兩個一元多項式相加（連結串列 || 順序表實現）
阿新 • • 發佈：2019-02-09
順序表實現：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <map> using namespace std; #define maxsize 1000 struct Data { double coef; int exp; }; struct List { Data* elem; int length; }; char s[100]; map <int, int> m; bool cmp(Data a, Data b) { return a.exp < b.exp; } void InitList(List &L) { L.elem = (Data* )new Data[maxsize]; L.length = 0; } void Derivative(int len, List L) { int n = 1,flag = 0; for(int i = 0;i < len;i ++) { double coef = L.elem[i].coef; int exp = L.elem[i].exp; if(exp == 0) { coef = 0; } else { coef *= exp; exp --; } if(fabs(coef) < 1e-6) 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 ++; } } if(!flag) printf("0"); printf("\n"); } int main() { freopen("in.txt", "r", stdin); List L; int op1 = 1, op2 = 1,op3 = 1,op4 = 1; double a; int b, cases = 1,flag; while(scanf("%s", s) != EOF) { printf("Case %d:\n",cases++); InitList(L); op1 = op2 = op3 = op4 = 1; flag = 0; int len = strlen(s); for(int i = -10000;i <= 10000;i ++) m[i] = -1; int k = 0; for(int i = 0;i < len;i ++) { a = 0; b = 0; op1 = op2 = 1; if(s[0] == '-' && i == 0) { op4 = -1; i ++; } if(s[i+1] == '(') { Derivative(k, L); flag = 1; } if(s[i] == '=') break; if(s[i] == '(') continue; if(s[i] == ')') { i ++; op4 = 1; if(s[i] == '=') { if(!flag) Derivative(k, L); break; } else { flag = 1; Derivative(k, L); if(s[i] == '+') op3 = 1; else if(s[i] == '-') op3 = -1; } } 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*op3*op4; } else if(s[i] == 'x') { a = op1*op3*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 --; } if(m[b] == -1) { L.elem[k].coef = a; L.elem[k].exp = b; m[b] = k; k ++; } else { L.elem[m[b]].coef += a; } } sort(L.elem, L.elem+k, cmp); int flag = 0; int n = 1; printf("%s", s); for(int i = 0;i < k;i ++) { if(fabs(L.elem[i].coef) < 1e-6) continue; else { flag = 1; if(n != 1 && L.elem[i].coef > 0) printf("+"); if(L.elem[i].exp == 0) printf("%g", L.elem[i].coef); else if(L.elem[i].coef != 1 && L.elem[i].exp != 0 && L.elem[i].coef != -1) { if(L.elem[i].exp == 1) printf("%gx",L.elem[i].coef); else printf("%gx^%d",L.elem[i].coef, L.elem[i].exp); } else if(L.elem[i].coef == 1) { if(L.elem[i].exp == 1) printf("x"); else printf("x^%d",L.elem[i].exp); } else if(L.elem[i].coef == -1) { if(L.elem[i].exp == 1) printf("-x"); else printf("-x^%d",L.elem[i].exp); } n ++; } } if(!flag) printf("0"); printf("\n"); } return 0; }
連結串列實現：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <string> #include <cmath> #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); //提取多項式中的係數和指數 printf("Case %d :\n",casen ++); Deriv(L1); //求第一項的導數 AddPolyn(L1, L2); //多項式相加 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; //係數
int exp; //指數
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) //比當前指數大
{
q = p;
p = p->next;
}
else if(flag == 1) //比當前指數小
{
s = (LNode* )new LNode;
s->coef = coef;
s->exp = exp;
s->next = p;
q->next = s;
return;
}
else if(flag == 0) //與當前指數相等
{
p->coef += coef;
return;
}
}
LNode*s = (LNode*) new LNode;
s->coef = coef;
s->exp = exp;
s->next = NULL;
q->next = s;
return;
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~讀取多項式
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);
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~讀第一個多項式
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);
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~讀第二個多項式
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);
}
}
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~兩個多項式相加
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 <<" "<<exp<<endl;
if(fabs(coef) < 1e-6)
{
p = p->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
測試資料：
(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)=
輸出結果（輸出第一行為第一項的導數值）：
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