关于整数域一元多项式的加减乘运算

关于整数域一元多项式的加减乘运算

编译环境：VC++6.0

运行截图：




源代码：

#include<stdio.h>
#include<windows.h>
#define POLYSIZE 10
/*******************************/
/*结构体定义*/
typedef struct term
{
	int coe;//coefficient(系数)
	int ind;//index(指数)
}Term;//term(项)
typedef struct polynomial
{
	int length;//存放多项式的长度
	Term* base;//存放多项式存储空间的基地址
}Poly;//polynomial(多项式)
/*******************************/
/*生成一个多项式*/
Poly makePoly(int n)
{
	Poly temp;
	temp.base = (Term*)malloc( sizeof(Term) * n );//为多项式分配存储空间，并保存基地址
	temp.length = n;
	return temp;
}
/*******************************/
/*生成多项式的一项*/
Term makeTerm(int c ,int i)
{
	Term t;
	t.coe = c;
	t.ind = i;
	return t;//返回给定系数与指数所生成的项
}
/*******************************/
/*对多项式进行初始化*/
void initPoly(Poly* p)
{
	for( int i=0;i < p->length;i++ )
	{
		int tc,ti;//要确保输入的数据合理
		while(1)
		{
			printf("请输入第%d项的系数：",i+1);
			if( scanf("%d",&tc) == 1 )
				break;
			else
				printf("请输入正确的数字\n");
			while( getchar() != '\n' );
		}
		while(1)
		{
			printf("请输入第%d项的指数：",i+1);
			if( scanf("%d",&ti) == 1 )
				break;
			else
				printf("请输入正确的数字\n");
			while( getchar() != '\n' );
		}
		*(p->base+i) = makeTerm(tc,ti);
	}
}
/*******************************/
/*按项的指数降序排序*/
void sort(Poly* p)
{
	Term* b = p->base;
	Term t;
	for( int i = 0; i < p->length-1; i++ )//冒泡排序法
	{
		for( int j = 0; j < p->length-1-i; j++ )
		{
			if( (b+j)->ind < (b+j+1)->ind )
			{
				t = *(b+j);
				*(b+j) = *(b+j+1);
				*(b+j+1) = t;
			}
			if( (b+j)->ind == (b+j+1)->ind )
			{
				(b+j)->coe += (b+j+1)->coe;
				(b+j+1)->coe = 0;
			}
		}
	}
	for( i = 0; i < p->length; i++ )//清理系数为0的项
	{
		if( (b+i)->coe == 0 )
		{
			int j = i;
			while( j != p->length-1 )
			{
				t = *(b+j);
				*(b+j) = *(b+j+1);
				*(b+j+1) = t;
				j++;
			}
			p->length--;
		}
	}
}//排序函数(降序)
/*******************************/
/*显示某个多项式*/
void display(Poly* p)
{
	sort(p);
	HANDLE hConsole = GetStdHandle(STD_OUTPUT_HANDLE);//获取句柄，目的是修改输出颜色
	Term* t = p->base;
	for( int i = 0; i < p->length; i++ )
	{
		SetConsoleTextAttribute(hConsole,0xF);
		if( (t+i)->coe > 0 )
		{
			if( i != 0 )	
				printf(" + ");
			SetConsoleTextAttribute(hConsole,0xb);
			if( (t+i)->coe != 1 )    
				printf("%d",(t+i)->coe);
		}
		else if( (t+i)->coe < 0 )
		{   printf(" - ");
			SetConsoleTextAttribute(hConsole,0xb);
			if( (t+i)->coe != -1 )
				printf("%d",-(t+i)->coe);
		}
		if( (t+i)->ind == 0 )
			continue;
		SetConsoleTextAttribute(hConsole,0xF);
		printf("x");
		if( (t+i)->ind == 1 )
			continue;
		printf("^");
		SetConsoleTextAttribute(hConsole,0xC);
		printf("%d",(t+i)->ind);
		
	}
	SetConsoleTextAttribute(hConsole,0xF);
	printf("\n");
}
/*******************************/
/*多项式加法*/
void addPoly(int m, int n, Poly* p)
{
	int ml = (p+m)->length;
	int nl = (p+n)->length;
	p->base = (Term*)malloc( sizeof(Term) * (ml+nl) );
	Term* mb = (p+m)->base;
	Term* nb = (p+n)->base;
	Term* pb = p->base;
	int i = 0, j = 0, k = 0;
	while(1)
	{
		if( i == ml )
		{
			while( j != nl )
			{
				*(pb+k) = *(nb+j);
				j++;
				k++;
			}
			break;
		}
		else if(j == nl)
		{
			while( i != ml )
			{
				*(pb+k) = *(mb+i);
				i++;
				k++;
			}
			break;
		}
		if( (mb+i)->ind > (nb+j)->ind )
		{
			*(pb+k) = *(mb+i);
			i++;
		}
		else if( (mb+i)->ind == (nb+j)->ind )
		{
			*(pb+k) = *(mb+i);
			(pb+k)->coe += (nb+j)->coe;
			i++;
			j++;
		}
		else if( (mb+i)->ind < (nb+j)->ind )
		{
			*(pb+k) = *(nb+j);
			j++;
		}
		k++;
	}
	p->length = k;
}
/*******************************/
/*多项式减法*/
void subPoly(int m, int n, Poly* p)
{
	int ml = (p+m)->length;
	int nl = (p+n)->length;
	p->base = (Term*)malloc( sizeof(Term) * (ml+nl) );
	Term* mb = (p+m)->base;
	Term* nb = (p+n)->base;
	Term* pb = p->base;
	int i = 0, j = 0, k = 0;
	while(1)
	{
		if( i == ml )
		{
			while( j != nl )
			{
				*(pb+k) = *(nb+j);
				(pb+k)->coe = -(pb+k)->coe;
				j++;
				k++;
			}
			break;
		}
		else if(j == nl)
		{
			while( i != ml )
			{
				*(pb+k) = *(mb+i);
				i++;
				k++;
			}
			break;
		}
		if( (mb+i)->ind > (nb+j)->ind )
		{
			*(pb+k) = *(mb+i);
			i++;
		}
		else if( (mb+i)->ind == (nb+j)->ind )
		{
			*(pb+k) = *(mb+i);
			(pb+k)->coe -= (nb+j)->coe;
			i++;
			j++;
		}
		else if( (mb+i)->ind < (nb+j)->ind )
		{
			*(pb+k) = *(nb+j);
			(pb+k)->coe = -(pb+k)->coe;
			j++;
		}
		k++;
	}
	p->length = k;
}
/*******************************/
/*多项式乘法*/
void mulPoly(int m, int n, Poly* p)
{
	int ml = (p+m)->length;
	int nl = (p+n)->length;
	p->base = (Term*)malloc( sizeof(Term) * (ml*nl) );
	Term* mb = (p+m)->base;
	Term* nb = (p+n)->base;
	Term* pb = p->base;
	int k = 0;
	for( int i = 0; i < ml; i++)
	{
		for( int j = 0; j < nl; j++)
		{
			*(pb+k) = *(mb+i);
			(pb+k)->coe *= (nb+j)->coe;
			(pb+k)->ind += (nb+j)->ind;
			k++;
		}
	}
	p->length = k;
}
/*******************************/
/*编号输入时的检查*/
int inCheck()
{
	int x;
	while(1)
	{
		printf("请输入多项式编号：");
		scanf("%d",&x);
		if( x >= 1 && x <= POLYSIZE )
			break;
		else
			printf("输入有误，请重新输入（编号应位于1-%d之间）\n",POLYSIZE);
		while( getchar() != '\n' );
	}
	return x;
}
/*******************************/
int main ()
{
	Poly* p;
	//分配一定空间存放Poly变量，其中第0个用于存放结果
	//其余用于存放输入的多项式
	p = (Poly*)malloc( sizeof(Poly) * (POLYSIZE+1) );
	while(1)
	{
		int m = 0, n = 0, c = 0;
		system("cls"); 
		printf("\n\n请按所选功能输入数字：\n");
		printf("\t1.添加新的多项式\n");
		printf("\t2.查看现有的多项式\n");
		printf("\t3.多项式相加并输出\n");
		printf("\t4.多项式相减并输出\n");
		printf("\t5.多项式相乘并输出\n");
		printf("\t0.退出程序\n");
		scanf("%d",&c);
		switch(c)
		{
		    case 0:
				exit(0);
			case 1:
				{
					int x = 0,n = 0;
					x = inCheck();
					while(1)
					{
						printf("请输入多项式的项数：");
						scanf("%d",&n);
						if( n >= 2 )
							break;
						else
							printf("输入有误，请重新输入（项数应大于等于2）\n");
						while( getchar() != '\n' );
					}
					*(p+x) = makePoly(n);
					initPoly(p+x);
					printf("添加成功!编号为%d的多项式：\n\n\ty%d = ",x,x);
					display(p+x);
				}
				break;
			case 2:
				{
					int x;
					x = inCheck();
					printf("编号为%d的多项式：\n\n\ty%d = ",x,x);
					display(p+x);
				}
				break;
			case 3:
				{
					m = inCheck();
					n = inCheck();
					addPoly(m,n,p);//add(加法)
					printf("运算的结果是：\n\tresult = ");
					display(p);
					free(p->base);//将存放结果的空间释放
				}
				break;
			case 4:
				{
					m = inCheck();
					n = inCheck();
					subPoly(m,n,p);//subtract(减法)
					printf("运算的结果是：\n\tresult = ");
					display(p);
					free(p->base);
				}
				break;
			case 5:
				{
					m = inCheck();
					n = inCheck();
					mulPoly(m,n,p);//multiply(乘法)
					printf("运算的结果是：\n\tresult = ");
					display(p);
					free(p->base);
				}
				break;
		}
		system("pause");
	}
	return 0;
}