大数加减法,主要处理进位就行。
大数乘法,按照普通的乘法计算方式,写程序便可。
大数除法,实际上就是转换为不断做减法,看从被除数里面最多能减去多少个除数,商就是多少。逐个减显然太慢,要判断一次最多能减少多少个(除数*10的n次方)。
以7546除23为例。
先减去23的100倍,就是2300,可以减3次,余下646。 此时商就是300;
然后646减去23的10倍,就是230,可以减2次,余下186。此时商就是320;
然后186减去23,可以减8次,此时商就是328.
源码如下
#include
#define MAX (33)
#define rint register int
char t1[MAX];
char t2[MAX];
int a[MAX];
int b[MAX];
int sum[MAX];
int cnt = 0;
bool mark = false;
char btn;
void clearChar(char *str)
{
if(str == NULL)
return;
while(*str != '\0')
*str++ = 0;
}
void clearInt(int *s)
{
for(rint i = 0; i < MAX; i++)
s[i] = 0;
}
void copyChar(char* from, char* to)
{
clearChar(to);
if(from == NULL || to == NULL)
return;
while(*from != '\0')
*to++ = *from++;
}
int lenChar(char* str)
{
int ret = 0;
if(str == NULL)
return ret;
while (*str++ != '\0')
ret++;
return ret;
}
int str2int(char* from, int *to)
{
if(from == NULL || to == NULL)
return 0;
clearInt(to);
int len = lenChar(from);
for(rint i = 0; i < len; i++)
{
to[len - 1 - i] = from[i] - '0';
}
return len;
}
void int2str(int *from, char* to, int len)
{
if(from == NULL || to == NULL)
return;
clearChar(to);
for(int i = 0; i < len; i++)
{
to[i] = from[len - 1 - i] + '0';
}
}
/*加法计算*/
void plusChar(char *s1, char *s2, char *s)
{
int len1 = str2int(s1, a);
int len2 = str2int(s2, b);
int len = len1;
if(len1 < len2)
{
len = len2;
}
for(rint i = 0; i < len1 || i < len2; i++)
{
sum[i] = a[i] + b[i];
}
for(int i = 0; i < len; i++)
{
if(sum[i] >= 10 )
{
sum[i + 1]++;
sum[i] %= 10;
if(i == len - 1)
{
len++;
break;
}
}
}
int2str(sum, s, len);
}
/*减法计算*/
void minusChar(char* s1, char* s2, char* s)
{
int len1 = str2int(s1, a);
int len2 = str2int(s2, b);
for(rint i = 0; i < len1; i++)
{
sum[i] = a[i] - b[i];
}
for(rint i = 0; i < len1 || i < len2; i++)
{
if(sum[i] < 0 )
{
sum[i + 1]--;
sum[i] += 10;
}
else if(sum[i] == 0 && i == len1 - 1)
{
len1--;
break;
}
}
for(int i = len1 - 1; i >= 0; i--)
{
if(sum[i] == 0)
len1 --;
else
break;
}
if(len1 == 0)
len1 = 1;
int2str(sum, s, len1);
}
/*减法计算*/
int minusChar(int* s1, int len1, int* s2, int len2)
{
if(len1 < len2)
{
return -1;
}
else if(len1 == len2)
{
for(int i = len1 - 1; i >= 0; i--)
{
if(s1[i] > s2[i])
break;
else if(s1[i] < s2[i])
return -1;
else
continue;
}
}
for(int i = 0; i < len1; i++)
{
s1[i] -= s2[i];
if(s1[i] < 0)
{
s1[i]+=10;
s1[i+1]--;
}
}
for(int i = len1 - 1; i >=0; i--)
{
if(s1[i])
return (i+1);
}
return 0;
}
/*除法计算,实际上就是转化为减法*/
void divideChar(char* s1, char* s2, char* s)
{
int len1 = str2int(s1, a);
int len2 = str2int(s2, b);
int times = len1 - len2;
int c[MAX] = {0};
int d[MAX] = {0};
int len = 1;
bool flag = false;
for(rint i = times; i >= 0; i--)
{
clearInt(d);
for(rint j = len2 - 1; j >= 0; j--)
d[j + i] = b[j];
for(rint j = 0; j < i; j++)
d[j] = 0;
int temp = 0;
while((temp = minusChar(a, len1, d, len2+i)) >= 0)
{
c[i]++;
len1 = temp;
}
if(c[i] != 0 && flag == false)
{
len = i+1;
flag = true;
}
}
int2str(c, s, len);
}
/*乘法计算*/
void multiply(char* s1, char* s2, char* s)
{
int len1 = str2int(s1, a);
int len2 = str2int(s2, b);
int c[MAX*2] = {0};
for(rint i = 0; i < len1; i++)
{
for(rint j = 0; j < len2; j++)
{
c[i+j] += a[i] * b[j];
}
}
for(rint i = 0; i < len1+len2; i++)
{
c[i+1] += c[i]/10;
c[i] = c[i]%10;
}
int len = 1;
for(rint i = len1+len2; i>=0; i--)
{
if(c[i])
{
len = i+1;
break;
}
}
int2str(c, s, len);
}
/*处理函数*/
void handle(char button,char* t1, char* t2, char* display)
{
if(lenChar(t1) == 0)
{
copyChar(t2, display);
btn = button;
return;
}
if(btn == '+')
{
plusChar(t1, t2, display);
}
else if(btn == '-')
{
minusChar(t1, t2, display);
}
else if(btn == '/')
{
divideChar(t1, t2, display);
}
else if (btn == '*')
{
multiply(t1, t2, display);
}
btn = button;
clearChar(t1);
copyChar(display, t2);
}
void Run(char button, char display[32 + 1])
{
if(button >= '0' && button <= '9')
{
if(cnt == 0)
{
copyChar(t2, t1);
clearChar(t2);
clearChar(display);
}
t2[cnt] = button;
display[cnt] = button;
cnt++;
return;
}
mark = false;
cnt = 0;
if(button == 'C')
{
clearChar(display);
clearChar(t1);
clearChar(t2);
clearInt(a);
clearInt(b);
clearInt(sum);
display[0] = '0';
return;
}
else
{
handle(button, t1, t2, display);
}
}