多項式結構運算 精簡版(C)

這跟前一篇是同樣的問題, 只是發問者不同; 而這次, 我寫了 int整數的精簡版...

//===============================================================
// 20090505 知識 +
// http://tw.knowledge.yahoo.com/question/question?qid=1609050507403
// 發問者 :(ajac) http://tw.knowledge.yahoo.com/my/my?show=AD01707585
//===============================================================

程式是以不定參數的方式來建立多項式, 項次的參數型態需為 int (參數使用方式不同前一篇)

#include <stdio.h>
#include <stdlib.h>
#include <stdarg.h>

typedef struct stMulti
{
  struct  stMulti * Next;
  int     Len;
  int     Num[1];
}Multi;
Multi * First = NULL;
Multi * Last = NULL;

// 產生新多項式
Multi * NewMulti(int n)
{
  Multi * m = (Multi *) malloc(sizeof(Multi)+sizeof(int)*n);
  m->Next = NULL;
  m->Len = n;
  while(--n>=0) m->Num[n] = 0;
  if (First==NULL) First = m;
  if (Last!=NULL) Last->Next = m;
  return Last = m;
}
// 建立多項式 : exp為最高冪值, 後為int型態係數
Multi * CreateMulti(int exp, ... )
{
  int n = (exp<0)? 1: exp+1;
  Multi * m = NewMulti(n);
  va_list arg;
  va_start( arg, exp);
  while(--n>=0) m->Num[n]=va_arg(arg, int);
  va_end( arg);
  return m;
}
// 多項式相加 A+B
Multi * MultiAdd(Multi *A, Multi *B)
{
  int n = (A->Len>B->Len)? A->Len: B->Len;
  Multi * m = NewMulti(n);
  while(--n>=0)
    m->Num[n]= (n<A->Len? A->Num[n]: 0) + (n<B->Len? B->Num[n]: 0);
  return m;
}
// 顯示多項式
void PrintMulti(char *s, char c, Multi *A)
{
  char f1[][10] = { "+%d", "%d", "+%d%c", "%d%c", "+%d%c^%d", "%d%c^%d" }; 
  char f2[][8] = { "+%c", "%c", "+%c^%d", "%c^%d" }; 
  int m, o=0;
  int n = A->Len;
  printf(s);
  while(--n>=0)
  {
    if (A->Num[n]==0) continue;
    m = n>1? 4: n*2;
    if (A->Num[n]<0 || o++==0) m++;
    if (n!=0 && A->Num[n]==1) printf( f2[m-2], c, n);
    else printf( f1[m], A->Num[n], c, n);
  }
  printf("\n");
}
// 清除配置體
void ClearAll()
{
   Multi * node = First;
   Multi * f;
   for( ;(f=node)!=NULL; free(f)) node=node->Next;
   First = Last = NULL;
}
// 主程式
int main()
{
  Multi *A, *B, *C; 
  //A(X)= 7x^5+3x^2+8   冪5 係數序: 7 0 0 3 0 +8 
  PrintMulti( "A(x) = ", 'x', A = CreateMulti( 5, 7, 0, 0, 3, 0, 8 ) ); 
  //B(X)= 5x^6+2x^5+9X^2+x+12 冪6 係數序: 5 2 0 0 9 1 +12
  PrintMulti( "B(x) = ", 'x', B = CreateMulti( 6, 5, 2, 0, 0, 9, 1, 12 ) ); 
  //C(X) = A(X)+B(X) = 5X^6+9X^5+12X^2+X+20
  PrintMulti( "C(x) = A(x)+B(x) = ", 'x', C = MultiAdd( A, B ) );
  ClearAll();
  system("pause");
  return 0;
}

~~~