[求助]请教各位大虾

lvyan

这是我的数据结构课程设计 实现一个多项式合并，字符串输入多项式，链表插入排序 4 A) o! H! ` \3 v: C' J/ {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 #include 3 Z* M: ?% P0 i5 \5 G0 [7 u#define MAXSIZE 400 8 d; q; V- |" ~: T7 n' fchar a[MAXSIZE]; typedef struct{char elem[MAXSIZE]; 0 A! B3 B$ h5 H4 l int top; }SqStack; ! _, h) f6 d' V r2 k! x# d, \% TSqStack S; 7 G( q+ m: t8 f9 {' _typedef struct poly 0 i) n- v) f, z8 b {int coef; /*xishu*/ int exp; /*zhishu*/ % E. ^; D# a; a3 [' h/ L6 r6 j \ struct poly *next; 3 t8 {4 W# V" ~8 |) o' ` }Lpoly; 3 l% M: I9 ?! \Lpoly *pa; - f3 S: Z4 |; ?" hint cf(int i) /*10^i*/ . }7 B% O1 Z% [{ int c=1,j; / M! |0 H8 O2 }$ f" j for(j=1;j<=i;j++) c=c*10; $ w( M8 D) r7 p: I% ~* T return c; } 7 d7 e6 X7 }7 z0 R8 t& p6 A f) Tvoid input(char a[],int j) { Lpoly *p,*h,*ptr; int i,k,flag=0,x=0; 1 @2 H7 z+ P/ n/ g% [7 \" t S.top=-1; % Y0 m7 n. ?5 M, C+ D pa=(Lpoly*)malloc(sizeof(Lpoly)); pa->coef=0;pa->exp=0; h=pa;h->next=NULL; 4 ~* K0 W# U1 s 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 ; C! ~, c8 p0 K+ ?7 g' x3 s, Z/ zchar a[MAXSIZE];

typedef struct { # }# s2 E1 p3 P* u |2 S char elem[MAXSIZE]; int top; }SqStack;

SqStack S;

typedef struct poly { int coef; /*xishu*/ . k" y; l0 x( |0 q1 o6 C& u3 i int exp; /*zhishu*/ 1 f% x0 n: E) i8 j& d struct poly *next; ! @- Q' K8 S( b1 S2 E) D, V, G}Lpoly;

Lpoly *pa;

int cf(int i) /*10^i*/ { int c = 1, j; for(j = 1; j <= i; j ++) & A( j; _- e# j) U- E, ^ V c = c * 10; return c; - x" q( f$ D& z! X& q8 I6 u7 C! Q}

void input(char a[],int j) { Lpoly *p, *h, *ptr; int i, k, flag = 0, x = 0; ' u F; S, W7 B) ^ S.top = -1; 9 B5 k5 T. X! v' A% E pa = (Lpoly*)malloc(sizeof(Lpoly)); pa->coef = 0; ( _/ h- b+ t$ g& b$ T2 ^ pa->exp = 0; h = pa; ; ~8 A# ?5 R/ }4 T( @( r$ c; H( t h->next = NULL; p = h; /*jianlipa*/ 8 v7 D0 A& T9 h2 A; h0 e: L for (i = 0; i <= j - 1; i ++) $ m) h) ^' @- a$ i# B: d4 `* a: x { ptr = (Lpoly*)malloc(sizeof(Lpoly)); * {1 [$ j- V" P- \4 O if(S.top) + W2 [5 d: P: p) ~, ^, p2 ? { S.top ++; 0 q) m2 D& k4 u8 |) k, j S.elem[S.top] = a; % t4 z0 Z! n- H& z0 q; F } if (a == '-') ' P) S. ?" [9 [0 M' _3 u flag=1; & }: g6 ]* a4 g4 E4 f' ?% @ / a' u7 l+ F! c if(a == 'x') /*fenlixishu*/ { if (S.top==0) { if (flag == 1) { / t4 x ~7 L) P2 r2 h ptr->coef = -1; ptr->next = NULL; : l2 H7 N( A m/ N+ z! t0 `8 a } else 0 l3 I8 w k8 T9 k" R5 S { ptr->coef = 1; ptr->next = NULL; } } ) Q- v' q* x8 m) U else 5 E# I& y n" J2 C; w4 m" p { 8 c' f7 \; Z2 a$ R4 i 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; + H( N( i0 V( i7 l! c( ` x = 0; } ! q ?0 S* ~5 [1 {2 D S.top = -1; flag = 0; /*clear*/ + C- n [- L: b3 k2 P% \# Q } if (a == '+' || a == '-') /*fenlizhishu*/ { 6 e L0 H, Y9 \0 J# R2 Z5 Y) F if(S.top == 0) 8 D' [% b, m8 i T2 W { Y7 {7 I: B, [" l' Z, M ptr->exp = 1; / o) U! U. W! j$ u& s5 k ptr->next = NULL; 8 u1 _8 x6 U' k+ u } # T& u* V/ Q; b4 m7 [ else { for (k = S.top - 1; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); : F: H2 C! R- H2 j3 \& J ptr->exp = x; ~) q' R7 M/ {: m ptr->next = NULL; x = 0; # E, M, h: v5 N1 E( H9 |9 e }/*tiquzhishu*/ 0 }+ t& e* L3 |; ^ S.top = -1; } 4 c. _0 |6 }7 T" W! A U N if (ptr->coef != 0 && ptr->exp != 0) { p->next = ptr; # U) ?: y; x$ V2 E' T p = ptr; 7 I* D+ x7 B% [2 R1 B+ P } u# L& ~. X4 t: c$ V if (i == j - 1) { if (ptr->coef == 0) / O5 G0 X' n0 t# n9 P { $ v0 @" s5 ?* Z! Q$ H5 W for (k = S.top; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - k); ptr->coef = x; % ~% p% Y' c3 P. P ptr->next = NULL; , k5 u3 \' d E9 M! H# B3 ~ } 9 e- W, F5 H5 K else " r$ Y& F: `; I: S { for (k = S.top; k >= 0; k --) % ^2 Y0 R- ]& M; I/ {$ W! F x = x + (S.elem[k] - '0') * cf(S.top - k); ptr->exp = x; ptr->next = NULL; } p->next = ptr; ptr->next = NULL; } } 2 [/ @& q2 u5 q/ ~}

void stinsort(Lpoly *pa) { Lpoly *p, *h, *q, *k; ; k; b! C. w. O3 A e h = pa; p = k = h->next; q = p->next; while (p->next != NULL) % @, {0 k' ~( [4 _% o, p { # ?. l9 s+ W! ]: ?; k if (q->exp) ! D! t- B" R& a" x$ r { $ c( L: V" r6 ?3 Q# x" U+ m" T while(q->exp > k->exp) { k = k->next; / ^6 j1 f9 N+ R! d& t6 y3 O h = h->next; } if(q->exp) { ! q$ t! S$ o9 @# t/ w# K* A p->next = q->next; h->next = q; ' }* |% k: a# Y Q8 s& ?2 u4 P2 S q->next = k; q = p->next; + O0 G: M# {" U+ H } if (q->exp = k->exp) & N, t' c3 s# B7 d { k->coef = k->coef + q->coef; / l5 R% k$ U4 C9 j4 B% r! N p->next = q->next; & }" O `/ \ ?9 Y: b$ G7 h e. [ q = p->next; } h=pa; k = h->next; 4 @8 T' u* s( m6 W: p) w( U } & @. Z: y$ ^; h) E4 M& S, w 6 l5 a+ y, n7 N if (q->exp = p->exp) { & f/ G, n: E: m, E- x9 G% g p->coef = p->coef + q->coef; p->next = q->next; q = p->next; ! j" i! P: t. O } if (q->exp > p->exp) { " X/ o; `$ b0 x# M/ Z6 H4 r& G) D q = q->next; ! f$ V0 X1 j! M2 ]% x! p8 Y p = p->next; } } }

main() % w% b' L5 W. `1 W+ ]6 g6 \$ j$ X+ o{ int i, j = 0; + X: k o/ W4 h K+ { _ printf("a="); 8 J% a: s( T9 M scanf("%s", a); ; ^7 D1 Y& o( e( J6 K3 H j = strlen(a);

input(a, j); stinsort(pa); printf("a="); for (i = 0; i <= j + 10; i ++) { # e- l( C( {3 Z5 i: \7 M printf("%dx%d+", pa->coef, pa->exp); pa = pa->next; } 3 V, |8 |# p: m- @} ) o, K' c/ d' G7 p/ _3 x0 x

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
3 K/ Q7 [& S# z4 c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
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!!~~~~~