大数相乘 快速算法

大数乘的最快算法是快速傅立叶变换法，这有一个，但不是我本人写的。
#include
#include
#include
#include
using namespace std;

const double PI = acos(-1);
typedef complex cp;
typedef long long int64;

const int N = 1 << 16;
int64 a[N], b[N], c[N << 1];

void bit_reverse_copy(cp a[], int n, cp b[])
{
int i, j, k, u, m;
for (u = 1, m = 0; u < n; u <<= 1, ++m);
for (i = 0; i < n; ++i)
{
j = i; k = 0;
for (u = 0; u < m; ++u, j >>= 1)
k = (k << 1) | (j & 1);
b[k] = a[i];
}
}

void FFT(cp _x[], int n, bool flag)
{
static cp x[N << 1];
bit_reverse_copy(_x, n, x);
int i, j, k, kk, p, m;
for (i = 1, m = 0; i < n; i <<= 1, ++m);
double alpha = 2 * PI;
if (flag) alpha = -alpha;
for (i = 0, k = 2; i < m; ++i, k <<= 1)
{
cp wn = cp(cos(alpha / k), sin(alpha / k));
for (j = 0; j < n; j += k)
{
cp w = 1, u, t;
kk = k >> 1;
for (p = 0; p < kk; ++p)
{
t = w * x[j + p + kk];
u = x[j + p];
x[j + p] = u + t;
x[j + p + kk] = u - t;
w *= wn;
}
}
}
memcpy(_x, x, sizeof(cp) * n);
}

void polynomial_multiply(int64 a[], int na, int64 b[], int nb, int64 c[], int &nc)
{
int i, n;
i = max(na, nb) << 1;
for (n = 1; n < i; n <<= 1);
static cp x[N << 1], y[N << 1];
for (i = 0; i < na; ++i) x[i] = a[i];
for (; i < n; ++i) x[i] = 0;
FFT(x, n, 0);
for (i = 0; i < nb; ++i) y[i] = b[i];
for (; i < n; ++i) y[i] = 0;
FFT(y, n, 0);
for (i = 0; i < n; ++i) x[i] *= y[i];
FFT(x, n, 1);
for (i = 0; i < n; ++i)
{
c[i] =(int64)(x[i].real() / n + 0.5);
}
for (nc = na + nb - 1; nc > 1 && !c[nc - 1]; --nc);
}

const int LEN = 5, MOD = 100000;
void convert(char *s, int64 a[], int &n)
{
int len = strlen(s), i, j, k;
for (n = 0, i = len - LEN; i >= 0; i -= LEN)
{
for (j = k = 0; j < LEN; ++j)
k = k * 10 + (s[i + j] - '0');
a[n++] = k;
}
i += LEN;
if (i)
{
for (j = k = 0; j < i; ++j)
k = k * 10 + (s[j] - '0');
a[n++] = k;
}
}

void print(int64 a[], int n)
{
printf("%I64d", a[--n]);
while (n) printf("%05I64d", a[--n]);
puts("");
}

char buf[N + 10];

int main()
{
int na, nb, nc;

while (scanf("%s", buf) != EOF)
{
bool sign = false;
if (buf[0] == '-')
{
sign = !sign;
convert(buf + 1, a, na);
}
else convert(buf, a, na);
scanf("%s", buf);
if (buf[0] == '-')
{
sign = !sign;
convert(buf + 1, b, nb);
}
else convert(buf, b, nb);
polynomial_multiply(a, na, b, nb, c, nc);
int64 t1, t2;
t1 = 0;
for (int i = 0; i < nc; ++i)
{
t2 = t1 + c[i];
c[i] = t2 % MOD;
t1 = t2 / MOD;
}
for (; t1; t1 /= MOD) c[nc++] = t1 % MOD;
if (sign) putchar('-');
print(c, nc);
}
return 0;
}

给你一个吧
速度还可以
自己读下代码
/**************************************
算法复杂度为：O(longhta*longthb）
longtha为乘数的位数
longhtb为被乘数的位数
***************************************/

#include
#include
#include
#define LEN 1000
void mult(char [],char [],char []);
main()
{
char op1[LEN],op2[LEN],op3[LEN*2-1];
scanf("%s%s",op1,op2);
mult(op1,op2,op3);
printf("%s\n",op3);
getch();
return 0;
}
void reverse(char a[])
{
int longth=strlen(a);
int i;
for(i=0;i char t;
t=a[i];
a[i]=a[longth-i-1];
a[longth-i-1]=t;
}
}
void mult(char op1[LEN],char op2[LEN],char ans[LEN*2-1])
{
char top1[LEN];
char top2[LEN];
strcpy(top1,op1);
strcpy(top2,op2);
reverse(top1);
reverse(top2);
int k;
int top1s=strlen(top1);
int top2s=strlen(top2);
for(k=0;k ans[k]='0';
}
int i,j;
int jw,ys;
int longth;
for(j=0;j jw=0;
for(i=0;i ys=((top1[i]-'0')*(top2[j]-'0')+jw+ans[i+j]-'0')%10;
jw=((top1[i]-'0')*(top2[j]-'0')+jw+ans[i+j]-'0')/10;
ans[i+j]=ys+'0';
}
if(jw>0){
ans[i+j]=jw+'0';
}
}
longth=i+j-1;
if(jw>0)
ans[longth++]=jw+'0';
ans[longth]='\0';
reverse(ans);
}

在长度不很大的情况下，FTT不能体现优势，其实只要减少乘法除法应该可以得到相当好的算法。不过大部情况下这些做法都比较复杂，其实简单的查表可能更好。

做一个0-9的乘法表结果m，每次计算时只要c[i +j] = m[ia[i]][ib[j]];就好。
进位也可以同理，0-99的进位表和余数表就可以解决除法和模的问题，不过这不是关键。

哪里的oj，怎么没听过