（趣学算法）分治算法：大整数乘法

 

 

#include
#include
#include
using namespace std;
#define M 100
char sa[1000];
char sb[1000];
typedef struct _Node{
	int s[M];
	int l;//代表字符串长度 
	int c;//幂 
} Node, *pNode;

void cp(pNode src, pNode des, int st, int l)
{
	int i, j;
	for(i = st, j = 0; i <= st+l; i++, j++)
	{
		des->s[j] = src->s[i];
	}
	des->l = l;
	des->c = st + src->c;//次幂 
}

void add(pNode pa, pNode pb, pNode ans)
{
	int i, cc, k, palen, pblen, len;
	int ta, tb;
	pNode temp;
	if((pa->c < pb->c)) //保证pa的次幂大
	{
		temp = pa;
		pa = pb;
		pb = temp;
	} 
	ans->c = pb->c;
	cc = 0;
	palen = pa->l + pa->c;
	pblen = pb->l + pb->c;
	if(palen > pblen)
		len = palen;
	else
		len = pblen;
	k = pa->c - pb->c;//k为左侧需要补零的个数 
	for(i = 0; i c; i++)//结果的长度为pa pb的最大长度-最低次幂 
	{
		if(i < k) 
			ta = 0;
		else
			ta = pa->s[i-k];
		if(i < pb->l)
			tb = pb->s[i];
		else
			tb = 0; 
		if(i >= pa->l + k)
			ta = 0;
		ans->s[i] = (ta+tb+cc)%10;
		cc = (ta + tb +cc)/10;
	}
	if(cc)
		ans->s[i++] = cc;
	ans->l = i;
} 

void mul(pNode pa, pNode pb, pNode ans)
{
	int i, cc, w;
	int ma = pa->l >> 1;
	int mb = pb->l >> 1;//长度除以2
	Node ah, al, bh, bl;
	Node t1, t2, t3, t4, z;
	pNode temp;
	if(!ma || !mb)//边界条件，如果其中一个数为1 
	{
		if(!ma)//如果a串的长度为1，pa,pb交换，pa的长度大于等于pb的长度
		{
			temp = pa;
			pa = pb;
			pb = temp;
		} //交换后pb的长度为1 
		ans->c = pa->c + pb->c;
		w = pb->s[0];//用w记录pb的长度，pb的长度为1
		cc = 0;//初始化进位为0
		for(i = 0; i < pa->l; ++i)
		{
			ans->s[i] = (w*pa->s[i] + cc) %10;
			cc = (w*pa->s[i] + cc)/10;
		} 
		if(cc)
		{
			ans->s[i++] = cc;
		}
		ans -> l = i;//记录结果的长度
		return ; 
	} 
	
	//分治的核心
	cp(pa, &ah, ma, pa->l - ma);
	cp(pa, &al, 0, ma);
	cp(pb, &bh, mb, pb->l - mb);
	cp(pb, &bl, 0, mb);
	
	mul(&ah, &bh, &t1);
	mul(&ah, &bl, &t2);
	mul(&al, &bh, &t3);
	mul(&al, &bl, &t4);
	
	add(&t3, &t4, ans);
	add(&t2, ans, &z);
	add(&t1, &z, ans); 
} 


int main()
{
	Node ans,a,b;
	cout << "输入大整数a: " << endl;
	cin >> sa;
	cout << "输入大整数b: " << endl;
	cin >> sb;
	a.l = strlen(sa);//sa,sb以字符串进行处理 
	b.l = strlen(sb);
	int z = 0, i;
	for(i = a.l - 1; i >= 0; --i)//将sa,sb字符串转为数字 
	{
		a.s[z++] = sa[i] - '0';
	} 
	a.c = 0;
	z = 0;
	for(i = b.l - 1; i >= 0; --i)
	{
		b.s[z++] = sb[i] - '0';
	}
	b.c = 0;
	mul(&a,&b,&ans);//乘法运算，将两个数进行相乘，不断的分解
	cout << "最终结果为： ";
	for(i = ans.l - 1; i >= 0; --i)//将最终结果逆序输出 
		cout << ans.s[i];
	cout << endl;
	return 0;
}