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Run Length EncodingDescription
% |5 P& O0 O) ~1 M9 `+ u8 F4 r " N. `1 s" l2 G/ @4 y# Z
Your task is to write a program that performs a ** form of run-length encoding, as described by the rules below. 9 b/ Q% V2 o6 F
4 X$ v& H8 ]& y" c
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. 4 P' p. k. K$ i8 l/ |
* o; D4 O9 e7 L T6 I% b- E M 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
J- y$ [9 w. [: ~7 Z2 E sequence, it is escaped with a 1, thus two 1 characters are output. 4 }4 i& Q @. R h {; X1 \
& [! N( J8 Z! H# g0 p& N Input
$ s v+ _+ p4 x, F# \ n
% g0 n' J) ~4 B- k- ~% P0 R 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. , D* X. I6 i. r' T
) x+ a7 p$ K k( } Output
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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. 6 h, w: V: z9 I% j/ A
$ E1 u! a& Q4 J+ R/ n 输入样例
0 n0 \ [7 F! R2 C# V' M6 q% d0 u5 Y AAAAAABCCCC
( U& c* i7 M' y+ T6 s 123444 O* W9 `, K* s
/ z/ E& m! r6 F5 t# s
; a, x( _9 b' R$ c" L( K x' {( e f 输出样例 * V- f. r; `' g r
6A1B14C
7 c2 u" j X; {9 V 11123124) A* `% A/ i. t* u+ ~
, H4 J7 V$ q. m8 ?
- o" \3 E" F4 B( z6 D& J/ T' J& F Source6 d. i; L4 g# B$ y' F# h
1 X z' p! b& h$ i$ M9 U
Ulm Local 2004
# k9 m' N Z1 A: D' S
: M! k4 p3 e8 h5 q( ^0 m' { example1:! a1 `. x s, t; l3 j/ \! L
#include<stdio.h>
. I8 R: N0 Z" @% `# ? #include<string.h>
, x! B3 @- I% ~; ?$ V1 j3 v void main(): }: U c& ?, S4 M: X( Q m
{ int i,j,k,n;
* |7 }" L6 ^) U8 S) x7 U char a[50];
' t9 \' M- ~7 O& a( K' _ gets(a);$ a5 O/ H8 f: s* a6 u
n=strlen(a);
$ N0 p& I- b6 j1 `4 y: n; E0 f0 X - l" c. s& `5 | W' N2 X
for(i=0;i<n-1; )" ?* ^3 h7 V5 ~" v; t$ A4 K
if(a==a[i+1])# g: ^6 k% i$ m B9 g8 f
{ for(j=i+1;a[j]==a[j+1];j++);
' z1 O+ ~$ X! T' c$ ?4 o6 g* T printf("%d%c",j-i+1,a);
2 R9 B9 W* ]% { t. z, K3 n0 c i=j+1;
4 M7 l9 Y& A4 N5 X( K }
- K* |4 n4 ]7 {+ v* \6 i( z else4 g4 B8 W7 a+ F7 C" @* c9 O9 `$ n
{ if(a==1)
; A; r" p8 X+ g4 V) m2 ~2 O { printf("11");
0 W' b( M0 t0 C' d# k6 a* n3 w i++;
9 W7 N% k7 h, e: {$ v2 ?6 I }
( d" p( r' W: M. O8 y; b6 v" @ else2 o; D8 Y4 E4 O+ P& P J* l
{ for(j=i+1;a[j]!=a[j+1];j++);1 t% Y" S; K6 r6 \
printf("1");7 G& n/ y* f5 v# S
if(j==n+1)# {$ `: i* R& d
j--;% ]4 ~2 P% d' y x; d6 V1 M
for(k=i;k<j;k++)1 G& Z9 j" C( a+ ?$ p! Y/ M* \
printf("%c",a[k]);
, S$ p( b h2 [* b, s# m printf("1");
' i) m' ~' |/ t1 o M i=j;
4 {7 b7 e7 z; u7 C }
3 j9 l3 g) j/ \: l3 \; ]/ X }
2 T/ }3 S1 {, {% S if(n==1)( t8 g* [" l* e! O$ D
if(a[0]=='1')6 {! S8 r/ f6 R i9 p, b/ }
printf("11");# Y! G! o% Y/ j5 G, }( C
else5 S, Q. k' N; _3 G1 a8 t
printf("1%c1",a[0]);
4 h ^1 S; \' _* o S; p: | printf("\n"); U/ O7 W5 V6 \# S3 I1 F- J
}
5 A9 f6 j; n& U- F5 N! S 评论人: Colby 发布时间: 2010-3-2 12:04:06 #include<stdio.h>! s$ O$ x2 q$ j) O
#include<string.h>7 `! h1 R' w& Y8 E3 N
void main()5 m$ s& K+ s, N* t2 u$ M
{ int i,j,k,n;: R+ f' T- A' H: B" ]; l3 [! {8 E
char a[50];) }) z8 | h" c: A2 X
gets(a);& f% n j5 @, O# N$ O; S2 j# n$ U
n=strlen(a);# T! `4 T* R, E8 w8 {$ ?' n
6 M* u# Y, @- I! G6 [
for(i=0;i<n-1; )
4 y: |8 r, } h if(a==a[i+1])
. d3 G0 d7 j' V" J% J, v% F { for(j=i+1;a[j]==a[j+1];j++);2 [( r" ~& l, E2 J9 }# p& i) {4 n
printf("%d%c",j-i+1,a);% }. j3 Z3 k( Q6 R: _" {
i=j+1;' E) p& V: k8 l4 w* W( X
}- B! ^; X! T1 S# f) v9 A0 C6 t: {
else- U4 ~! I* m1 a$ B
{ if(a==1)
1 A4 _2 r/ d+ F { printf("11");
' A+ i% q; h2 ~5 S i++;+ ]* e1 G1 n4 D" M0 k
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else
( B+ @! m9 n; U/ l! v6 | { for(j=i+1;a[j]!=a[j+1];j++);
7 y5 v4 j. K8 U7 b printf("1");: X9 L, y! A/ i* o
if(j==n+1)
+ Q" U+ P( t, `4 ], V j--;
6 O, D f9 W% j for(k=i;k<j;k++)
8 j0 G1 ~: q# F9 e8 z' T o printf("%c",a[k]);
; z( Y& K/ v$ H; p. L& e printf("1");$ N+ G) e4 v6 T. K' r# \
i=j;6 V% r9 u/ G- }% D. D
}
/ Q! `" [" k) S } w% j; z, _6 N( Y
if(n==1)
0 g/ C( w. r0 c t$ A# O3 ~: { if(a[0]=='1')
- O5 r7 v$ V, O" ]$ e3 ~. }$ v7 o printf("11");1 t7 _, }* E8 R1 V) C
else% u# m4 K8 M2 t' s7 b3 s
printf("1%c1",a[0]);# |6 z+ m" y6 K" h7 g( f
printf("\n");- h+ Y4 y& N W9 O b0 d* X) J7 }
} example2:#include<stdio.h>
# q( v( J6 S# W1 N+ Y #include<string.h>! }* q% {2 O* @1 H# q- j
void main()0 `" p# @% X- `. C) h& Y
{ int i,j,k,n;, p" E/ ]: i# z( W) x
char a[50];
6 p9 x5 x, Z+ c# T8 x# k* O gets(a);
7 ~, i. W$ y% Y: e6 r, r6 j n=strlen(a);
7 o o, N( Z! `1 h9 [
! M1 I4 B7 H' c3 x9 Z! p" x3 L2 g' }. P0 { for(i=0;i<n-1; )
5 x' s4 z ^, ?4 y2 B. v if(a==a[i+1])
8 ~& |. U7 S0 T4 { { for(j=i+1;a[j]==a[j+1];j++);% a- F V# j; m- H3 M* o
printf("%d%c",j-i+1,a);
. T. b7 N j8 ]( A i=j+1;9 Y% B7 Q& h9 p
}* I9 p1 v6 ]% A+ \ u
else
, v% Y( _3 I3 H- J7 h2 d { if(a==1), j8 t# `' `2 J2 m# {/ f
{ printf("11");
, Q0 E- C- b6 ?' G- L# i$ u% Z i++;2 p8 J/ {+ d' k8 N- r" W' W
}
' k" x3 Q) ^8 _" n# M+ ~9 t else
. Z3 U+ ~! n% m. Y { for(j=i+1;a[j]!=a[j+1];j++);
, |) Y" y3 @5 y/ f6 c printf("1");
: Q* N* R/ f& {8 A. d9 W if(j==n+1)
) j7 K$ M; [- P; h6 g+ ?, I7 R j--;
9 s' m- e9 v y1 g4 V: p for(k=i;k<j;k++)+ |) o& ?2 l$ e9 l* r6 t+ }3 z
printf("%c",a[k]);# W7 P2 O/ M9 U1 X2 _0 H0 _1 s( U/ l
printf("1");+ x6 l3 T! r* m1 t0 ]1 _+ z
i=j;5 ~4 k4 Y; @' h
}
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if(n==1): B1 p* M; M6 E( H: _! P5 S( i' X1 e
if(a[0]=='1')4 M4 T- [3 [2 Q. M
printf("11");
' J' d5 F, s: c1 e! q# d" S else
' F E, I+ @& E# U7 l* { printf("1%c1",a[0]);) G( |0 m& e# H8 K+ j$ s/ {" | N
printf("\n");; T$ o' `4 |* ?& z1 B
}
* _( ]5 c6 F2 V) o: ~- v example3:#include<stdio.h>7 y. P1 m5 L: P4 _$ m$ @
#include<string.h>. \$ T$ O- Y1 @4 z- H# G1 h$ d: ?2 m+ L t
void main()1 B. B$ c# _/ W
{ int i,j,k,n;2 j* \0 Y8 k" E; L: A
char a[50];
; n! R8 [) k/ Q+ {7 u& n& S0 U gets(a);
7 ]8 p" ?' \2 Y% O ] n=strlen(a);
`1 W7 K& A: t5 w3 _' |1 ] , n2 |7 N; F7 F2 B0 t$ B+ w! a& n
for(i=0;i<n-1; )1 {3 I8 } m) u* |$ P
if(a==a[i+1])! `, c' k* P& ?. p: X9 B: u
{ for(j=i+1;a[j]==a[j+1];j++);
) `* w) E* }& f) L; {5 S- L: F% H& s printf("%d%c",j-i+1,a);) [% [1 G4 f# \$ a7 K% m
i=j+1;
/ Z% x% l, a3 E% z5 ` } h0 K5 e) J+ U4 F2 g2 O6 K
else3 D @; P9 L+ O" ]2 @# k
{ if(a==1)
# ?6 \/ A" v8 D/ B( x% _ { printf("11");( O i4 c! V- ^2 |2 p
i++;$ [, u7 s' {1 V; a9 o' U6 d* f( e7 b9 M
}5 d4 O$ A1 }/ R/ s3 `( N
else
( s2 M0 j( C# q9 T1 o" n7 ~ { for(j=i+1;a[j]!=a[j+1];j++);3 {; O4 ?. Z6 A/ X3 E0 c
printf("1");
* [& A0 |9 w# m o- Q9 R5 F if(j==n+1)5 m+ i/ p4 Q, j0 V) V$ U6 \
j--;
+ C. W; ]; B, w' h for(k=i;k<j;k++)
% e: q' x7 I- Q3 n0 x printf("%c",a[k]);( F! J; |+ V3 r' T d0 H
printf("1");) c# g+ s2 U. |$ e. w
i=j;
( j% y \8 ?" t. | |$ ~7 u4 u, J1 }5 v9 O }
* U6 w* ? L6 R( Y& |8 E }
2 @9 k$ C5 D6 X* d if(n==1)
9 B* V7 y9 e% Q" ~ if(a[0]=='1')2 c" H8 n) E$ V$ w- t6 ]
printf("11");8 c4 i$ t+ B. p+ `
else
! r [# g" h6 u6 e printf("1%c1",a[0]);
. g2 q2 V5 o9 X( Z# ?( a printf("\n");
, [# l6 a. J) A1 n, X- ^ }
% k- c: P" g# W, G, ] 来源:编程爱好者acm题库
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