m1.c

用c语言实现的

//高精度整数除法是高精度加减乘的综合！
#include <stdio.h> 
#include <stdlib.h> 
#include <string.h> 
#include <ctype.h> 

int cchkdig(char *r) 
{ 
	int i=0; 
	while(r[i]!='\0') 
	{ 
		if(isdigit(r[i++])==0) 
			return (0); 
	} 
	return (1); 
} 
//去掉整数串表示前面多余的零，最后结果为空串时置为"0" 
void cdel0(char *r) 
{
	unsigned int lr; 
	int i=0, j; 
	lr=strlen(r); 
	while(r[i]=='0') 
		++i;
	if(i>0) 
	{
		for(j=0; j<(lr-i); ++j)
			r[j]=r[j+i];
		for(j=lr-i; j<lr; ++j) 
		{
			r[j]='\0'; 
		}
	}

	if(r[0]=='\0')
	{
		r[0]='0';
	} 
} 

int scmp(char *r, char *u) 
{ 
	unsigned int lr, lu; 
	cdel0(r);
	cdel0(u); 

	lr=strlen(r); 
	lu=strlen(u); 

	if(lr>lu) 
	{
		return 1;
	}
	else if (lr<lu) 
	{
		return -1;
	}
	return (strcmp(r, u)); 

}//end scmp() 

//两个串表示数的减法 
char *ssub(char *r, char *u) 
{
	unsigned int i,lr, lu, lp,c=0; 
	char h,hc; 
	char *p;
	if(scmp(r, u)<0) 
		return NULL; 
	lr=strlen(r); 
	lu=strlen(u); 
	p=(char *)malloc((unsigned int)(lr+1)*sizeof(char)); 
	for(i=0; i<lu; ++i) 
	{
		h=r[lr-i-1]-u[lu-i-1]-c; 
		if(h<0) 
		{
			c=1; 
			h=h+10;
		} 
		else 
			c=0; 
		p[i]=h+'0';
		hc=h+'0';
	} 
	for (i=lu; i<lr; ++i) 
	{ 
		h=r[lr-i-1]-'0'-c;
		if(h<0) 
		{
			c=1; 
			h=h+10; 
		} 
		else 
			c=0; 
		p[i]='0'+h; 
		hc='0'+h;
	}
	p[i]='\0';
	lp=i-1;

	while(p[lp]=='0'&&lp!=0)
	{
		p[lp]='\0';
		lp--; 
	}

	for(i=0; i<(lp+1)/2; ++i) 
	{ 
		hc=p[i]; 
		p[i]=p[lp-i]; 
		p[lp-i]=hc; 
	} 
	return (p); 
}//end ssub() 

//两个串表示数的加法 
char *sadd(char *r, char *u) 
{ 
	unsigned int lr, lu, lp; 
	int i, h, c=0; 
	char hc, *p; 
	lr=strlen(r); 
	lu=strlen(u); 
	if(lu>lr) 
	{ 
		p=r; 
        r=u; 
		u=p; 
		h=lr; 
		lr=lu; 
		lu=h; 
	} 
	p=(char *)malloc((unsigned int)(lr+2)*sizeof(char)); 
	for(i=0; i<lu; ++i) 
	{ 
		h=r[lr-i-1]-'0'+u[lu-i-1]-'0'+c; 
		if(h>9) 
		{ 
			c=1;
			h=h-10; 
		} 
		else 
			c=0; 
		p[i]=h+'0'; 
	} 
	for(i=lu; i<lr; ++i) 
	{ 
		h=r[lr-i-1]-'0'+c; 
		if(h>9) 
		{ 
			c=1; 
			h=h-10; 
		} 
		else 
			c=0; 
		p[i]='0'+h; 
	} 
	if(c>0) 
	{ 
		p[i]=c+'0'; 
		lp=i; 
	} 
	else 
		lp=i-1; 
	for(i=lp+1; i<lr+2; ++i) 
		p[i]='\0'; 
	for(i=0; i<(lp+1)/2; ++i) 
	{ 
		hc=p[i]; 
		p[i]=p[lp-i]; 
		p[lp-i]=hc; 
	} 
	return (p); 
}//end sadd() 

//两个串表示数的乘法 
char *smut(char *r, char *u) 
{ 
	unsigned int lr, lu, lp; 
	int i, j, c, h; 
	char *p; 
	lr=strlen(r); 
	lu=strlen(u); 
	p=(char *)malloc((unsigned int)(lr+lu+1)*sizeof(char)); 
	for(i=0; i<lr+lu; ++i) 
		p[i]='0'; 
	p[lr+lu]='\0'; 
	for(i=lr-1; i>=0; --i) 
	{ 
		c=0; 
		for(j=lu-1; j>=0; --j) 
		{ 
			lp=i+j+1; 
			h=(r[i]-'0')*(u[j]-'0')+p[lp]-'0'+c; 
			c=h/10; 
			h=h%10; 
			p[lp]=h+'0'; 
		} 
		if(c>0)p[i+j+1]=c+'0'; 
	} 
	cdel0(p); 
	return p; 
}//end smut() 

//两个串表示数的除法，结果精确到小数点后第n位 
char *sdivf(char *u, char *v, int n) 
{ 
	char *p, *f, *r,*q; 
	unsigned int i, lu, lv, lr, iw, c, h; 
	int kh, j; 
	lu=strlen(u); 
	lv=strlen(v); 
	f=(char *)malloc((unsigned int)(lu+n+3)*sizeof(char)); 
	q=(char *)malloc(sizeof(char)); 
	for(i=0; i<lu+n+3; ++i) 
		f[i]='\0'; 
	r=(char *)malloc((unsigned int)(lv+2)*sizeof(char)); 
	for(i=0; i<lv+2; ++i) 
		r[i]='\0'; 
	for(iw=0; iw<lu+n+2; ++iw) 
	{ 
		if(iw<lu) 
		{ 
			cdel0(r); 
			lr=strlen(r); 
			r[lr]=u[iw]; 
			r[lr+1]='\0';
		} 

		else if(iw>lu) 
		{ 
			cdel0(r);
			q[0]='0';
			if(scmp(r, q)==0)
			{
				break; 
			}
			lr=strlen(r); 
			r[lr]='0';
			r[lr+1]='\0'; 
		} 
		else 
		{ 
			f[lu]='.'; 
			continue; 
		}
		kh=0; 
		while(scmp(r, v)>=0) 
		{ 
			p=r;
			r=ssub(p, v); 
			++kh; 
		} 
		f[iw]=kh+'0'; 
	} 
	if(iw==lu+n+2) 
	{ 
		if(f[lu+n+1]>='5') 
		{ 
			f[lu+n+1]='\0'; 
			c=1; 
			for(j=lu+n; j>=0; --j) 
			{ 
				if(c==0) 
				{
					break; 
				}
				if(f[j]=='.')
				{
					continue; 
				}
				h=f[j]-'0'+c; 
				if(h>9) 
				{ 
					h=h-10; 
					c=1; 
				} 
				else 
					c='\0'; 
				f[j]=h+'0'; 
			} 
		} 
		else 
			f[lu+n+1]='\0'; 

	} 
	free(r); 
	free(p);
	q=NULL;
	free(q);
	cdel0(f); 
	return(f); 
}//end sdivf() 

//两个串表示数的除法，结果分别用整商与余数表示 
char *sdivkr(char *u, char *v, char **rout) 
{ 
	char *f, *r; 
	unsigned int i, lu, lv, lr, iw; 
	int kh; 
	lu=strlen(u); 
	lv=strlen(v); 

	f=(char *)malloc((unsigned int)(lu+1)*sizeof(char)); 
	for(i=0; i<lu+1; ++i) f[i]='\0'; 
	r=(char *)malloc((unsigned int)(lv+2)*sizeof(char)); 
	for(i=0; i<lv+2; ++i) r[i]='\0'; 

	for(iw=0; iw<lu; ++iw) 
	{ 
		cdel0(r); 
		lr=strlen(r); 
		r[lr]=u[iw]; 
		r[lr+1]='\0'; 
		kh=0; 
		while(scmp(r, v)>=0) 
		{ 
			r=ssub(r, v); 
			++kh; 
		} 
		f[iw]=kh+'0'; 
	} 
	cdel0(r); 
	*rout=r; 
	cdel0(f); 
	return(f);
}//end *sdivkr() 

//调用上述函数实现两任意长正整数任意指定精度的算术计算器程序 
void main(int argc, char *argv[]) 
{ 
	char *p, *r; 
	int n; 
	if(argc!=4)
	{ 
		if(argc!=3)
			printf("\n>>\"order n1 op n2\" or n ! "); 
		exit(0); 
	} 
	cdel0(argv[1]); 
	if(cchkdig(argv[1])==0)
	{ 
		printf("Input data error, Input again!"); 
		exit(0); 
	} 
	cdel0(argv[3]); 
	if(cchkdig(argv[3])==0) 
	{ 
		printf("Input data error, Input again!"); 
		exit(0); 
	} 

	if(strcmp(argv[2], "+")==0) 
	{ 
		printf("%s", p=sadd(argv[1], argv[3])); 
		free(p); 
	} 
	else if(strcmp(argv[2], "-")==0) 
	{ 
		printf("%s", p=ssub(argv[1], argv[3])); 
		free(p); 
	} 
	else if(strcmp(argv[2], "*")==0) 
	{ 
		printf("%s", p=smut(argv[1], argv[3])); 
		free(p); 
	} 
	else if(argv[2][0]=='/' && strlen(argv[2])==1) 
	{ 
		if(argv[3][0]=='0')
		{
			printf("error!devided by zero!!\n");
			exit(0);
		}
		p=sdivkr(argv[1], argv[3], &r); 
		printf("k=%s r=%s", p, r); 
		free(p); 
		free(r); 
	} 

	else if(argv[2][0]=='/'&&strlen(argv[2])>1) 
	{ 
		if(argv[3][0]=='0')
		{
			printf("error!devided by zero!!\n");
			exit(0);
		}

		argv[2][0]='\0'; 
		cdel0(argv[2]); 
		if(cchkdig(argv[2])==0) 
		{ 
			printf("Input data error, Input again!"); 
			exit (0); 
		} 
		n=atoi(argv[2]); 
		printf("%s", p=sdivf(argv[1], argv[3], n));
		free(p); 
	} 
}