[求助]请教各位大虾

lvyan

这是我的数据结构课程设计 ( q5 N H: H, h实现一个多项式合并，字符串输入多项式，链表插入排序 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~~~ ; H( } t# E% a1 d6 H#include #include 9 {! p: p2 R" q6 B0 P6 h" ^6 d4 P#define MAXSIZE 400 char a[MAXSIZE]; $ [, A l5 ^. x4 vtypedef struct{char elem[MAXSIZE]; 9 t& z: l( G2 }7 v& R5 T6 ? int top; }SqStack; / U/ m$ \3 D" J. BSqStack S; typedef struct poly + k0 X& o! H" x* o! `/ i {int coef; /*xishu*/ . P1 W: L+ {9 Y' l1 D3 _, w int exp; /*zhishu*/ 6 V& X! M ~. M7 A) s struct poly *next; }Lpoly; Lpoly *pa; $ r$ L U' n bint cf(int i) /*10^i*/ { int c=1,j; for(j=1;j<=i;j++) c=c*10; ' Y# G) {) }, f+ S: h; l* [ return c; } : m. g. q. U M+ J& v3 P7 W. Uvoid input(char a[],int j) & K+ |4 L' C1 z4 {0 k, P{ Lpoly *p,*h,*ptr; int i,k,flag=0,x=0; 4 r& H3 A$ S/ M" J9 \ S.top=-1; * u. J( W5 O6 f; } pa=(Lpoly*)malloc(sizeof(Lpoly)); pa->coef=0;pa->exp=0; ! q: m1 e8 }: Q/ \ h=pa;h->next=NULL; p=h; /*jianlipa*/ for(i=0;i<=j-1;i++) 6 o4 L' Z, g/ w6 x: P {ptr=(Lpoly*)malloc(sizeof(Lpoly)); ! n0 R' x9 V3 d2 K 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" $ p) o+ c, b' P& f$ v* H#include "string.h" #include "stdio.h"

#define MAXSIZE 400 ) m) t( N3 ]' l6 [% I" F) jchar a[MAXSIZE];

typedef struct 8 ^- l. B% A; G; { P: q4 s{ char elem[MAXSIZE]; 0 q. x Q- H+ ^# Q% \ Y) n int top; }SqStack;

SqStack S;

typedef struct poly 3 [- T. {. @: z. a5 F6 G" {1 Z1 a5 K{ * k7 T3 |) ?4 w4 ~# I int coef; /*xishu*/ int exp; /*zhishu*/ 6 V+ ~( U7 j' k. l, V3 M struct poly *next; 5 z7 g, ~# b: b}Lpoly;

Lpoly *pa;

int cf(int i) /*10^i*/ { int c = 1, j; for(j = 1; j <= i; j ++) $ e( `" T: k8 _ c = c * 10; ! I3 e" z$ ^, b$ v; j return c; }

void input(char a[],int j) { + [! Q `7 v; Q9 b5 [6 I Lpoly *p, *h, *ptr; ) i- a$ N. Z* @: S1 b2 b ]4 X int i, k, flag = 0, x = 0; S.top = -1; pa = (Lpoly*)malloc(sizeof(Lpoly)); pa->coef = 0; % \- z2 A% y2 ]( X, D8 m+ ~ pa->exp = 0; 2 l# O' I2 ^: I0 K( J3 h$ I h = pa; . L7 d* p: j' q% W6 A. X5 p q, | h->next = NULL; p = h; /*jianlipa*/ for (i = 0; i <= j - 1; i ++) ) }) m6 _+ c& r, |; W { % Z$ P- L% g6 Y4 r, m; s0 T% n ptr = (Lpoly*)malloc(sizeof(Lpoly)); - H0 h( ^) e6 { [ if(S.top) " o5 ?" z; [' e& m: o" {9 o { S.top ++; + N B2 a: m3 m3 g" ^! g S.elem[S.top] = a; } if (a == '-') flag=1; 4 Z) s& _/ r) j4 H9 ~" e; [ if(a == 'x') /*fenlixishu*/ { r! D5 @( I' V, q if (S.top==0) ' D! O9 f& k/ W" C: q7 ?4 M { ! u# u' l6 w% F- f; w/ t if (flag == 1) { ptr->coef = -1; 0 M {$ G" Y, w- y- T1 C6 O# e ptr->next = NULL; & h8 ?5 v4 ~7 R0 ?6 g } 3 Y! k5 l& j p7 B: Y% ^; u1 c else { @+ Z" f/ Z' c2 G7 x- Q) R& A ptr->coef = 1; ( r& J' x* O. @/ f) s ptr->next = NULL; , b6 L2 D8 p6 a% d } ) x- d2 b5 h: I5 u } 7 q2 ^2 N0 }4 [) ?. |# M else { 0 ^1 K- U* R/ G# S: z2 w for (k=S.top - 1; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); ptr->coef = x; ptr->next = NULL; 6 A% q( y" B8 J7 l& |! s x = 0; $ k& U3 x y3 S } S.top = -1; flag = 0; /*clear*/ } if (a == '+' || a == '-') /*fenlizhishu*/ { if(S.top == 0) { ptr->exp = 1; * T% p) i2 b* C) k ptr->next = NULL; } else ( W6 h; |8 I8 W: S/ _5 o7 v { for (k = S.top - 1; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); 7 d6 E1 |: J8 o6 }& a ptr->exp = x; ) ~/ o) g2 u4 \' e- A ptr->next = NULL; x = 0; }/*tiquzhishu*/ S.top = -1; } ' r; L% x! z* k3 a6 y6 r* x* ^ if (ptr->coef != 0 && ptr->exp != 0) 3 q) ?- O1 z( m h, b" c1 d, ^$ G { 1 }, {3 {7 e% v5 F p->next = ptr; p = ptr; } if (i == j - 1) { 9 d6 O, I' P4 n" B, m A4 `# G" a, S if (ptr->coef == 0) { for (k = S.top; k >= 0; k --) $ ?0 P/ N$ X4 r) u9 ` s0 p x = x + (S.elem[k] - '0') * cf(S.top - k); ptr->coef = x; ptr->next = NULL; } , E3 ^* q& }! b0 s _ else { 1 M3 Q! R! M. z W. C7 a for (k = S.top; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - k); ptr->exp = x; # f# t( |5 o+ L3 G ptr->next = NULL; $ n+ b r& q% E" L } p->next = ptr; ptr->next = NULL; 3 p4 Z* u0 l' p ]( m% Z: s3 x" C } ; x& q5 q, L1 l! r, P- O! n% O( e( j } # v6 m; Z3 Y s. A0 c}

void stinsort(Lpoly *pa) { Lpoly *p, *h, *q, *k; % M- D# @: |0 C$ }8 I& z h = pa; p = k = h->next; q = p->next; while (p->next != NULL) ( d: x0 C: O# o4 m" \* D; G { if (q->exp) 7 a6 B: j$ F/ k! }5 Y- ]& `# H* [ { " j$ D* {6 a, ^+ q+ q" H! r while(q->exp > k->exp) 5 _5 |1 p% q8 z/ F { + l9 y% O- f# X4 y8 P% a k = k->next; * l; R7 g& O0 f( B1 ^ h = h->next; } ! F5 b" i( N0 P if(q->exp) { ) S% z; s+ `9 G& B. g p->next = q->next; h->next = q; 9 a2 D F) ?5 \ q->next = k; 7 {1 ?% L3 m7 X( d# Y" C0 I q = p->next; 8 V1 p* F4 p5 J0 x! @% k } / w: n; X* H1 d8 m2 _# r if (q->exp = k->exp) , P2 J7 P) k. L( |' z { k->coef = k->coef + q->coef; p->next = q->next; 1 U8 J2 V- C& G$ k: a, E" A+ O q = p->next; } 5 W7 N" O* j( [. K! J h=pa; k = h->next; ( x1 U- R' Q9 Y) j; e } 7 y& g- t1 C. j8 C+ t- p3 P 9 `$ |. U2 ]0 \4 G if (q->exp = p->exp) ; p# b3 M _8 H* r1 b { p->coef = p->coef + q->coef; $ ?/ u) M+ B$ E1 X- { p->next = q->next; q = p->next; } if (q->exp > p->exp) { " ]. h6 s) T F' F' Q9 f q = q->next; ; c9 B5 I+ p, T y5 A; Q7 h p = p->next; ! d' R, X' V1 w% o } & t2 r, E1 @" s$ @, t/ W& h } & Q/ K! A' _" ^! R/ J}

main() { $ c7 @$ g" w* S int i, j = 0; printf("a="); ! {6 U& N3 G" G) ]2 J) n* G scanf("%s", a); & _9 o/ r3 u' q( I' N4 i j = strlen(a);

input(a, j); " m9 Y% j- N6 d5 g1 a% D a3 ^ stinsort(pa); printf("a="); ; c- L" a; h; S8 G9 [ i. r for (i = 0; i <= j + 10; i ++) { printf("%dx%d+", pa->coef, pa->exp); ) h9 y6 Z0 \3 ^( u# b/ m% S& } pa = pa->next; } : b7 b" B5 m. A/ g) p} ! U3 o# n3 f1 C

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
) k5 o8 M# Y% n" i8 R+ Z3 Y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!!~~~~~