[求助]请教各位大虾

lvyan

这是我的数据结构课程设计 ! N- m L9 Q \1 ?8 j- t- b实现一个多项式合并，字符串输入多项式，链表插入排序 6 ?# f! _$ p3 d可是我运行出来结果不一样啊，大家帮我看下哪里错了，小的在此谢过啦~~~ #include 0 n" v! C- E4 h" E3 {# K#include #define MAXSIZE 400 char a[MAXSIZE]; typedef struct{char elem[MAXSIZE]; 2 y w. A& v3 f+ ^" z; c H int top; }SqStack; 4 {" N- R- Y& ]' T* J6 _ XSqStack S; " v2 }& l% C- }; _+ H( R1 P# gtypedef struct poly {int coef; /*xishu*/ ; U G* _+ C- W) x$ `/ O int exp; /*zhishu*/ , B. }& F, d' k( T1 O struct poly *next; $ b$ B7 [" ^" J! A* @ }Lpoly; 9 O2 o8 F; |, H6 c+ b0 GLpoly *pa; int cf(int i) /*10^i*/ { int c=1,j; ) H# @( |6 }3 {9 |8 i for(j=1;j<=i;j++) c=c*10; # H9 V- {8 M; D n9 l return c; } % C: R+ |! X0 i) ] Qvoid input(char a[],int j) % z" J* r6 P* A1 \- s+ S: N{ Lpoly *p,*h,*ptr; int i,k,flag=0,x=0; S.top=-1; pa=(Lpoly*)malloc(sizeof(Lpoly)); 1 D) e9 @ j) P6 d9 _! B pa->coef=0;pa->exp=0; 3 z- A- }" d r( Q" f h=pa;h->next=NULL; p=h; /*jianlipa*/ 4 A) g1 R2 a9 n, l& q( ? for(i=0;i<=j-1;i++) 4 p, } I. p9 u {ptr=(Lpoly*)malloc(sizeof(Lpoly)); : y3 F& t: \. w v# z Q if(S.top

游侠无极限

猛然发现我已经受不了C的那种模式了……尤其是别人写的……

还有搂主最好打个包，代码发到BBS上全走样了，有些符号没了，没办法帮你看

夜编程

不用对齐的编程风格也就算了，可是把几行挤到一块儿的风格也太小家子气了吧？改改啦，不改成不了程序员的。

夜编程

还要注释要么用英语，要么就用汉字，不要用拼音，恶心死了，不是每个人都习惯看拼音的。

夜编程

还有，楼猪这段代码错误满天飞，居然也可以运行？

夜编程

#include "stdlib.h" #include "string.h" #include "stdio.h"

#define MAXSIZE 400 / M2 Q7 s1 c: u) achar a[MAXSIZE];

typedef struct { / c) i y9 M/ O+ L4 n2 _$ ?: N char elem[MAXSIZE]; int top; }SqStack;

SqStack S;

typedef struct poly 9 j" ~, {! a/ H9 \1 Q{ int coef; /*xishu*/ int exp; /*zhishu*/ 6 v. q6 n% A/ m1 s& D. i- C struct poly *next; }Lpoly;

Lpoly *pa;

int cf(int i) /*10^i*/ { . k5 |3 a0 P) d- |; { int c = 1, j; for(j = 1; j <= i; j ++) ; S2 I) m5 b& ^: D: m% B c = c * 10; `% j7 e% ^* A8 Z return c; }

void input(char a[],int j) * U d0 m1 q$ ~1 r{ # U0 A( X/ J8 }6 G# D. L1 d Lpoly *p, *h, *ptr; , Z- _6 j) Z3 c, l) A" g1 G) j int i, k, flag = 0, x = 0; S.top = -1; 0 H$ A. [! O3 @ ] pa = (Lpoly*)malloc(sizeof(Lpoly)); - A4 z& Y6 _- m: S pa->coef = 0; 1 r y5 @/ X+ S- E" {4 m pa->exp = 0; % l! ~! G. H! p h = pa; h->next = NULL; 0 O/ Z, v6 ~; S7 o p {3 L; z p = h; /*jianlipa*/ for (i = 0; i <= j - 1; i ++) { ptr = (Lpoly*)malloc(sizeof(Lpoly)); if(S.top) ~- n; H0 ]! p: a { S.top ++; S.elem[S.top] = a; ( ]5 g r: ~ w0 X } if (a == '-') flag=1; 9 D9 g0 E* f0 }9 {0 H1 n if(a == 'x') /*fenlixishu*/ { 7 Q* ^% Z3 o1 L if (S.top==0) 7 [. e# B6 ~; O5 | { # t8 ~* F1 s+ h) i if (flag == 1) { # K0 A2 K* J6 N( S5 j ptr->coef = -1; 5 D6 Z* ?" k$ w) h1 W5 Y; ^ ptr->next = NULL; } else { 9 O4 i! ?8 \) Z* I* ] ptr->coef = 1; * z8 l3 P' g/ i# T+ i ptr->next = NULL; } - P5 @0 P5 F7 A+ J3 L' { } 3 E$ n; T3 [) `# S; [! y1 L else { for (k=S.top - 1; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); ptr->coef = x; : {0 }2 l, u! x" p2 \# J3 _ ptr->next = NULL; 8 b7 T1 ?1 G( k- o1 _$ Q x = 0; 1 P3 K% J: j6 _' q1 a3 ^4 e4 g } ( c& Q# C: W$ H5 g5 H3 l- c S.top = -1; ' i! l9 r# W/ z+ F/ U# F+ A flag = 0; /*clear*/ } - A: {# i ?. k& c if (a == '+' || a == '-') /*fenlizhishu*/ { ) w3 X; p0 w, l, i# m2 ] if(S.top == 0) & R3 j8 W/ c1 h4 b& W* u6 [, t: z { ' n y4 K8 O3 @: m ptr->exp = 1; ptr->next = NULL; } else { for (k = S.top - 1; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); ptr->exp = x; ptr->next = NULL; x = 0; }/*tiquzhishu*/ S.top = -1; - w, O' k6 s0 Z6 d* F" ^; W } if (ptr->coef != 0 && ptr->exp != 0) { 2 s3 T8 P/ w0 x& T/ v* G p->next = ptr; p = ptr; } ' K0 c8 a, ^% o& T7 a if (i == j - 1) - ?( n% _$ T6 N" s( x7 y% _ { . I6 h8 s5 S/ V8 M' o2 n9 ?' z: M if (ptr->coef == 0) { ( G: K6 ~- P+ x! n* j for (k = S.top; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - k); " {0 g( J$ V$ V" ?5 v+ k ptr->coef = x; ptr->next = NULL; } ! ~- z/ k4 {' k6 U* X* c6 w2 p$ ~ else { for (k = S.top; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - k); ptr->exp = x; ptr->next = NULL; } " Q8 u6 M3 r |/ U p->next = ptr; & l9 Q8 \' c, c: A8 p1 `& I0 L6 L ptr->next = NULL; } : A* f8 t' c5 h% z* f& r5 Q6 Z } }

void stinsort(Lpoly *pa) ' _' j1 Z8 f( y S$ Q& a9 B{ Lpoly *p, *h, *q, *k; . w; S2 W/ |- e8 d. ] h = pa; p = k = h->next; 1 V( f$ Q8 Y/ S8 `( I q = p->next; + R& O) Z2 r! }* Y) I: M( l while (p->next != NULL) { : i3 C- B4 r5 Y0 K if (q->exp) # ~' b, Q8 [+ d { 3 c5 ^- A# c, B5 P- T( \" D$ p# t while(q->exp > k->exp) { k = k->next; h = h->next; 4 n+ F4 Z0 c" I" e" y5 H0 N0 q } : i# V# y( B. t if(q->exp) { p->next = q->next; h->next = q; q->next = k; 7 U' u. H, L; o% B q = p->next; } if (q->exp = k->exp) { & o, M/ L" l/ ? k->coef = k->coef + q->coef; p->next = q->next; 3 w4 c5 x3 S1 V! l( I) e q = p->next; ! i- e1 S4 A0 g% G% P8 _' Z } * n, q- Z: P) F" n7 o h=pa; : C, | L( z# _( |; t& X4 k" R( y k = h->next; } if (q->exp = p->exp) 5 b" \7 U$ g- h { p->coef = p->coef + q->coef; p->next = q->next; 0 V/ ~1 S$ j( M" @% X( B q = p->next; } 3 c, w$ W: z7 v" {0 D1 O: W if (q->exp > p->exp) . N4 ]4 T! _# H5 e/ G! M# n { q = q->next; $ M0 z4 f9 d! D/ { p = p->next; } } 0 w0 J, ]4 n& n6 p7 ~}

main() 3 @9 M$ q I' F- D+ M4 O3 ^- E6 ]{ int i, j = 0; , U3 ]0 ^$ f* P# S/ e3 t' z6 t; ~ printf("a="); 3 k S+ `$ E, A scanf("%s", a); j = strlen(a);

input(a, j); stinsort(pa); 3 T' V1 b" e, p- c printf("a="); for (i = 0; i <= j + 10; i ++) 2 R6 l; o) U& P9 r { printf("%dx%d+", pa->coef, pa->exp); pa = pa->next; } }

我把楼猪的代码弄了下，把里面大小写错误排除了，这段代码会有死循环，谁有空谁研究去。

lvyan

这是上学期的数据结构课程设计
结果我已经弄出来了，得了个优秀,没跟大家说不好意思
多谢大家给的意见,我现在水平还很差
4 ?0 `7 n- l. {+ \至于注释都是给自己看的,发到网上的时候没改,呵呵~~~~~~

Q7

全还给老师了- -想当初为考试忙这忙那的

Eagle

老帖，一口血就出来了……

406082496

it is too hard for me!!~~~~~