大整数的乘法

int gethead(char *str)

{
int n = strlen(str);

// string heada = "";

char * head = new char[n / 2];

for (int i = 0; i < n / 2; i++)
{
head[i] = str[i];
}

int result;

sscanf_s(head, "%d", &result);

return  result;

}

int gettail(char *str)

{
int n = strlen(str);

// string heada = "";

char * tail = new char[n-n/ 2];

for (int i = n/2; i < n ; i++)
{
tail[i-n/2] = str[i];
}
int result;

sscanf_s(tail, "%d", &result);

return  result;

}

int ridevalue(int a,int an,int b,int bn)

{

if (an == 1)
{
return a*b;
}

else if (bn == 1)
{
return a*b;
}

//对数字进行切分。
char   *stra=new char[an];
//sprintf(stra, "%d", a);
_itoa_s(a,stra,10,10);

    char *strb = new char[bn];

    //sprintf(strb, "%d", b);
_itoa_s(b,strb,10,10);

//对字符串进行分割。sscanf(str,"%d",&a);

    

int heada = gethead(stra);
int taila = gettail(stra);

int headb = gethead(strb); 

int tailb = gettail(strb);

cout << heada << "," << taila << endl;
cout << tailb << "," << headb << endl;

// 

int c2 = ridevalue(heada,an/2,headb,bn/2);

int c0= ridevalue(taila,(an-an/2),tailb,(bn-bn/2));

int c1 = (heada + taila)*(headb + tailb) - (c2 + c0);

cout << c2 << "," << c1 << "," << c0 << endl;

return c2*pow(10,bn) + c1*pow(10,bn/2) + c0;

}