[求助]请教各位大虾

lvyan

这是我的数据结构课程设计 * ]! ^$ D8 e& {0 J0 ?5 F# _实现一个多项式合并，字符串输入多项式，链表插入排序 8 A* b c: D; E" B9 f" H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 ( ]3 [( }3 E5 Q8 Z1 \#include & e6 N' p9 j+ w0 a#define MAXSIZE 400 7 F. {7 O$ a9 h: Z7 ?: \ [; u uchar a[MAXSIZE]; typedef struct{char elem[MAXSIZE]; ( T# e" {; ^ n3 ~4 y int top; }SqStack; SqStack S; typedef struct poly - D" X, k+ ]3 n6 A a; z {int coef; /*xishu*/ int exp; /*zhishu*/ struct poly *next; " e: x" I, l! U) v }Lpoly; Lpoly *pa; 6 Q; V5 _) Q; D2 l5 p7 Kint cf(int i) /*10^i*/ ' ~+ @ t" {; X{ int c=1,j; 2 p+ j: F: O( ?% d for(j=1;j<=i;j++) c=c*10; return c; } , F! }, v8 a8 K+ _+ h7 O; D3 ovoid input(char a[],int j) : b* b. a8 ]7 H4 V c0 {2 _{ Lpoly *p,*h,*ptr; int i,k,flag=0,x=0; S.top=-1; pa=(Lpoly*)malloc(sizeof(Lpoly)); 7 f, j2 a5 J, ?- ^' n pa->coef=0;pa->exp=0; & J( k" W. y+ Q1 H- d$ h h=pa;h->next=NULL; p=h; /*jianlipa*/ for(i=0;i<=j-1;i++) {ptr=(Lpoly*)malloc(sizeof(Lpoly)); 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 char a[MAXSIZE];

typedef struct 9 e, h. j" J2 C2 f: p; p{ char elem[MAXSIZE]; int top; + z4 T9 a3 X6 q1 Q}SqStack;

SqStack S;

typedef struct poly " \4 J) G: h0 L4 n6 h2 [' \4 l{ int coef; /*xishu*/ ) e" `: ?5 K6 R. x9 J+ K3 [: K int exp; /*zhishu*/ struct poly *next; }Lpoly;

Lpoly *pa;

int cf(int i) /*10^i*/ 8 d5 {2 z, w, X3 |1 k9 `; @{ int c = 1, j; $ Q9 T9 _. }/ g for(j = 1; j <= i; j ++) * j& o( F& g* L: R }5 d; {& @ c = c * 10; return c; }

void input(char a[],int j) { 0 T# M; P+ |4 E Lpoly *p, *h, *ptr; - a2 h, U- R+ J0 X; S: H, T6 x int i, k, flag = 0, x = 0; S.top = -1; pa = (Lpoly*)malloc(sizeof(Lpoly)); pa->coef = 0; / }0 E5 g- Q4 T; w5 ?$ E5 W pa->exp = 0; 2 V4 P1 X, H6 f. T h = pa; h->next = NULL; p = h; /*jianlipa*/ . k' T9 ]5 I: P7 O* G% x9 y8 ?5 N p4 A V/ j" E* O( G! U) p for (i = 0; i <= j - 1; i ++) + i {$ f0 K' C9 p% |) | { 8 Y$ [' h; N! Z6 {) ~( C& U, A ptr = (Lpoly*)malloc(sizeof(Lpoly)); if(S.top) $ i" c. E1 B7 j F+ m1 h { S.top ++; ; a A, U; \9 W1 Q0 Q/ l) J& I b S.elem[S.top] = a; 5 ]3 Q3 B$ M( F; @% P- {+ S } - b4 U/ G% B( \ V/ @3 B5 b; T6 e7 \ if (a == '-') flag=1; if(a == 'x') /*fenlixishu*/ ( {. q7 s& w6 q; q; K { if (S.top==0) { if (flag == 1) { ptr->coef = -1; ptr->next = NULL; } else { . S' o: }3 U$ K& p& n ptr->coef = 1; 1 {# q0 w9 j7 v" C# V! K ptr->next = NULL; } 5 L: g! C! o% N, u" n } 3 U; ~7 k; r$ J, V+ r6 [5 ^ else " C6 Q8 u8 T" o3 V/ { { for (k=S.top - 1; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); ptr->coef = x; 9 Y+ U/ d% l) i. ]$ _4 j% a ptr->next = NULL; 8 L' |, {7 f t _- z. R+ L9 m x = 0; } S.top = -1; F. m: G# ]4 B3 Q7 r flag = 0; /*clear*/ } * n) i4 w" w+ O1 J0 [ if (a == '+' || a == '-') /*fenlizhishu*/ 2 V7 w, T% P: y! R; ~ { if(S.top == 0) 7 E) G8 d9 t# B; @' Z { ; L y$ ~( l2 V7 n1 g g3 Y' [ ptr->exp = 1; ; D8 Z' T1 ^7 C Y ptr->next = NULL; 6 w6 }/ J1 k5 o4 _0 G# _) C) X. y } else { for (k = S.top - 1; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); ) ?" W6 o) Y4 ^$ S! B ptr->exp = x; ptr->next = NULL; x = 0; }/*tiquzhishu*/ 7 A2 S+ Z5 t0 U+ d5 n5 v4 A S.top = -1; ; m: A& d( e. M8 h7 M } if (ptr->coef != 0 && ptr->exp != 0) # q) v N6 Q& J3 | { p->next = ptr; # \* g. q' K$ ]0 D* i2 { p = ptr; 2 O3 {0 p5 Y+ {* g } if (i == j - 1) % V& l. S! O, h0 x/ R8 g0 X4 w { , V5 [& a" K" |0 z if (ptr->coef == 0) { for (k = S.top; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - k); 7 L2 z" j4 u5 a ptr->coef = x; " i; _( ?9 {9 \7 D ptr->next = NULL; $ u( R" B- l A } else 0 r7 w7 d9 F9 s/ F) @$ u+ J" g { for (k = S.top; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - k); ptr->exp = x; ptr->next = NULL; + e1 I. `1 L0 {5 k } p->next = ptr; ptr->next = NULL; 7 K. n# s* J5 L2 _* ~ } } }

void stinsort(Lpoly *pa) { Lpoly *p, *h, *q, *k; h = pa; p = k = h->next; . u0 \# d7 F% W W+ A$ ?8 K q = p->next; while (p->next != NULL) / S6 L0 v& Y5 B* X# a { if (q->exp) { while(q->exp > k->exp) { k = k->next; h = h->next; ; B, q: J v# Q- i- ? } if(q->exp) { ' S' p" ?# p* n' o p->next = q->next; h->next = q; q->next = k; 0 E6 _, Q1 S1 O8 Q7 Y9 @: Y3 v q = p->next; # P" W: z& i9 V } if (q->exp = k->exp) { 8 t& M2 d- l) S) \ k->coef = k->coef + q->coef; / Z7 h4 X$ r# K" d. V2 O' h9 G p->next = q->next; q = p->next; E* ^: P5 k; i5 \) b } h=pa; $ I. j& W3 [6 r k = h->next; ; a6 r' t, e; ^/ | } 2 m& V( X& O) n; Z; Y- a/ b if (q->exp = p->exp) . u0 w9 H; j/ O) {- N2 `6 ]/ n { 1 ^6 }8 {6 V$ Z p->coef = p->coef + q->coef; p->next = q->next; - m, ? Y/ ~' @9 S7 X% j) P q = p->next; } if (q->exp > p->exp) { q = q->next; + J# R% V! Q3 b! j1 N5 Y) F p = p->next; } } }

main() $ W8 {- q0 W& I& J{ 5 N' _' Y# ]1 E: ~ int i, j = 0; " W. ~. a- `! s/ q printf("a="); 5 b# e4 ~! h, P. S, {/ F. C scanf("%s", a); j = strlen(a);

input(a, j); 0 m' z2 t7 j) A* \ stinsort(pa); ) X! F3 ~* R6 n2 M& g. J printf("a="); for (i = 0; i <= j + 10; i ++) { + @3 I& S/ b2 @) E printf("%dx%d+", pa->coef, pa->exp); u% F9 K/ b E. P pa = pa->next; 3 D( w: {7 w' h( k/ ~8 U } ( }, Q* B' }! a}

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
* {+ }6 f5 _- F; Z2 }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!!~~~~~