两个一元多项式相加（链表 || 顺序表实现）
顺序表实现：
#include
#include
#include
#include
#include 链表实现：
#include
#include
#include
#include
#include
#include
#include "AddPolyn.h"
#include "Derivative.h"
using namespace std;
#define maxlen 1000
int main()
{
char s[maxlen];
int casen = 1;
freopen("in.txt", "r", stdin);
freopen("out.txt","w",stdout);
while(scanf("%s", s) != EOF)
{
LNode* L1;
init(L1);
LNode* L2;
init(L2);
Inputpolyn(s, L1, L2); //提取多项式中的系数和指数
printf("Case %d :\n",casen ++);
Deriv(L1); //求第一项的导数
AddPolyn(L1, L2); //多项式相加
printf("%s", s);
print(L1);
printf("\n");
}
return 0;
}
//Derivative.h
#ifndef DERIVATIVE_H_INCLUDED
#define DERIVATIVE_H_INCLUDED
void Deriv(LNode* L1)
{
LNode* p = L1->next;
double coef;
int exp, n = 1, flag = 0;
while(p)
{
coef = p->coef;
exp = p->exp;
if(exp == 0)
{
coef = 0;
}
else
{
coef *= exp;
exp --;
}
//Êä³öµ¼Êý~~~~~~~~~~~~~~~~~~~~~~
if(fabs(coef) < 1e-6)
{
p = p->next;
continue;
}
else
{
flag = 1;
if(n != 1 && coef > 0)
printf("+");
if(exp == 0)
printf("%g", coef);
else if(coef!= 1 && exp != 0 && coef != -1)
{
if(exp == 1)
printf("%gx",coef);
else
printf("%gx^%d",coef, exp);
}
else if(coef == 1)
{
if(exp == 1)
printf("x");
else
printf("x^%d",exp);
}
else if(coef == -1)
{
if(exp == 1)
printf("-x");
else
printf("-x^%d",exp);
}
n ++;
}
p = p->next;
}
if(!flag) printf("0");
printf("\n");
}
#endif // DERIVATIVE_H_INCLUDED
//AddPolyn.h
#ifndef INPUTPOLYN_H_INCLUDED
#define INPUTPOLYN_H_INCLUDED
struct LNode
{
double coef; //系数
int exp; //指数
LNode *next;
};
void init(LNode* &L)
{
L = (LNode*) new LNode;
L -> next = NULL;
}
int compare(int x, int y)
{
if(x == y) return 0;
if(x < y) return 1;
return -1;
}
void Insert(double coef, int exp, LNode* &L)
{
LNode* p = L->next;
LNode* q = L;
while(p != NULL )
{
LNode *s;
int flag = compare(exp, p->exp);
if(flag == -1) //比当前指数大
{
q = p;
p = p->next;
}
else if(flag == 1) //比当前指数小
{
s = (LNode* )new LNode;
s->coef = coef;
s->exp = exp;
s->next = p;
q->next = s;
return;
}
else if(flag == 0) //与当前指数相等
{
p->coef += coef;
return;
}
}
LNode*s = (LNode*) new LNode;
s->coef = coef;
s->exp = exp;
s->next = NULL;
q->next = s;
return;
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~读取多项式
void Inputpolyn(char s[], LNode* &L1, LNode* &L2)
{
int op1 = 1,op2 = 1,op4 = 1,len,p;
double a = 0;
int b = 0;
len = strlen(s);
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~读第一个多项式
for(int i = 0; i < len; i ++)
{
a = 0;
b = 0;
op1 = op2 = 1;
if(s[i] == '=')
break;
if(s[0] == '-' && i == 0)
{
op4 = -1;
i ++;
}
if(s[i] == '(') continue;
if(s[i] == ')')
{
p = i;
break;
}
else if(s[i+1] == '(')
{
p = i-1;
break;
}
if(s[i] == '-')
{
i ++;
op1 = -1;
}
else if(s[i] == '+')
{
i ++;
op1 = 1;
}
if(s[i] >= '0' && s[i] <= '9')
{
while(s[i] >= '0' && s[i] <= '9' && i < len)
{
a = a*10+(s[i]-'0');
i ++;
}
if(s[i] == '.')
{
i ++;
double x = 0.1;
while(s[i] >= '0' && s[i] <= '9' && i < len)
{
a += x*(s[i]-'0');
i ++;
x *= 0.1;
}
}
a = a*op1*op4;
}
else if(s[i] == 'x')
{
a = op1*op4;
}
if(s[i] == 'x')
{
i ++;
if(s[i] == '^')
{
i ++;
if(s[i] == '-')
{
op2 = -1;
i ++;
}
else if(s[i] == '+')
{
op2 = 1;
i ++;
}
while(s[i] >= '0' && s[i] <= '9' && i < len)
{
b = b*10 + s[i] - '0';
i ++;
}
b = b*op2;
i --;
}
else
{
b = op2;
i --;
}
}
else
{
b = 0;
i --;
}
//i --;
//printf("%lf %d", a, b);
Insert(a, b, L1);
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~读第二个多项式
op4 = 1;
for(int i = p+1; i < len; i ++)
{
a = 0;
b = 0;
op1 = op2 = 1;
if(s[i] == '-' && i == p+1)
{
op4 = -1;
i ++;
}
else if(s[i] == '+' && i == p+1)
{
op4 = 1;
i++;
}
if(s[i] == '=') break;
if(s[i] == '(') continue;
if(s[i] == ')')
{
i ++;
if(s[i] == '=')
break;
}
if(s[i] == '-')
{
i ++;
op1 = -1;
}
else if(s[i] == '+')
{
i ++;
op1 = 1;
}
if(s[i] >= '0' && s[i] <= '9')
{
while(s[i] >= '0' && s[i] <= '9' && i < len)
{
a = a*10+(s[i]-'0');
i ++;
}
if(s[i] == '.')
{
i ++;
double x = 0.1;
while(s[i] >= '0' && s[i] <= '9' && i < len)
{
a += x*(s[i]-'0');
i ++;
x *= 0.1;
}
}
a = a*op1*op4;
}
else if(s[i] == 'x')
{
a = op1*op4;
}
if(s[i] == 'x')
{
i ++;
if(s[i] == '^')
{
i ++;
if(s[i] == '-')
{
op2 = -1;
i ++;
}
else if(s[i] == '+')
{
op2 = 1;
i ++;
}
while(s[i] >= '0' && s[i] <= '9' && i < len)
{
b = b*10 + s[i] - '0';
i ++;
}
b = b*op2;
i --;
}
else
{
b = op2;
i --;
}
}
else
{
b = 0;
i --;
}
Insert(a, b, L2);
}
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~两个多项式相加
void AddPolyn(LNode* &L1, LNode* &L2)
{
LNode* p1 = L1->next, *p2 = L2->next, *r = L1, *q;
//LNode* pre1 = L1,*pre2 = L2;
while(p1 && p2)
{
int flag = compare(p1->exp, p2->exp);
if(flag == 1)
{
r->next = p1;
r = p1;
//pre1= p1;
p1 = p1->next;
}
else if(flag == -1)
{
r->next = p2;
r = p2;
//pre2 = p2;
p2 = p2->next;
}
else if(flag == 0)
{
double sum = p1->coef + p2->coef;
if(fabs(sum) < 1e-5)
{
if(L1->next == p1)
L1->next = p1->next;
q = p1;
p1 = p1->next;
delete q;
}
else
{
p1->coef = sum;
r->next = p1;
r = p1;
p1 = p1->next;
}
//if(pre2->next == p2)
//pre2->next = p2->next;
q = p2;
p2 = p2->next;
delete q;
}
}
if(p1)
r->next = p1;
if(p2)
r->next = p2;
delete L2;
L2 = q = p1 = p2 = r = NULL;
}
void print(LNode *L)
{
int flag = 0,n = 1;
LNode *p = L->next;
while(p)
{
double coef = p->coef;
int exp = p->exp;
//cout << coef <<" "<next;
continue;
}
else
{
flag = 1;
//if(n == 1) printf("(");
if(n != 1 && coef > 0)
printf("+");
if(exp == 0)
printf("%g", coef);
else if(coef!= 1 && exp != 0 && coef != -1)
{
if(exp == 1)
printf("%gx",coef);
else
printf("%gx^%d",coef, exp);
}
else if(coef == 1)
{
if(exp == 1)
printf("x");
else
printf("x^%d",exp);
}
else if(coef == -1)
{
if(exp == 1)
printf("-x");
else
printf("-x^%d",exp);
}
n ++;
}
p = p->next;
}
if(!flag) printf("0");
//else printf(");");
return;
}
#endif // INPUTPOLYN_H_INCLUDED
测试数据：
(2x+5x^8-3.1x^11)+(7-5x^8+11x^9)=
(6x^-3-x+4.4x^2-1.2x^9+1.2x^9)-(-6x^-3+5.4x^2-x^2+7.8x^15)=
(1+x+x^2+x^3+x^4+x^5)+(-x^3-x^4)=
(x+x^3)+(-x-x^3)=
(x+x^100)+(x^100+x^200)=
(x+x^2+x^3)+0=
(7-5x^8+11x^9)+(2x+5x^8-3.1x^11)=
-(-6x^-3+5.4x^2-x^2+7.8x^15)+(6x^-3-x+4.4x^2-1.2x^9+1.2x^9)=
(-x^3-x^4)+(1+x+x^2+x^3+x^4+x^5)=
(-x-x^3)+(x+x^3)=
(x+x^100)+(x^100+x^200)=
0+(x+x^2+x^3)=
(x+x)=
(1+x-x)+x=
(x-9+x^100)+(x-x^8)=
(-x+x)=
(23+10)=
输出结果（输出第一行为第一项的导数值）：
Case 1 :
2+40x^7-34.1x^10
(2x+5x^8-3.1x^11)+(7-5x^8+11x^9)=7+2x+11x^9-3.1x^11
Case 2 :
-18x^-4-1+8.8x
(6x^-3-x+4.4x^2-1.2x^9+1.2x^9)-(-6x^-3+5.4x^2-x^2+7.8x^15)=12x^-3-x-7.8x^15
Case 3 :
1+2x+3x^2+4x^3+5x^4
(1+x+x^2+x^3+x^4+x^5)+(-x^3-x^4)=1+x+x^2+x^5
Case 4 :
1+3x^2
(x+x^3)+(-x-x^3)=0
Case 5 :
1+100x^99
(x+x^100)+(x^100+x^200)=x+2x^100+x^200
Case 6 :
1+2x+3x^2
(x+x^2+x^3)+0=x+x^2+x^3
Case 7 :
-40x^7+99x^8
(7-5x^8+11x^9)+(2x+5x^8-3.1x^11)=7+2x+11x^9-3.1x^11
Case 8 :
-18x^-4-8.8x-117x^14
-(-6x^-3+5.4x^2-x^2+7.8x^15)+(6x^-3-x+4.4x^2-1.2x^9+1.2x^9)=12x^-3-x-7.8x^15
Case 9 :
-3x^2-4x^3
(-x^3-x^4)+(1+x+x^2+x^3+x^4+x^5)=1+x+x^2+x^5
Case 10 :
-1-3x^2
(-x-x^3)+(x+x^3)=0
Case 11 :
1+100x^99
(x+x^100)+(x^100+x^200)=x+2x^100+x^200
Case 12 :
0
0+(x+x^2+x^3)=x+x^2+x^3
Case 13 :
2
(x+x)=2x
Case 14 :
0
(1+x-x)+x=1+x
Case 15 :
1+100x^99
(x-9+x^100)+(x-x^8)=-9+2x-x^8+x^100
Case 16 :
0
(-x+x)=0
Case 17 :
0
(23+10)=33