C语言 加法 减法 乘法 的高精度模版

#include<stdio.h>
#include<string.h> 
int a[10086];
int b[10086];
int c[10086];
char s1[10086];
char s2[10086];
//additon
void High_precision_addition(char *n,char *m)
{
	int l1=strlen(n);
    int l2=strlen(m);
         
    int len;
         
    if(l1>l2)
        len=l1;
    else
        len=l2;
             
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
         
    for(int i=l1-1;i>=0;i--)
        a[l1-i-1]=n[i]-'0';
    for(int i=l2-1;i>=0;i--)
        b[l2-i-1]=m[i]-'0';
             
    for(int i=0;i<len;i++)
    {
        a[i]+=b[i];
        a[i+1]+=a[i]/10;
        a[i]%=10;
    }
        
    if(a[len]!=0) len++;
    
    for(int i=len-1;i>=0;i--)
            printf("%d",a[i]);
        printf("\n");
}
//difference
void High_precision_subtraction(char *n,char *m)
{
	     int l1=strlen(n);
        int l2=strlen(m);
        int flag=0;
         
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));  
         
        if (l1<l2||(strcmp(n,m)<0&&l1==l2) )
        {
            flag = 1;
            for (int i=l2-1;i>=0;i--)
                a[l2-i-1]=m[i]-'0';
            for (int i=l1-1;i>=0;i--)
                b[l1-i-1]=n[i]-'0';
        }
        else
        {
            for (int i=l1-1;i>=0;i--)
                a[l1-i-1]=n[i] - '0';
            for (int i=l2-1;i>=0;i--)
                b[l2-i-1]=m[i]-'0';
        }
        int len;
        if (l1>l2) 
            len=l1;
        else
            len=l2;
 
 
        for (int i=0;i<len;i++)
        {
            a[i]=a[i]-b[i];
            if(a[i]<0)
            {
                a[i+1]-=1;
                a[i]+=10;
            }
        }
         
        while (a[len-1]==0&&len>1) len--;
         
        if (flag==1) printf("-");
        for (int i=len-1;i>=0;i--)
            printf("%d",a[i]);
        printf("\n");	
}
//multiply
void High_precision_multiplication(char *n,char *m)
{
    int l1=strlen(n),l2=strlen(m);
    int i,j;
    
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    
    for(i=0,j=l1-1;i<l1;i++,j--) 
        a[i]=n[j]-'0';
        
    for(i=0,j=l2-1;i<l2;i++,j--)
        b[i]=m[j]-'0';
    
    for (i=0;i<l1;i++) 
        for (j=0;j<l2;j++) 
            c[i+j]+=a[i]*b[j];  
	
    for (i=0;i<l1+l2;i++) 
	{
        if (c[i]>=10) 
		{
            c[i+1]+=c[i]/10;
            c[i]%=10;
        }
    }
    
    for (j=10085;j>0;j--) 
	{
        if (c[j]!=0)
        break;
    }
    
    for (i=j;i>=0;i--) 
	{
    	printf("%d",c[i]);
    }
    printf("\n");
}
//main
int main(void)
{
	char Arithmetic_symbol;
	while(1)
	{
		printf("Enter two numbers and the Arithmetic symbol which you wanna make it opreation:\n");
		printf("Tips:Arithmetic symbol you can enter:+,-,*\n"); 
		scanf("%s%s",s1,s2); getchar(); scanf("%c",&Arithmetic_symbol);
		if(Arithmetic_symbol=='+')	High_precision_addition(s1,s2);
		else if(Arithmetic_symbol=='-')	High_precision_subtraction(s1,s2);
		else if(Arithmetic_symbol=='*')  High_precision_multiplication(s1,s2);
		else printf("Please enter the correct value!\n");
	}
	return 0;
}