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升级   19% TA的每日心情 | 奋斗 2016-7-14 13:54 |
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签到天数: 219 天 [LV.7]常住居民III
群组: 2015年美赛冲刺 群组: 高等数学串讲 群组: 2014第三期英语写作 群组: SAS数据分析大赛冲刺 群组: 2015美赛优秀论文解析 |
#include<stdio.h>8 Y j5 R; D: `1 Z1 G
#include<malloc.h>7 W. ]1 u' Z4 n! o
typedef struct node" T% e( \ I+ a% h, [
{) f4 } j {! H/ K2 D$ ^
float xi;
0 g! p2 ^1 ?* T6 E int n;" ]( \1 z' ]# |# q, R
struct node *next;
1 K: v7 _- D* O6 ~}term;
. ^+ H& r9 I' X9 ~2 K- lvoid sort(term *head)- ?6 S$ Y0 m- C) U
{) g/ z& u1 u/ Y- Z" c: N/ O
term *p,*q,*s;
+ X* Z: } c7 w" z) D1 E1 b p=(term *)malloc(sizeof(term));3 ^* D6 Z+ x* X* j K7 U
q=(term *)malloc(sizeof(term));
7 U7 d0 A3 ^! q4 }$ c& y/ U! e s=(term *)malloc(sizeof(term));8 R' k$ Q6 ~; u- ^# e
p=head;2 ]! g$ ^# l3 t
+ k! F& N# D% e4 o& s a" {* b while(p->next!=NULL)
( i: x9 T$ m+ J( Z0 _4 s {
; O: |0 \ w" x8 F6 V* c2 B s=p->next;, t: v8 F! N+ K# ?4 \% I9 U
while(s!=NULL)
1 _! q; O# A/ W: y# ~7 w( ? { 1 o# q* B4 w, p) g
if(p->n>s->n), E4 q p- k, l* z
{
( u/ z& W, @8 ] D q->n=p->n;
( V7 j2 ^# F( Y1 W/ R# I, e ] p->n=s->n;
2 K+ o3 t! K% H! b: p2 { s->n=q->n;) }" r% c1 p' g$ Q$ l9 o8 I
q->xi=p->xi;
8 z* L$ D0 Q# m p->xi=s->xi;: ~+ Y8 Q& c% e0 X6 c
s->xi=q->xi;
/ h3 G: I2 {, X/ ]/ b }8 d- z u K) ~* @# S ~
s=s->next;
& M7 f. U5 g% _* u: { }
( c% g7 B& R+ O4 x/ F p=p->next;- a2 g4 j2 T3 P
}
% b/ O' G6 i; |: ^% h}
' A( L% D" B* i6 s- G* Bterm *polynadd(term *head1,term *head2)1 ~' C1 d8 x0 g
{" B7 X# v' I- B( i
term *p,*q,*s,*r,*head3;
. x% P5 L4 m5 v7 V% i K4 } float x;
( ^; c7 U; b. q( w5 @ p=head1->next;% c+ S) W% V8 w2 m
q=head2->next;' s0 P* S0 \& f- K) H
head3=(term *)malloc(sizeof(term)); x3 Z8 l! k" E0 p1 s
r=head3;
( i4 y4 ?1 O# @ while(p!=NULL&&q!=NULL)
$ t' c' P" a+ k8 P5 [ {
) _: y0 O' A0 k; y T; n# n. Y if(p->n==q->n)
2 L% T; y1 z' Z( x {9 n R6 |( Z: I7 [- t# l' Z' S
x=p->xi+q->xi;
8 P8 e& X, a2 ~7 s7 b$ H. p if(x!=0)
5 z ~$ L) b! U! j& b# t {
: _" X) y) |4 l' ~( W5 ]6 i4 | s=(term *)malloc(sizeof(term));
2 b4 h# F+ n k: W0 v4 L7 E) \3 `+ G s->xi=x;0 B# _) N# n# `4 m7 f; T
s->n=p->n;
. _3 d. V% r: T& e$ R! }# }1 l! |' x r->next=s;
C8 d4 G+ G: K6 m) I/ d r=s;
( X8 y8 E- L, s {6 p7 G1 P2 e }' C8 A5 e/ D8 h- k7 s8 Y, }7 z
p=p->next;
) C! A5 d B2 _' Z q=q->next;
1 K! ? i6 }. v! B }
4 \5 z% a/ K) ^9 | Y) o2 ], p else if(p->n>q->n)
6 v+ T9 K' f5 T6 C/ O8 \0 @5 r {
$ b" J$ X% ]$ R: \1 |' G s=(term *)malloc(sizeof(term));+ n! D2 R! g5 C! }3 E
s->n=q->n;4 T5 x/ a- ?6 g w
s->xi=q->xi;
1 a- O/ y" X" ?3 O' _$ _! D9 S) ^ r->next=s;
K7 ?, K% O4 E6 H" [1 ]; p r=s;0 S* H t# ~3 e- q: I
q=q->next;
" F/ E- p5 r4 g3 }9 v7 y- I }
) `" I& U. h. M: H else
. m+ }0 R, l k9 i0 T8 ^3 f* } {% C {& @5 q! j) \1 s6 o2 ~( p
s=(term *)malloc(sizeof(term));2 k: ?/ [- u3 m r
s->n=p->n;
; c& R/ U+ {7 h+ `, `" t9 z( w s->xi=p->xi;8 n' v" W! d$ Z
r->next=s;* Y5 n, ?4 ]( Q3 ~
r=s;
6 T+ Z; \6 d* x7 m+ ? p=p->next;
8 v2 O7 f2 x0 n. |! X: F9 f }
! t, Z: v5 f, x }
8 ~: V9 J) i) M while(p!=NULL)' p. d0 a5 q: d. {& P- C5 g: {
{% N9 ^( ^. |& \( e6 t8 F
s=(term *)malloc(sizeof(term));
% V$ [6 n/ w4 s' D' I& h s->n=p->n;
9 _# e% ~; u i( d2 ]+ n# @ s->xi=p->xi;
" f0 \; V1 C7 u r->next=s;
! J2 u0 @; n# I, V& |+ q- i% T r=s;8 v S) Q( t5 Q2 R
p=p->next;3 y5 N9 O+ q0 E1 _( s3 R
}, h5 `* v3 v3 |, }$ u
while(q!=NULL)0 y- w$ V) H1 H" \
{, _7 H1 a1 \0 K
s=(term *)malloc(sizeof(term));; t& y @( m6 h1 f& U& i: P) T
s->n=q->n;- d: K5 z$ n4 t; Z/ f. l2 M! j- r9 l
s->xi=q->xi;$ ~- L: E( e5 t5 [. ~0 p& I
r->next=s;1 `& G6 ~9 w. h( R
r=s;
5 e7 B; `& s, w q=q->next;# k' T% A9 D5 F0 b* S; Z X
}2 B4 }- k1 e- t. }" f# i
r->next=NULL;: `' ^# L* r& T4 z* h* I8 m( L
return head3;' L& U4 j; i* W0 j( i0 u+ f
5 q0 b# E9 @, |, w}3 c7 q" q3 I/ J( d% L
term *createpolyn(int m)! O: O& e, n: v
{
) Q e' E( x6 ?" l9 m" @- S) \% C8 j* T' d( p6 L% [- D# K
term *p,*head,*q;5 ]3 [! p! X. _7 r7 F0 t2 V
int i;7 d. k* V) s$ z. k5 N
head=(term *)malloc(sizeof(term));
' ^, R( v! |! H9 F& ]' W4 E4 l q=head;
; P# R4 O; N* I) w7 H. { for(i=0;i<m;i++)8 j. _1 y" K& ]
{
/ q! f' U4 d- A p=(term *)malloc(sizeof(term));$ |/ J+ r4 l' }. b
printf("请输入第%d项数的系数和指数\n",i+1);% n4 Y1 V! ?0 T/ I7 h
scanf("%f%d",&p->xi,&p->n);0 m: T$ l) H: b& K$ B. l
q->next=p;! o1 E# d4 T6 l8 ~7 Y; z
q=p;
# z1 n2 x$ O% ~ }! {1 q% L' N2 u1 A1 `+ W. X
p->next=NULL;
1 O/ P& \5 B1 l5 `; [ return head;3 H$ P i1 U1 {: q1 ~" A) h6 a5 \
* D% Z7 g F1 }( }5 R: A
}
3 Y% r3 Y/ F( Z( zdayin(term *head)
4 m# d- @. T7 {% b" [& V, m{' ]- `4 i! D; Y- }) L4 U" q
term *p,*q;
0 S5 a* L2 a. p6 g' G q=head->next;5 y% w* u) o( a1 P" e* t: E* ^1 E5 S
if(q->xi==0)
* j. I7 G7 m; n! A: \% T printf(" ");. u; H) h4 J" y
if(q->n==0&&q->xi>0&&q->xi!=1)
- |8 D4 \8 z% D1 p/ H printf("%0.2f",q->xi);' n2 a/ A1 e5 W
if(q->n==0&&q->xi==1)6 s, e+ P, {1 |8 g, t5 Y
printf("1");
7 x8 }/ c4 ]7 y3 D* @9 o1 q$ A4 C if(q->n==0&&q->xi<0): p- {' s+ p3 z E2 Y" G
printf("%0.2f",q->xi);6 O5 @7 f5 y: ?. H. W
if(q->n==1&&q->xi==1)
# i! D% r7 d: c- g, h1 O printf("x");1 i) `, j5 I( ~& _/ h) j1 w! l, u5 M. P
if(q->n==1&&q->xi!=1&&q->xi>0)
+ ^$ f. S- S( W7 p0 Q' h# s7 ?6 h printf("%0.2fx",q->xi);
1 N5 Y3 @' k2 m! I x3 `) s if(q->n==1&&q->xi!=1&&q->xi<0)" S% u( E' e- ?
printf("%0.2fx",q->xi);
" F/ {4 W- [* h7 x2 u if(q->n!=1&&q->n!=0&&q->xi>0&&q->xi!=1)
- L6 D/ p9 p4 U1 h* y1 R* T printf("%0.2fx^%d",q->xi,q->n);# Q/ m ?/ I/ R9 P) `( v* m( x; F9 _
if(q->n!=1&&q->n!=0&&q->xi<0)
& T$ Z( C# }. ]; z( b printf("%0.2fx^%d",q->xi,q->n);* _" o% S6 d* L* h9 R+ H( Y
if(q->n!=0&&q->n!=1&&q->xi==1)
/ p9 w' d; U7 v+ _, M+ ?* u* c9 L. f printf("x^%d",q->n);3 Y- g5 N0 f9 p2 z' q( q6 n/ j
p=q->next;
4 |; K- k8 }, B* M) j0 _ while(p!=NULL)
3 H( V& \+ |0 b {
/ g- g3 ?. G4 T8 H if(p->xi==0)
; V9 x8 x" G6 V, h; Z* M printf(" ");% Z; I+ P% D* l# u' _1 u8 |
if(p->n==0&&p->xi>0&&p->xi!=1)
0 b4 |+ ~ N6 J0 u u printf("+%0.2f",p->xi);
. B( m" V4 k$ c1 [* ] if(p->n==0&&p->xi==1)
4 _- ]- V7 L7 W5 q& q printf("+1");, P+ Y# x P- F$ S6 |5 J
if(p->n==0&&p->xi<0)
5 u, V- \* b( ~. B/ `; s3 J printf("%0.2f",p->xi);* O' H; c' s2 j/ W$ e- c( P2 _4 F# z
if(p->n==1&&p->xi==1)/ O& m# ]1 ` e7 X
printf("+x");% ]: x- f' I1 y# c
if(p->n==1&&p->xi!=1&&p->xi>0)' ?6 m- }7 \& u" D9 Q. E4 S6 a/ y
printf("+%0.2fx",p->xi);4 p7 M7 O, {' T+ X0 z
if(p->n==1&&p->xi!=1&&p->xi<0)
: C" [1 x7 y( l printf("%0.2fx",p->xi);
4 |" i7 a9 E. u if(p->n!=1&&p->n!=0&&p->xi>0&&p->xi!=1)4 x( j3 i, P8 O8 v$ C
printf("+%0.2fx^%d",p->xi,p->n);
0 [. C/ \, t$ G( u if(p->n!=1&&p->n!=0&&p->xi<0): U9 ?5 k- E# e1 g+ U- m
printf("%0.2fx^%d",p->xi,p->n);1 J: L* u" F4 t0 F+ A( v# z
if(p->n!=0&&p->n!=1&&p->xi==1)6 X9 p, y7 P7 d7 g9 i/ ]! s
printf("+x^%d",p->n);
! P9 S7 y! f( G0 v# s% p/ ^; N p=p->next;- g% z+ \3 O H( S
}
* R, V5 M# I' Z6 { printf("\n");9 G0 H g5 W- B8 ]
}+ V8 u% q, q- K7 U7 @
main()) T* v0 h* z6 d, m3 ]& X, j4 A
{
2 T7 [4 n/ J" W/ C" c' A7 E int i,j;
* k8 N( s; G/ m! U) V3 I+ U term *head1,*head2,*head3;
& Q( I2 `* X) q( R! R printf("请输入第一个多项式的项数:\n");
/ F' |' {+ N6 t4 p1 F9 m0 e' J scanf("%d",&i);2 P$ P! l: S+ s
head1=createpolyn(i);
5 B9 H1 H j2 ^& I" u- s7 R sort(head1);+ c9 ]: a4 w7 _: u/ Y6 k
printf("\n");
+ w5 X' H' h8 E3 E/ e) I. d dayin(head1);) h* K9 T2 q `
printf("\n");
& Z6 [, g5 R4 ~2 K7 O+ w3 u9 R printf("请输入第二个多项式的项数:\n");7 q$ I: n: j" ^/ e3 g
scanf("%d",&j);/ [% L6 \9 A" b {' I. o
head2=createpolyn(j);% Y( D6 g3 Q6 S$ E2 E
sort(head2);
, @; d) O- m$ s' M; M2 g printf("\n");
6 h: h( e2 I. B" o" \$ T dayin(head2);( t8 n5 C5 U* p+ C+ a9 _' X
printf("\n\n");
& H, ?' Z7 J6 H# o- S7 o head3=polynadd(head1,head2);
, b- k+ Y6 G4 C- E V sort(head3);
( K: _: r% t# v# N1 h printf("\n");
. o2 \( s% d9 ^( o! e6 [3 { dayin(head3);
. d( v' |& t1 g' z" \ printf("\n");
* y1 r5 f# F" Z; y} |3 U" G4 k* l/ }. a
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