[求助]请教各位大虾

lvyan

这是我的数据结构课程设计 - _+ ?& \' A4 b8 d8 k实现一个多项式合并，字符串输入多项式，链表插入排序 " @9 A# U0 F. `7 Q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 #define MAXSIZE 400 # T# Q5 X$ ]: ?char a[MAXSIZE]; typedef struct{char elem[MAXSIZE]; 5 s0 `0 P! b: F. Q, C% d int top; + W7 d% o: |, T j: E" \& g& N- K }SqStack; ' R& ?, A+ D5 lSqStack S; typedef struct poly {int coef; /*xishu*/ int exp; /*zhishu*/ 8 Y9 y% Q0 m* Y struct poly *next; }Lpoly; Lpoly *pa; $ l+ b1 ]6 n2 z; p, s8 w. i4 H$ J7 bint cf(int i) /*10^i*/ $ N0 U+ A5 g' u, p' c8 K8 {{ int c=1,j; for(j=1;j<=i;j++) c=c*10; , e8 \" @% T8 h5 f$ K return c; & ?/ p1 e3 G5 i- g1 M7 W} ) Z3 N( V I2 ?& U* ?9 Kvoid input(char a[],int j) " L5 {7 E- x2 n1 n$ v, `+ d{ Lpoly *p,*h,*ptr; int i,k,flag=0,x=0; : Z/ h% `3 J6 f( l+ O6 [" t' o# h4 x( ^ S.top=-1; pa=(Lpoly*)malloc(sizeof(Lpoly)); pa->coef=0;pa->exp=0; 4 J/ i p* ?, i, \5 ?5 ^ h=pa;h->next=NULL; ( S8 c9 t8 E8 L1 K p=h; /*jianlipa*/ L7 l- [7 R0 Y& [+ e/ D for(i=0;i<=j-1;i++) 7 M. e+ |$ O9 _8 G {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" ! f+ c# v5 q! j% A$ X#include "string.h" #include "stdio.h"

#define MAXSIZE 400 char a[MAXSIZE];

typedef struct - H7 `7 E% E$ P! t{ ) b* Z0 q0 T( M5 | char elem[MAXSIZE]; int top; : S$ y! W4 B8 w1 L& r}SqStack;

SqStack S;

typedef struct poly + l F. o( r0 m, x4 N4 ?{ int coef; /*xishu*/ + Y# B' S% g: y& v int exp; /*zhishu*/ struct poly *next; }Lpoly;

Lpoly *pa;

int cf(int i) /*10^i*/ { int c = 1, j; for(j = 1; j <= i; j ++) * w2 h6 M: h I8 j/ v3 k+ N9 W c = c * 10; ( ]2 c- W n+ F' v return c; }

void input(char a[],int j) 0 Y% Y8 `; U' j& ^+ C z1 a{ Lpoly *p, *h, *ptr; 1 m c& ]( |6 g int i, k, flag = 0, x = 0; S.top = -1; - x8 ]) L. M. V' c8 d- O! c pa = (Lpoly*)malloc(sizeof(Lpoly)); ; o; D8 i' z1 G. D" I. { pa->coef = 0; pa->exp = 0; / c1 o. s- x0 I; M' X5 O5 f5 j h = pa; h->next = NULL; p = h; /*jianlipa*/ $ X' e* Q6 s- h3 O( a& u$ I7 J+ c for (i = 0; i <= j - 1; i ++) * b, k) w" c- J { ptr = (Lpoly*)malloc(sizeof(Lpoly)); _7 B5 c i) o8 \$ F f9 W2 V+ z { if(S.top) " y. h3 n5 O4 o1 s' [) p8 t/ { { 4 D5 y0 p" _# q S.top ++; S.elem[S.top] = a; } 7 `! H; {4 h2 U+ C if (a == '-') flag=1; 3 A# Z) q! x! q! U& K if(a == 'x') /*fenlixishu*/ # W3 P4 ?3 y! _3 v { if (S.top==0) { if (flag == 1) / \7 q7 C' I; F5 M! [ { ptr->coef = -1; $ k& `) w- d- x8 w! { ptr->next = NULL; } else { ptr->coef = 1; ptr->next = NULL; } ' A6 u+ N! X$ [, p } s3 c8 v, X' x7 @ else { ! j8 a4 N' Y* a# a! g# F X) ] for (k=S.top - 1; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); + c& T0 p4 |* k3 V ptr->coef = x; ptr->next = NULL; 6 V" Z X7 o( y$ S! b% e( V7 T. h1 D9 F x = 0; . u: E) s" e2 w9 F } S.top = -1; flag = 0; /*clear*/ } / ~8 o( B, }& O* }) n& [( p if (a == '+' || a == '-') /*fenlizhishu*/ { if(S.top == 0) ; W* V1 R+ O* @0 p4 C8 S { # S4 D3 f% j# p( L ptr->exp = 1; 1 W3 J) U% c& p& x% Q ptr->next = NULL; } else { 2 V/ P5 R* A# a2 e/ k. p/ S for (k = S.top - 1; k >= 0; k --) ' G- ]5 n5 d! q1 o x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); ptr->exp = x; ptr->next = NULL; / C/ f! a( P0 ?' K+ I& z x = 0; }/*tiquzhishu*/ S.top = -1; } if (ptr->coef != 0 && ptr->exp != 0) ' l! w! s" _) S { p->next = ptr; p = ptr; } if (i == j - 1) ' W( z/ h! m- f1 v3 E0 W4 U { ( o8 n( Y- u. B; d" s if (ptr->coef == 0) { for (k = S.top; k >= 0; k --) * i$ J* @- D. ~: D. B x = x + (S.elem[k] - '0') * cf(S.top - k); % F6 z9 N" j3 x+ [) C; I ptr->coef = x; ' q& I0 B$ V- F( Z2 U ptr->next = NULL; ; ~ t7 W5 X, Z/ i* ~# R- C" ?8 l } else ; h% F8 _1 S7 M% t% O! m { * i2 W# o& ]# {# y e for (k = S.top; k >= 0; k --) " `- t6 X) ?0 [4 }) z( r4 `: g' W x = x + (S.elem[k] - '0') * cf(S.top - k); 7 y/ y! q" V" k0 w1 | ptr->exp = x; $ V4 s4 K* n* ~. G: ` ptr->next = NULL; } p->next = ptr; * p8 |" p7 a# b3 R# b" H/ @; b# S" M ptr->next = NULL; } } }

void stinsort(Lpoly *pa) { 7 y) h8 |* D: u3 t" `4 z Lpoly *p, *h, *q, *k; 0 E9 \4 Z9 P4 e4 h! U h = pa; p = k = h->next; q = p->next; while (p->next != NULL) 5 n* ]! E$ T4 }- K; T { if (q->exp) ( d: S. }+ ]! h { - U( z7 X2 W" d; g# J7 H while(q->exp > k->exp) * r) W1 o9 b, q+ G. U: b { b3 _8 U; D) u k = k->next; & x9 i1 n8 E( J/ Z- m1 D h = h->next; 0 H8 h, ]" a8 x: a9 y6 ` } if(q->exp) { 4 s7 `6 U+ P" h2 z4 H/ | p->next = q->next; & f. {, M% x7 u h->next = q; 0 U+ ]# P/ J! F7 r- K q->next = k; q = p->next; 6 X0 {- B4 z+ K } if (q->exp = k->exp) { k->coef = k->coef + q->coef; p->next = q->next; q = p->next; 2 V5 o+ U( s! L0 Y; V2 v9 o# ` } % X' O) u! G7 z6 h5 `, { D h=pa; k = h->next; , |' J. X- R+ B P } - S! i: J! ~9 T if (q->exp = p->exp) 0 p5 R/ u/ ?4 F; t$ z { p->coef = p->coef + q->coef; 8 Q# x6 s( F4 v) w$ B6 l p->next = q->next; 1 S* j* @2 A2 A7 h q = p->next; " l6 i t# @: L: Y" S } if (q->exp > p->exp) 2 h i( Y* ]9 c { q = q->next; p = p->next; } } 5 j% w& ~; R; `# D5 I# h}

main() % n# v. C1 E L$ y1 [{ int i, j = 0; printf("a="); " W# f3 ~7 ^# b4 N) ]# }/ h' u scanf("%s", a); j = strlen(a);

input(a, j); w$ G- ^" ]6 C) w/ o stinsort(pa); printf("a="); * I4 H( p2 q5 \, r% } for (i = 0; i <= j + 10; i ++) 3 h: h9 Y" k9 r6 R( b, g( T { printf("%dx%d+", pa->coef, pa->exp); pa = pa->next; } , q8 B& n: o& [; j% _, ]}

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
4 y; U  J5 e- V结果我已经弄出来了，得了个优秀,没跟大家说不好意思
! y  z. _, B0 B+ P6 R: u多谢大家给的意见,我现在水平还很差
" g9 S  ~9 ?7 `3 H5 m+ E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!!~~~~~