Question

题目：用单链表实现任意两个一元多项式的加、减法运算

Answer 1

#include

#include

#include

#define N 7

typedef enum { add, nul, sub, div1, yu, l, r }OP;

int a[N][N] = {

{ 0, 0, -1, -1, -1, 1, 2 },

{ 0, 0, -1, -1, -1, 1, 2 },

{ 1, 1, 0, 0, 0, 1, 2 },

{ 1, 1, 0, 0, 0, 1, 2 },

{ 1, 1, 0, 0, 0, 1, 2 },

};

int top;

OP beg;

int b[1024];

OP op[1024];

void init_nu( ){ top = 0; }

void push_nu( int term ){ b[top++] = term; }

int pop_nu( ){ return b[--top]; }

int is_empty_nu( ){ return top == 0; }

void destory_nu( ) { top = 0; }

void init_sign( ){ beg = 0; }

void push_sign( OP sign ){ op[beg++] = sign; }

void destory_sign( ){ beg = 0; }

OP pop_sign( ){ return op[--beg];}

OP get_sign( ){ return op[beg - 1]; }

int is_empty_sign( ){return beg == 0; }

int eval()

{

int i, j;

i = pop_nu();

j = pop_nu();

switch( pop_sign() )

{

case '+': push_nu( j + i ); break;

case '-': push_nu( j - i ); break;

case '*': push_nu( j * i ); break;

case '/': push_nu( j / i ); break;

case '%': push_nu( j & i ); break;

defult: break;

}

}

int change( char *s )

{

int i;

int n = strlen( s );

for( i = 0; i < n; i++ )

{

if( s[i] >= '0' && s[i] = '0' && s[i]