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acm编程竞赛题目2(Run Length Encoding）

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发表于 2010-5-6 18:38 |只看该作者 |正序浏览

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Run Length EncodingDescription

Your task is to write a program that performs a ** form of run-length encoding, as described by the rules below.

Any sequence of between 2 to 9 identical characters is encoded by two characters. The first character is the length of the sequence, represented by one of the characters 2 through 9. The second character is the value of the repeated character. A sequence of more than 9 identical characters is dealt with by first encoding 9 characters, then the remaining ones.

Any sequence of characters that does not contain consecutive repetitions of any characters is represented by a 1 character followed by the sequence of characters, terminated with another 1. If a 1 appears as part of the
sequence, it is escaped with a 1, thus two 1 characters are output.

Input

The input consists of letters (both upper- and lower-case), digits, spaces, and punctuation. Every line is terminated with a newline character and no other characters appear in the input.

Output

Each line in the input is encoded separately as described above. The newline at the end of each line is not encoded, but is passed directly to the output.

输入样例

4 ^* Z8 |. u5 u3 X$ m
AAAAAABCCCC
$ m; _6 P- E! c12344
4 Y7 X* }% F4 z6 I' G5 o

输出样例

* e6 o9 e7 C1 w7 E' f. w/ K" h" O6A1B14C
5 k. ~7 M' o0 u* y$ }9 Z111231244 r1 ` X7 ]* w: p+ \: n

Source

Ulm Local 2004

example1:
#include<stdio.h>
#include<string.h>
void main()
{  int i,j,k,n;
char a[50];
gets(a);
n=strlen(a);

for(i=0;i<n-1; )
   if(a==a[i+1])* F* Q: R3 j4 F! H  Y5 z
   {  for(j=i+1;a[j]==a[j+1];j++);
5 v% Q+ Z: `: b/ q       printf("%d%c",j-i+1,a);
5 q' l2 b" S0 A' J: ]       i=j+1;
) D( \" y. {' p+ i    }
, p" V9 \4 u4 p; m    else
: [" x4 f4 T" Z5 Q8 o    {  if(a==1)# v% L# a1 u+ X. @
         { printf("11");
$ [9 u2 ]/ j, |: Q                i++;8 ?5 U. ~. v! t2 E0 ?3 p* A
         }$ @* E7 z6 l3 R( K" F8 y
      else' Z' R: e7 r2 k- S
      {  for(j=i+1;a[j]!=a[j+1];j++);6 ^( U, m/ n/ O1 J; v4 B- A
            printf("1");
) o4 W7 j% c6 H* L1 s9 @+ ?             if(j==n+1)
8 t# r) L0 b0 M0 A                j--;
5 N% N" s6 Y" @! [/ k7 r+ c             for(k=i;k<j;k++): H5 e, _5 `8 Z+ k- u
               printf("%c",a[k]);6 Y) [9 f, e6 s8 t& A$ t4 ~3 T
            printf("1");$ p( e& ]0 N5 p$ p- h9 @; T7 R
            i=j;2 q! ?, m( Y7 d, r  _) X
      }1 A3 U, S1 @( l5 T5 N$ t: N- c
   }: t2 r& |9 X1 R% h7 H" B0 M
   if(n==1)
* c" s$ v% V: d       if(a[0]=='1')
$ T1 p6 @& s0 B& I% ~$ e% I, N             printf("11");
0 T/ S, M" V( I1 ^       else* [: w' c7 S8 e, ?1 j$ U; H
            printf("1%c1",a[0]);" V5 @& p) E0 i) v: a9 R, o1 ^, q
printf("\n");
+ b) ^2 |1 d6 n' \0 q' z9 ^ }  A: x, I! @" G' y& ?4 ^
评论人: Colby  发布时间: 2010-3-2 12:04:06 #include<stdio.h>7 ?3 ^1 I5 ~- i: _
#include<string.h>, a, K/ D/ @* s# n# O  R, ]0 c$ i
void main()
/ N+ d, V( ~8 L! V{  int i,j,k,n;: Y' Q  \) ~; g% {+ S8 f; U/ y
char a[50];, `/ A6 w8 y5 W  ^8 w) a$ L: V
gets(a);
; L$ S4 m1 n: j7 e* S1 N6 L6 Z n=strlen(a);
! p' \- q, R+ p# E( j, r3 y1 s( B2 A+ S4 o1 W# M
for(i=0;i<n-1; )0 n1 R- d3 B* k# B: N- i8 p: I
   if(a==a[i+1])4 \' q, m" l) e/ a" C. d
   {  for(j=i+1;a[j]==a[j+1];j++);
" O% U8 a  @: }& \! {       printf("%d%c",j-i+1,a);: O( P+ e" a# ~2 J* O
      i=j+1;
2 F% O# e& a# Y0 \3 n: _    }& \2 L( F( }- {2 c% C) X* g; A) J! u
   else: E- v3 D' m, m+ L
   {  if(a==1)$ ^) {; G1 }, Q& S9 u' v/ R1 ~% e
         { printf("11");5 l. w% A0 u; A/ T5 g1 w) x; l
               i++;
0 ?% ]/ z8 }2 _5 t          }( b3 n2 _3 k4 `- m
      else  q9 ^$ G4 s; j4 A8 d
      {  for(j=i+1;a[j]!=a[j+1];j++);
5 R) ^% @9 o" M$ v9 i: y" s             printf("1");+ X9 W8 I( `( D: u% D; V
            if(j==n+1)' @4 M5 P% _9 R" v: \  ^' k2 M1 W
               j--;) p! D* W$ a' w& t' Z' ^6 }
            for(k=i;k<j;k++)
& g6 D5 [  R  T9 }) ~5 d                printf("%c",a[k]);
! d4 \+ B2 m, I/ f             printf("1");% G" z4 ~( A9 i. d
            i=j;4 a7 E$ P6 N$ r8 Z2 O0 R8 M" v) W
      }7 W' J1 l0 W6 {6 L3 e
   }
% x# a: F3 f' i0 U4 C4 r    if(n==1)2 z/ V! T7 N. F9 z( g" d2 d
      if(a[0]=='1')
3 a  V* k3 S3 Q& Q             printf("11");
& _) G  Y: h1 A- r+ T# x  S       else
; q% h  X9 t2 ^8 w& p) R! h7 C/ t             printf("1%c1",a[0]);
( R' t' h2 W* [$ R0 ?9 z6 P  T3 N; F printf("\n");% e* \# ~$ a9 x. y
} example2:#include<stdio.h>
( Z& B4 R) x3 m#include<string.h>2 x6 x  p4 m* @5 x' ]& H
void main()
' {) b+ ]6 X* k1 C' U6 G{  int i,j,k,n;+ q% `7 C6 f% W- @/ r
char a[50];2 l* O5 [- p, ?
gets(a);- {) G% _  G- E" ?/ m; `0 d
n=strlen(a);
" j4 H" b8 Y5 U' \  _  p) ~; n, m9 E9 L) ?* t/ Z5 m3 }' @
for(i=0;i<n-1; )
8 m" F6 U# W% u2 l0 ]( A. o    if(a==a[i+1])
  t: T, E6 W/ l; v* c    {  for(j=i+1;a[j]==a[j+1];j++);
. S4 m) J5 s' s$ W% ?, K. Y; C       printf("%d%c",j-i+1,a);. n: @! C& i2 O( X5 O2 {) e
      i=j+1;
! V/ l. I7 Z2 v' s3 @    }
1 j% q% ]6 R  \    else
6 u4 }/ p0 y7 e8 ]8 x" o    {  if(a==1)3 u* P! O% E! M4 `. w2 C* c+ v
         { printf("11");  b7 J& i9 w* L$ Z6 C- D
               i++;
/ ^$ K! R* i/ l' r' I* s# P2 O          }) |( w& U9 Y9 R" q$ ?/ _3 O
      else
7 p9 k8 k. F  M/ a" \       {  for(j=i+1;a[j]!=a[j+1];j++);
( T! J  h$ n9 e1 d0 F             printf("1");2 m6 }2 f" ], t% `. T2 D! t4 \
            if(j==n+1)1 H9 p3 f7 s. i6 _
               j--;
2 O* z5 ?. z; M( h2 b             for(k=i;k<j;k++)
% l+ R* i6 a' E+ x5 z0 R                printf("%c",a[k]);
3 H5 w3 e$ Z7 O1 g             printf("1");
! u7 }" L; F: c8 _  S3 G( Q             i=j;; N( M' ^7 G5 E9 R, @/ w$ D/ C
      }. ^/ m, b) @) m: u* L; _
   }6 ]# G4 l; A" Z7 L( C; q
   if(n==1)
& o0 n7 v- ?( g; e5 u       if(a[0]=='1')
$ f# J' B- f- [" j4 A2 N* H" u             printf("11");
0 J9 ?3 S: \1 u# v       else
% j9 W) A! D2 M             printf("1%c1",a[0]);
- ?2 t+ r" _4 ^8 G9 P" N$ j printf("\n");
$ R: U6 {# r1 P" U& r }
% G8 b4 g0 T8 f0 t9 U( `. c example3:#include<stdio.h>
# j& g( l; Z" s( B1 u# W/ D#include<string.h>
# k: X- g$ Y( Avoid main()
% S5 ]1 W. d4 t{  int i,j,k,n;4 k0 A% H, d0 ^- q
char a[50];; i% }' _% e4 y, L) C
gets(a);
& x( k* m' i* x( Z n=strlen(a);
* E! P/ C  \4 m$ t7 u4 l0 h2 L) g8 e# M% T7 E; }2 f* {
for(i=0;i<n-1; )% j$ v, U$ m6 C( N) C2 N
   if(a==a[i+1]): x3 Z$ m5 X+ W5 k
   {  for(j=i+1;a[j]==a[j+1];j++);; I1 A/ ^9 |4 d( w7 J3 ~
      printf("%d%c",j-i+1,a);- v! ?  C* ]- [1 {: I' o1 n# @
      i=j+1;6 x2 c' c' X, ^8 l
   }
  P* t( H2 {' a* J: ]    else/ Z% D/ U/ N* ?% N
   {  if(a==1)
# g) ^0 T7 Y) O, x. C          { printf("11");
5 T2 I+ y( r5 q, }; S1 r. G                i++;. y3 d4 c" A; p% n, u* I$ d$ }
         }
1 T$ k6 T' G/ a, z1 ~  @       else
$ M& O! q/ M. [% z( l- l3 a       {  for(j=i+1;a[j]!=a[j+1];j++);
5 D4 ?6 M+ E$ m  S! `( [             printf("1");
" }/ T% S2 P+ ~3 a- ]/ E" a* U             if(j==n+1)
) j+ p6 A' M- U) `5 w                j--;
' j& J* J. p+ K$ C- I             for(k=i;k<j;k++)
1 h: ]8 {( D& x% X: U                printf("%c",a[k]);, ^7 O8 Z* Y( u) _! V% s
            printf("1");
; x  L/ |% \$ G7 x6 M) H             i=j;3 J& X1 I! u0 U# d( ]" u/ ]
      }
! m' B# J# r5 }$ N6 f    }' y' _, k  G3 S, l
   if(n==1)
% n$ J1 n$ j6 @0 @* i% E       if(a[0]=='1')
3 S5 K! F* Q8 h" h             printf("11");# i* f: f9 E3 p  Z! a
      else! @% z) A6 [6 I  T& Z4 h
            printf("1%c1",a[0]);
+ G/ O# k2 s; M7 f! ]. B printf("\n");( c$ C3 Q' R7 B( v8 q% ]
}& k) p* O6 z* m
   来源：编程爱好者acm题库