ q
   
4. using namespace std;
   
5. # s$ m/ Q& X! z/ `6 U4 [. d4 W' R% u
6. #define PH      0
   
7. 5 }2 R) ~! S5 y' D& y
8. #define EM      1
   
9. ' K: `9 \* t8 t9 z% V
10. #define IN      2
    
11. \\" B, b9 ^; H' f- E1 r
12. int get_Top(int days, int kind);
    
13. # v  F! C; n- u3 Q$ K6 y/ {: }
14. int triple(int ptop, int etop, int itop, int cur);
    
15. 6 p! e3 a( x! |& _: n# j
16. bool is_Integer(float n);
    
17. 2 v: K. R; E, ^. T
18. int main(), a. t- U( p7 B; P5 D8 p( B) [
    
20. {
    
21. : L. J. P$ l0 a9 W+ y
22. int ptop,  etop,  itop;4 G* {: q# T: F7 n4 Q4 C
    
24. int pdays, edays, idays, curdays;+ W* Q' ]+ v2 Q6 w( B: B
    
26. int count =0;
    
27. 4 T\\" V7 x+ U# o* M4 J
28. 
    
29. % T9 h' q8 P; d, ~
30. while(cin >> pdays >> edays >> idays >> curdays)% y& {0 p: [; j6 z) j$ \# e2 @6 w
    
32. {% j: P) D% m. u
    
34.   if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)
    
35. 6 p* M8 C& m4 v( d! n- s
36.    break;
    
37. + O( M6 O# T9 d2 U* i& e, h
38.   count++;
    
39. 9 }/ }' \! f* j6 {
40.   ptop = get_Top(pdays,PH);
    
41. 9 W5 k7 L* ~+ b% v: V' V
42.      etop = get_Top(edays,EM);9 P  q\\" v* M) i7 `; E
    
44.      itop = get_Top(idays,IN);) n7 r- N$ _7 g' Y  A
    
46.   cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";- A\\" R; k) [) |. s0 p, d$ U
    
48. }5 x% k/ A/ t2 V
    
50. ' N9 S; A2 B* {5 ?! h
    
52. return 0;
    
53. # w2 Z\\" V% H' N, w9 K, E, C, ?
54. }/ \$ {+ u1 [! _2 e9 B! d* [\\" p
    
56. int get_Top(int days, int kind)
    
57. 1 a! T7 W* o+ F! _: m/ k% ~- ]
58. {  ?0 X5 K( G% m9 U8 z& |! z' M. `
    
60. switch(kind)
    
61.   |- ]5 t9 `; E1 W% R
62. {
    
63. 8 {% ~9 W. g: N  c\\" f, l
64. case 0:\\" h5 p7 y' ^9 o6 u
    
66.   return days%23;
    
67. 7 B+ y0 m5 z1 Q
68. case 1:
    
69. 2 u2 r6 n8 B. _: B$ q
70.   return days%28;
    
71. ; n3 Z: p  |9 k; d4 k
72. case 2:0 z3 n/ U( h/ V8 R6 c2 n$ S
    
74.   return days%33;& B' y0 F  z. T# V
    
76. }
    
77. 8 Y\\" p% ^) {+ k# T( `! C\\" P
78. return 0;5 I$ b/ Q: k, y2 ]0 g& d1 m
    
80. }: u- I1 ~5 }1 v% _
    
82. * @8 _+ s* }! z' n) Y1 M
    
84. int triple(int ptop, int etop, int itop, int cur)
    
85. & z4 f' I4 A, H3 E5 ^/ Q
86. {
    
87. 8 x/ v) A/ e' a) B7 ]7 f! [
88. float x, y, z;; \$ `8 `3 k; U( f/ _( m
    
90. int m1   = etop-ptop-5,, {4 a4 g6 a\\" H
    
92.   m2   = itop-etop-5;! ]. A3 m, o3 l
    
94. int temp = 0;
    
95. ; D% x9 u9 b' ^3 o7 ^& ~7 d
96. for(z=2;  (itop+(z-1)*33) <= 21252;  z++)
    
97. . q3 h2 U2 R% C; E
98. {
    
99. 1 u- v6 K! t* a# i. _\\" q9 A  ~
100.   temp = m2 + 33*z;9 l1 k8 n/ T! {5 V
     
102.   y = (float)temp/28;
     
103. & a; n( n/ |3 v
104.   x = (float)(m1+temp)/23;
     
105. ( I6 T3 v' M5 f. J8 S
106.   if(x>0 && y>0 && is_Integer(x) && is_Integer(y))
     
107. ; G) k. a7 k1 j& d5 u+ p+ p
108.    return itop + (z-1)*33 -cur;
     
109. 6 R5 s4 Z7 l- V9 @& r5 |$ a# x
110. }
     
111.   u3 G- A! j0 q2 C1 W
112. return 0;
     
113. ; m& a5 M5 X; ~$ {\\" V
114. }: g- F- ^% e. F  r: k8 S% l
     
116. 
     
117. 6 s; M0 }% O# q* K* ~1 Y( E
118. bool is_Integer(float n)
     
119. % W- ~& G7 W: A$ R+ T( J# {, Z9 G- y; j
120. {1 g$ q  G( T* X+ S3 l1 g
     
122. if( (int)n - n ==0)' `$ x; _2 B: M
     
124.   return true;
     
125. + g- ^, A* W# j; h; M- M# q
126. else( U, w' s) n: |  G6 q; r
     
128.   return false;4 V8 t) _* q4 n- o8 m( j
     
130. }( e( S/ q$ }0 M
     
132. ' U( ^, W+ h( C* q0 B. }
     

#include <iostream><font class=\\"jammer\\">4 i% Z) K1 D\\" y3 f. A1 Z$ Y1 R+ q</font><br /> using namespace std;<br /> <span style=\\"display:none\\"># s$ m/ Q& X! z/ `6 U4 [. d4 W' R% u</span> #define PH      0<br /> <span style=\\"display:none\\">5 }2 R) ~! S5 y' D& y</span> #define EM      1<br /> <span style=\\"display:none\\">' K: `9 \* t8 t9 z% V</span> #define IN      2<br /> <span style=\\"display:none\\">\\" B, b9 ^; H' f- E1 r</span> int get_Top(int days, int kind);<br /> <span style=\\"display:none\\"># v  F! C; n- u3 Q$ K6 y/ {: }</span> int triple(int ptop, int etop, int itop, int cur);<br /> <span style=\\"display:none\\">6 p! e3 a( x! |& _: n# j</span> bool is_Integer(float n);<br /> <span style=\\"display:none\\">2 v: K. R; E, ^. T</span> int main()<font class=\\"jammer\\">, a. t- U( p7 B; P5 D8 p( B) [</font><br /> {<br /> <span style=\\"display:none\\">: L. J. P$ l0 a9 W+ y</span> int ptop,  etop,  itop;<font class=\\"jammer\\">4 G* {: q# T: F7 n4 Q4 C</font><br /> int pdays, edays, idays, curdays;<font class=\\"jammer\\">+ W* Q' ]+ v2 Q6 w( B: B</font><br /> int count =0;<br /> <span style=\\"display:none\\">4 T\\" V7 x+ U# o* M4 J</span> <br /> <span style=\\"display:none\\">% T9 h' q8 P; d, ~</span> while(cin >> pdays >> edays >> idays >> curdays)<font class=\\"jammer\\">% y& {0 p: [; j6 z) j$ \# e2 @6 w</font><br /> {<font class=\\"jammer\\">% j: P) D% m. u</font><br />   if(pdays==edays&&pdays==idays&&pdays==curdays&&pdays==-1)<br /> <span style=\\"display:none\\">6 p* M8 C& m4 v( d! n- s</span>    break;<br /> <span style=\\"display:none\\">+ O( M6 O# T9 d2 U* i& e, h</span>   count++;<br /> <span style=\\"display:none\\">9 }/ }' \! f* j6 {</span>   ptop = get_Top(pdays,PH);<br /> <span style=\\"display:none\\">9 W5 k7 L* ~+ b% v: V' V</span>      etop = get_Top(edays,EM);<font class=\\"jammer\\">9 P  q\\" v* M) i7 `; E</font><br />      itop = get_Top(idays,IN);<font class=\\"jammer\\">) n7 r- N$ _7 g' Y  A</font><br />   cout<<"Case "<<count<<": "<<"the next triple peak occurs in "<<triple(ptop, etop, itop, curdays)<<" days.";<font class=\\"jammer\\">- A\\" R; k) [) |. s0 p, d$ U</font><br /> }<font class=\\"jammer\\">5 x% k/ A/ t2 V</font><br /> <font class=\\"jammer\\">' N9 S; A2 B* {5 ?! h</font><br /> return 0;<br /> <span style=\\"display:none\\"># w2 Z\\" V% H' N, w9 K, E, C, ?</span> }<font class=\\"jammer\\">/ \$ {+ u1 [! _2 e9 B! d* [\\" p</font><br /> int get_Top(int days, int kind)<br /> <span style=\\"display:none\\">1 a! T7 W* o+ F! _: m/ k% ~- ]</span> {<font class=\\"jammer\\">  ?0 X5 K( G% m9 U8 z& |! z' M. `</font><br /> switch(kind)<br /> <span style=\\"display:none\\">  |- ]5 t9 `; E1 W% R</span> {<br /> <span style=\\"display:none\\">8 {% ~9 W. g: N  c\\" f, l</span> case 0:<font class=\\"jammer\\">\\" h5 p7 y' ^9 o6 u</font><br />   return days%23;<br /> <span style=\\"display:none\\">7 B+ y0 m5 z1 Q</span> case 1:<br /> <span style=\\"display:none\\">2 u2 r6 n8 B. _: B$ q</span>   return days%28;<br /> <span style=\\"display:none\\">; n3 Z: p  |9 k; d4 k</span> case 2:<font class=\\"jammer\\">0 z3 n/ U( h/ V8 R6 c2 n$ S</font><br />   return days%33;<font class=\\"jammer\\">& B' y0 F  z. T# V</font><br /> }<br /> <span style=\\"display:none\\">8 Y\\" p% ^) {+ k# T( `! C\\" P</span> return 0;<font class=\\"jammer\\">5 I$ b/ Q: k, y2 ]0 g& d1 m</font><br /> }<font class=\\"jammer\\">: u- I1 ~5 }1 v% _</font><br /> <font class=\\"jammer\\">* @8 _+ s* }! z' n) Y1 M</font><br /> int triple(int ptop, int etop, int itop, int cur)<br /> <span style=\\"display:none\\">& z4 f' I4 A, H3 E5 ^/ Q</span> {<br /> <span style=\\"display:none\\">8 x/ v) A/ e' a) B7 ]7 f! [</span> float x, y, z;<font class=\\"jammer\\">; \$ `8 `3 k; U( f/ _( m</font><br /> int m1   = etop-ptop-5,<font class=\\"jammer\\">, {4 a4 g6 a\\" H</font><br />   m2   = itop-etop-5;<font class=\\"jammer\\">! ]. A3 m, o3 l</font><br /> int temp = 0;<br /> <span style=\\"display:none\\">; D% x9 u9 b' ^3 o7 ^& ~7 d</span> for(z=2;  (itop+(z-1)*33) <= 21252;  z++)<br /> <span style=\\"display:none\\">. q3 h2 U2 R% C; E</span> {<br /> <span style=\\"display:none\\">1 u- v6 K! t* a# i. _\\" q9 A  ~</span>   temp = m2 + 33*z;<font class=\\"jammer\\">9 l1 k8 n/ T! {5 V</font><br />   y = (float)temp/28;<br /> <span style=\\"display:none\\">& a; n( n/ |3 v</span>   x = (float)(m1+temp)/23;<br /> <span style=\\"display:none\\">( I6 T3 v' M5 f. J8 S</span>   if(x>0 && y>0 && is_Integer(x) && is_Integer(y))<br /> <span style=\\"display:none\\">; G) k. a7 k1 j& d5 u+ p+ p</span>    return itop + (z-1)*33 -cur;<br /> <span style=\\"display:none\\">6 R5 s4 Z7 l- V9 @& r5 |$ a# x</span> }<br /> <span style=\\"display:none\\">  u3 G- A! j0 q2 C1 W</span> return 0;<br /> <span style=\\"display:none\\">; m& a5 M5 X; ~$ {\\" V</span> }<font class=\\"jammer\\">: g- F- ^% e. F  r: k8 S% l</font><br /> <br /> <span style=\\"display:none\\">6 s; M0 }% O# q* K* ~1 Y( E</span> bool is_Integer(float n)<br /> <span style=\\"display:none\\">% W- ~& G7 W: A$ R+ T( J# {, Z9 G- y; j</span> {<font class=\\"jammer\\">1 g$ q  G( T* X+ S3 l1 g</font><br /> if( (int)n - n ==0)<font class=\\"jammer\\">' `$ x; _2 B: M</font><br />   return true;<br /> <span style=\\"display:none\\">+ g- ^, A* W# j; h; M- M# q</span> else<font class=\\"jammer\\">( U, w' s) n: |  G6 q; r</font><br />   return false;<font class=\\"jammer\\">4 V8 t) _* q4 n- o8 m( j</font><br /> }<font class=\\"jammer\\">( e( S/ q$ }0 M</font><br /> <font class=\\"jammer\\">' U( ^, W+ h( C* q0 B. }</font><br />

* P9 a$ |% V5 `8 B) v3 R

ultra1989

5 A1 z' ]) d( h( ^) w6 r) S! l% C
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>
* ^) E- y, z& l1 Xint main(){
- k' u' P% D& R* P+ s7 K: o4 Q   int a,b,c,m,n,i,d=1;
   while(scanf("%d%d%d%d",&a,&b,&c,&m)){

0 p4 `5 m" f8 Gif(a==-1&&b==-1&&c==-1&&m==-1) break;
9 e& o8 E# w. w$ q4 U( I
n=0;
( l7 y, u& v6 N" A
6 b- @' b' v4 E7 D: I* Fi=(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);
$ Q3 ^0 ]# |: Z1 Q3 G
   }
+ p3 K5 T4 Y1 z3 b
return 0;
/ O  ]& G4 R; `" f1 J: }3 W}

5 R+ N: f: G0 X1 o. ]

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


" h+ W0 O$ f1 _8 i  C- |1 a5 F5 i    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

/ j. b+ X. M. z3 |/ U两个程序一对比，一下就体现出数学的强大力量！
$ t4 y- N+ d$ F

39133120

好复杂啊，得仔细

鸿光满面

好复杂啊