如下为网上搜索到的快速乘法的c语言程序,请老师看看能不能运行?能不能翻译成vb程序?
迭代型:(共119行,乱了,再整理一下)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <conio.h>
#define N 150010const double pi = 3.141592653;
char s1[N>>1], s2[N>>1];
double rea[N], ina[N], reb[N], inb[N];
int ans[N>>1];
void Swap(double *x, double *y)
{
double t = *x;
*x = *y;
*y = t;
}
int Rev(int x, int len)
{
int ans = 0;
int i;
for(i = 0;
i < len; i++)
{
ans <<= 1;
ans |= (x & 1);
x >>= 1;
}
return ans;
}
void FFT(double *reA, double *inA, int n, bool flag)
{
int s;
double lgn = log((double)n) / log((double)2);
int i;
for(i = 0; i < n; i++)
{
int j = Rev(i, lgn);
if(j > i)
{
Swap(&reA[i], &reA[j]);
Swap(&inA[i], &inA[j]);
}
}
for(s = 1; s <= lgn; s++)
{
int m = (1<<s);
double reWm = cos(2*pi/m), inWm = sin(2*pi/m);
if(flag) inWm = -inWm;
int k;
for(k = 0; k < n; k += m)
{
double reW = 1.0, inW = 0.0;
int j;
for(j = 0; j < m / 2; j++)
{
int tag = k+j+m/2;
double reT = reW * reA[tag] - inW * inA[tag];
double inT = reW * inA[tag] + inW * reA[tag];
double reU = reA[k+j], inU = inA[k+j];
reA[k+j] = reU + reT;
inA[k+j] = inU + inT;
reA[tag] = reU - reT;
inA[tag] = inU - inT;
double rew_t = reW * reWm - inW * inWm;
double inw_t = reW * inWm + inW * reWm;
reW = rew_t;
inW = inw_t;
}
}
}
if(flag)
{
for(i = 0;
i < n; i++)
{
reA[i] /= n;
inA[i] /= n;
}
}
}
int main()
{
#if 0
freopen("in.txt","r",stdin);
#endif
while(~scanf("%s%s", s1, s2))
{
memset(ans, 0 , sizeof(ans));
memset(rea, 0 , sizeof(rea));
memset(ina, 0 , sizeof(ina));
memset(reb, 0 , sizeof(reb));
memset(inb, 0 , sizeof(inb));
int i, lent, len = 1, len1, len2;
len1 = strlen(s1);
len2 = strlen(s2);
lent = (len1 > len2 ? len1 : len2);
while(len < lent) len <<= 1;
len <<= 1;
for(i = 0;
i < len; i++)
{
if(i < len1) rea[i] = (double)s1[len1-i-1] - '0';
if(i < len2) reb[i] = (double)s2[len2-i-1] - '0';
ina[i] = inb[i] = 0.0;
}
FFT(rea, ina, len, 0);
FFT(reb, inb, len, 0);
for(i = 0; i < len; i++)
{
double rec = rea[i] * reb[i] - ina[i] * inb[i];
double inc = rea[i] * inb[i] + ina[i] * reb[i];
rea[i] = rec; ina[i] = inc;
}
FFT(rea, ina, len, 1);
for(i = 0; i < len; i++)
ans[i] = (int)(rea[i] + 0.4);
for(i = 0; i < len; i++)
{
ans[i+1] += ans[i] / 10;
ans[i] %= 10;
}
int len_ans = len1 + len2 + 2;
while(ans[len_ans] == 0 && len_ans > 0)
len_ans--;
for(i = len_ans; i >= 0; i--)
printf("%d", ans[i]);
printf("\n");
}
return 0;
}
————————————————
迭代型比递归型稍快一点。
发帖时间 昨天 09:18