[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 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 " D( V) U* Z" ?0 q, O7 F#define MAXSIZE 400 char a[MAXSIZE]; 9 ?; A! C& ^( b! ]; p/ jtypedef struct{char elem[MAXSIZE]; / w, W: s5 @9 g* B- L9 r int top; / M6 Y, J7 v( H }SqStack; $ K: \" r- c6 r/ f; w; |0 s/ q: f/ h9 LSqStack S; - T# H4 X! R6 @, ?# h# Y. F. U5 |typedef struct poly \8 j/ @& x: x$ C# z) L( Y {int coef; /*xishu*/ 8 a! G: R5 \' ?. P: [8 L7 R int exp; /*zhishu*/ struct poly *next; }Lpoly; $ N5 l4 A% o0 E- X s nLpoly *pa; w0 W! j' ?0 h9 O* S5 tint cf(int i) /*10^i*/ 1 ^. \* x5 R& U3 [: [" s% w/ [' {{ int c=1,j; for(j=1;j<=i;j++) c=c*10; 4 D$ x6 G; c/ T d return c; } void input(char a[],int j) 3 d4 U! t1 P' {8 F6 h* T1 d) y{ Lpoly *p,*h,*ptr; int i,k,flag=0,x=0; S.top=-1; 7 o' C" {4 L r7 N pa=(Lpoly*)malloc(sizeof(Lpoly)); pa->coef=0;pa->exp=0; h=pa;h->next=NULL; p=h; /*jianlipa*/ for(i=0;i<=j-1;i++) {ptr=(Lpoly*)malloc(sizeof(Lpoly)); " o. S9 ^/ J, E' J, C 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" ( S* u! L) c& E# d$ }0 A#include "string.h" ! n( V# ~$ s4 E4 L$ ~#include "stdio.h"

#define MAXSIZE 400 ) U9 i7 Y% D! |; w& l7 [char a[MAXSIZE];

typedef struct ' x# Y/ o% y# @ `{ char elem[MAXSIZE]; int top; }SqStack;

SqStack S;

typedef struct poly { int coef; /*xishu*/ int exp; /*zhishu*/ struct poly *next; }Lpoly;

Lpoly *pa;

int cf(int i) /*10^i*/ 3 V% _7 R0 H% @{ int c = 1, j; for(j = 1; j <= i; j ++) c = c * 10; ' F5 n; x$ L0 K return c; }

void input(char a[],int j) : c% o5 @8 k! z0 B4 x4 |{ Lpoly *p, *h, *ptr; 3 Y" ]2 [; v( x. q5 i int i, k, flag = 0, x = 0; * Z0 ]( ~' y1 m: Q S.top = -1; pa = (Lpoly*)malloc(sizeof(Lpoly)); " @2 N% t0 J2 s: k pa->coef = 0; pa->exp = 0; # u5 k2 f& a7 {9 n, ]# Z: A h = pa; h->next = NULL; p = h; /*jianlipa*/ 9 W" j. h4 V+ N0 E- v" R! j 5 _5 Z0 @% H8 H4 j7 \% H for (i = 0; i <= j - 1; i ++) ' @3 ]2 |# h6 ] [ s8 ? { . F* z/ e3 u) \5 x ptr = (Lpoly*)malloc(sizeof(Lpoly)); if(S.top) { . u( [4 g6 x0 n+ Z0 V S.top ++; ! I; n2 _, t+ |1 b) B2 x S.elem[S.top] = a; } & B8 p1 i. M( R7 n if (a == '-') flag=1; if(a == 'x') /*fenlixishu*/ ' R/ I. u; @ I { if (S.top==0) , |2 _7 G7 q# h2 V$ ` { if (flag == 1) { ptr->coef = -1; ptr->next = NULL; $ B# t' g4 o; a } else { ptr->coef = 1; " @! `5 r; o R, f ptr->next = NULL; } } ( L; q. s/ N& J+ b else { for (k=S.top - 1; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); 5 I$ K+ G1 j1 ?3 K1 `. m1 } ptr->coef = x; ptr->next = NULL; * E" o6 h5 M1 D. I1 a) r# S. i0 {8 k x = 0; } # ?9 T; B3 `5 [) s- l S.top = -1; flag = 0; /*clear*/ / n% w$ h0 }; Z v1 R8 L' _+ g } if (a == '+' || a == '-') /*fenlizhishu*/ 9 s5 Z: N4 d, r { if(S.top == 0) ' e: L9 W' e Y {6 E, P { ptr->exp = 1; ; Q( `- U/ V2 o0 C ptr->next = NULL; } else # d. ?0 }6 \% W- {* Y/ `: u- v { for (k = S.top - 1; k >= 0; k --) % Z/ G2 k8 n' o) j6 K x = x + (S.elem[k] - '0') * cf(S.top - 1 - k); ptr->exp = x; ( P. s3 S- o2 I: |# w r( F4 N4 V5 @3 N9 Q( R ptr->next = NULL; x = 0; }/*tiquzhishu*/ S.top = -1; * t; ]" f$ h5 G; N } ) I/ J, U2 v* `. k" U3 d1 O; D if (ptr->coef != 0 && ptr->exp != 0) { p->next = ptr; p = ptr; } + w6 M* H; R y# b if (i == j - 1) { if (ptr->coef == 0) { : u7 q M/ N( x' Q* \ for (k = S.top; k >= 0; k --) : ~' g8 z+ u! R/ |6 u* E/ M x = x + (S.elem[k] - '0') * cf(S.top - k); ptr->coef = x; ' V3 B5 P2 S) q8 ?! \8 a ptr->next = NULL; } ; |: Y( m+ g1 P4 t( C7 v" A else { ; R2 `. G5 V" }8 Z ^" U for (k = S.top; k >= 0; k --) x = x + (S.elem[k] - '0') * cf(S.top - k); 0 \7 h; p4 o+ T$ K$ O0 Q- V) n2 s ptr->exp = x; ptr->next = NULL; & I9 d$ i8 _ X& F8 O4 C } p->next = ptr; 1 U g+ H# f, A( }8 p; @3 h' Z ptr->next = NULL; 8 q2 w+ Y1 Q+ f, d+ ]( S } + a& N% Q6 t3 L: u2 o; F7 v } }

void stinsort(Lpoly *pa) . M1 Z {2 G( }4 T' h6 V{ % q+ U4 s+ l* @% @0 J Lpoly *p, *h, *q, *k; 0 j+ l( m4 B# _9 {0 Q' f h = pa; p = k = h->next; ) J6 j, r/ S5 O, O% i q = p->next; while (p->next != NULL) { if (q->exp) 0 `% x% g4 Y/ d( O) A+ u$ Y! Z0 A { while(q->exp > k->exp) { k = k->next; 8 R2 }1 Q( `& z. X h = h->next; } if(q->exp) { ; K( E. U% ?4 U9 B p->next = q->next; 0 Z1 r) `+ g1 q! V; O+ b h->next = q; / B! \' s3 w5 S4 t5 E- Z B2 J0 J6 V: S q->next = k; % ~- N/ Y3 f& b. |: m q = p->next; } % }/ @8 \( E8 H2 k% b+ T if (q->exp = k->exp) { k->coef = k->coef + q->coef; p->next = q->next; q = p->next; 4 E3 |4 P+ d3 L } & [& U5 f2 _! q* c6 c/ r* F: ?9 R h=pa; k = h->next; } 1 ~9 A5 T4 Z G8 P if (q->exp = p->exp) { p->coef = p->coef + q->coef; p->next = q->next; q = p->next; & f5 U" k2 O/ r x9 _8 K } if (q->exp > p->exp) 9 }9 O9 a3 d, Y8 H6 M* f. y { q = q->next; 9 Y6 w0 z2 t7 O% S- {" p! U p = p->next; ' [9 T" u* w2 q } 3 t/ W' _7 m/ \5 O } }

main() / @5 l2 J( y5 {7 d/ t% O1 y{ int i, j = 0; 3 y8 c2 d- ]0 u9 K# U; p1 X! t printf("a="); scanf("%s", a); / F+ S6 K e2 U) g8 Z9 J. { j = strlen(a);

input(a, j); stinsort(pa); * T y8 {, n4 k printf("a="); , D8 L: C6 V% P% i; [- e for (i = 0; i <= j + 10; i ++) $ K5 H, v! W' E* E z { $ L; C9 t# }1 Q6 _5 O8 p8 I printf("%dx%d+", pa->coef, pa->exp); 9 g* e% n( m2 J" C 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
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!!~~~~~