/*程序使用范例：
cal n1 op n2
cal 123123123 + 8762548
cal 123123123 - 8762548
cal 123123123 * 8762548
cal 123123123 / 8762548
cal 123123123 /200 8762548  ##精确到200位
*/

#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';
}

//比较两个串表示整数的大小
int scmp(char *r, char *u)
{ unsigned int lr, lu;
char hc, *p;
cdel0(r);
cdel0(u);
lr=strlen(r);
lu=strlen(u);
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 lr, lu, lp;  
  char hc, *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';
}
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;
}
lp=i-1;
for(i=lp+1; i<lr+1; ++i) p[i]='\0';
for(i=0; i<(lp+1)/2; ++i)
{ hc=p[i]; p[i]=p[lp-1]; 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
int *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;  
  unsigned int i, lu, lv, lr, iu, iw, c, h;  
  int mh, kh, j;  
  lu=strlen(u);  
  lv=strlen(v);  

  f=(char *)malloc((unsigned int)(lu+n+3)*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); if(scmp(r, "0")==0) break;  
      lr=strlen(r; r[lr]='0'; r[lr+1]='\0';  
    }
    else { f[lu]='.'; continue; }
    kh=0;  
    while(scmp(r, v)>=0)
    { r=ssub(p=r, v); free(p); ++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);  
  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 *p, *f, *r;  
  unsigned int i, lu, lv, lr, iu, iw, c, h;  
  int mh, kh, j;  
  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(p=r, v); free(p); ++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) { printf("\n>>order n1 op n2"); 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(strcmp0(argv[2], "+")==0)
  { printf("%s", p=sadd(argv[1], argv[3])); free(p); }

  else if(strcmp0(argv[2], "-")==0)
  { printf("%s", p=sadd(argv[1], argv[3])); free(p); }

  else if(strcmp0(argv[2], "*")==0)
  { printf("%s", p=smut(argv[1], argv[3])); free(p); }

  else if(argv[2][0]=='/' && strlen(argv[2])==1)
  { 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)
  { 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);  
  }
}