cal_big_soft.cpp

这是一个用于计算的算法

#include <MATH.h> 
#include <TU.h> 
#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);
					}

}