求个流程图、真的急需啊 哪位大神帮忙画一个 有题 有代码。。

?e枕黄粱]

顺序结构，,动态链表结构下的一元多项式的加、减、乘法的实现
设有一元多项式Am(x)和Bn(x).
Am(x)=A0+A1x1+A2x2+A3x3+…+Amxm
Bn(x)=B0+B1x1+B2x2+B3x3+…+Bnxn
请实现求M(x)=Am(x)+Bn(x)、M(x)=Am(x)-Bn(x)和M(x)=Am(x)×Bn(x)。
要求：
1)首先判定多项式是否稀疏
2)分别采用顺序和动态存储结构实现；
3)结果M(x)中无重复阶项和无零系数项；
4)要求输出结果的升幂和降幂两种排列情况

代码：
#include<stdlib.h>
#include<stdio.h>
#include<ctype.h>

typedef struct term {  //项的表示，多项式的项作为LinkList的数据元素
float coef;        //系数
int expn;          //指数

   struct term *next;
}term;

term* CreatPolyn(term *P,int m) {  // 算法2.22
  // 输入m项的系数和指数，建立表示一元多项式的有序链表P
  if(m <= 0) return NULL;
  term *h = P = (term*)malloc(sizeof(term)), *q;
  P->coef = 0.0;
  int i;
  printf("依次输入%d个非零项\n",m);
  for (i = 1; i <= m; ++i) {  // 依次输入m个非零项
   scanf("%f%d",&P->coef,&P->expn);
   if(P->coef)
    q = P;
    P = P->next = (term*)malloc(sizeof(term));
  }
  q->next = NULL;
  free(P);
  return h;
} // CreatPolyn

term* selsort(term *h) {
term *g, *p, *q;
if(!h) return NULL;
float f;
int i, fini = 1;
    for(g = h;g->next&&fini;g = g->next) {
  fini = 0;
  for(p = h,q = h->next;q;p = p->next,q = q->next)
   if (p->expn < q->expn) {
    f = p->coef;i = p->expn;
       p->coef = q->coef;p->expn = q->expn;
       q->coef = f;q->expn = i;
    fini = 1;
   }
}
for(g = h,p = g->next;p;)
  if(g->expn==p->expn) {
   g->coef += p->coef;
   g->next = p->next;
   q = p;
   p = p->next;
   free(q);
  }
  else if(g->next) {
   g = g->next;
   p = p->next;
  }
return h;
}

void PrintfPoly(term *P) {
  term *q = P;
  if(!q) {
   putchar('0');
      return;
  }
  if(q->coef!=1) {
     printf("%g",q->coef);
     if(q->expn==1) putchar('X');
     else if(q->expn) printf("X^%d",q->expn);
  }
  else if(!q->expn) putchar('1');
      else if(q->expn==1) putchar('X');
            else printf("X^%d",q->expn);
  q = q->next;
  while (q) {
     if(q->coef > 0) putchar('+');
     if(q->coef!=1) {
        printf("%g",q->coef);
        if(q->expn==1) putchar('X');
        else if(q->expn) printf("X^%d",q->expn);
   }
     else if(!q->expn) putchar('1');
          else if(q->expn==1) putchar('X');
                else printf("X^%d",q->expn);
     q = q->next;
  }
}

Compare(term *a, term *b) {
  if (a->expn < b->expn) return -1;
  if (a->expn > b->expn) return 1;
  return 0;
}

term* APolyn(term *Pa, term *Pb) {  // 算法2.23
  // 多项式加法：Pa = Pa＋Pb，利用两个多项式的结点构成"和多项式"。
  term *h, *qa = Pa, *qb = Pb, *p, *q;
  float sum;
  h = p = (term*)malloc(sizeof(term));
  p->next = NULL;
  while (qa && qb) {     // Pa和Pb均非空
    switch (Compare(qa,qb)) {
      case -1:  // 多项式PA中当前结点的指数值小
          p->next = qb;
          p = qb;
    qb = qb->next;
          break;  
      case 0:   // 两者的指数值相等
          sum = qa->coef + qb->coef;
          if (sum != 0.0) {  // 修改多项式PA中当前结点的系数值
     p->next = qa;
     qa->coef = sum;
              p = qa;
     qa = qa->next;
          }
    else {  // 删除多项式PA中当前结点
     q = qa;
     qa = qa->next;
              free(q);
          }
    q = qb;
    qb = qb->next;
    free(q);
          break;
      case 1:   // 多项式PB中当前结点的指数值小
          p->next = qa;
          p = qa;
    qa = qa->next;
          break;
    } // switch
  } // while
  if (Pa) p->next = qa;   // 链接Pa中剩余结点
  if (Pb) p->next = qb;   // 链接Pb中剩余结点
  q = h;
  h = h->next;
  free(q);
  return h;
} // APolyn

term* A(term *Pa, term *Pb) {
int n;
puts("再输入一一元多项式的项数");
scanf("%d",&n);
Pb = CreatPolyn(Pb,n);
Pb = selsort(Pb);
PrintfPoly(Pa);
if(Pb && Pb->coef>0) printf(" + ");
PrintfPoly(Pb);
    Pa = APolyn(Pa,Pb);
printf(" = ");
Pa = selsort(Pa);
PrintfPoly(Pa);
return Pa;
}

term* BPolyn(term *Pa, term *Pb) {  // 算法2.23
  // 多项式减法：Pa = Pa-Pb，利用两个多项式的结点构成"差多项式"。
term *p = Pb;
while(p) {
  p->coef *= -1;
  p = p->next;
}
  return APolyn(Pa,Pb);
} // BPolyn

term* B(term *Pa, term *Pb) {
int n;
puts("再输入一一元多项式的项数");
scanf("%d",&n);
Pb = CreatPolyn(Pb,n);
Pb = selsort(Pb);
PrintfPoly(Pa);
printf(" - ");
putchar('(');PrintfPoly(Pb);putchar(')');
    Pa = BPolyn(Pa,Pb);
printf(" = ");
Pa = selsort(Pa);
    PrintfPoly(Pa);
return Pa;
}

term* CPolyn(term *Pa, term *Pb) {  // 算法2.23
  // 多项式乘法：Pa = Pa*Pb，利用两个多项式的结点构成"积多项式"。
if(!Pb) return NULL;
term *pa = Pa, *p, *q, *r, *s, *t;
r = p = (term*)malloc(sizeof(term));
while(pa) {
  p->coef = pa->coef;
  p->expn = pa->expn;
  q = p;
  p = p->next = (term*)malloc(sizeof(term));
  pa = pa->next;
}
q->next = NULL;
free(p);
pa = Pa;
t = s = (term*)malloc(sizeof(term));
while(pa) {
  q = s;
  s = s->next = (term*)malloc(sizeof(term));
  pa = pa->next;
}
q->next = NULL;
free(s);
pa = Pa;
while(pa) {
  pa->coef *= Pb->coef;
  pa->expn += Pb->expn;
  pa = pa->next;
}
Pb = Pb->next;
while(Pb) {
  p = r;
  s = t;
  while(p) {
   s->coef = p->coef * Pb->coef;
      s->expn = p->expn + Pb->expn;
   p = p->next;
   s = s->next;
  }
  Pa = APolyn(Pa,t);
  Pb = Pb->next;
}
return Pa;
} // CPolyn

term* C(term *Pa, term *Pb) {
int n;
puts("再输入一一元多项式的项数");
scanf("%d",&n);
Pb = CreatPolyn(Pb,n);
Pb = selsort(Pb);
putchar('(');PrintfPoly(Pa);putchar(')');
printf(" * ");
putchar('(');PrintfPoly(Pb);putchar(')');
printf(" = ");
Pa = CPolyn(Pa,Pb);
Pa = selsort(Pa);
PrintfPoly(Pa);
return Pa;
}

void main() {
term *M,*N;
char s[2];
int i,n;
puts("一元多项式计算:\n输入一一元多项式的项数");
scanf("%d",&n);
M = CreatPolyn(M,n);
M = selsort(M);
PrintfPoly(M);
p: puts("\n1:加\n2:减\n3:乘\n4:退出");
getchar();
q: gets(s);
if(s[1]!='\0' || !isdigit(*s)) {
  puts("输入有误,请重新输入!");goto q;
}
    i = *s-48;
switch(i) {
case 1:M = A(M,N);goto p;;
case 2:M = B(M,N);goto p;;
case 3:M = C(M,N);goto p;
case 4:break;
default:puts("输入有误,请重新输入!");goto q;
}
}

?e枕黄粱]

:'(难道没人么。。。

?e枕黄粱]

小甲鱼老师呢！   来帮帮忙啊{:7_169:}

?e枕黄粱]

在线坐等啊    急需！各种求。。。

番茄

没明白你的意思，需要什么流程图？  作业？

猎鹰三国

为什么这个程序一编译就有许多错误啊

紫炎

数据结构的课程设计么

y290176346

我是来领鱼币的