HDU1402 A * B Problem Plus 大数乘法 FFT(快速傅里叶变换)优化

题目

长度不超过5000，据称高精度会TLE，必须 O(nlogn) ，FFT首敲。

代码

bit_reverse_swap(a,n)参考自算法导论30.3的迭代实现，非递归方式完成下图过程。

#include <cstdio>
#include <cmath>
#include <complex>
#include <cstring>

using namespace std;
const double PI(acos(-1.0));
typedef complex<double> C;

const int N = (1<<17);
int ans[N];
C a[N],b[N];
char s[N],t[N];

void bit_reverse_swap(C *a,int n)
{
    for(int i = 1,j = n>>1,k; i < n-1; ++i)
    {
        if(i < j) swap(a[i],a[j]);
        //tricky
        for(k = n>>1; j >= k; j -= k,k >>= 1) //inspect the highest "1"
            ;
        j+=k;
    }
}

void FFT(C* a,int n,int t)
{
    bit_reverse_swap(a,n);
    for(int i = 2; i <= n; i <<= 1)
    {
        C wi(cos(2.0*t*PI/i),sin(2.0*t*PI/i));
        for (int j = 0; j < n; j += i)
        {
            C w(1);
            for (int k = j,h = i >> 1; k < j + h; ++k)
            {
                C t = w*a[k+h],u = a[k];
                a[k] = u + t;
                a[k+h] = u - t;
                w *= wi;
            }
        }
    }
    if (t == -1)
    {
        for (int i = 0; i < n; ++i)
        {
            a[i] /= n;
        }
    }
}

int trans(int x)
{
    return 1 << int(ceil(log(x)/log(2) - 1e-9));    //math.h/log() 以e为底
}

int main()
{
    int n,m,l;
    for(; ~scanf("%s%s",s,t); )
    {
        n = strlen(s);
        m = strlen(t);
        l = trans(n + m - 1); //n次*m次不超过n+m-1次
        for (int i = 0; i < n; ++i)
            a[i] = C(s[n-1-i]-'0');
        for (int i = n; i < l; ++i)
            a[i] = C(0);
        for (int i = 0; i < m; ++i)
            b[i] = C(t[m-1-i]-'0');
        for (int i = m; i < l; ++i)
            b[i] = C(0);

        FFT(a,l,1);//把A和B换成点值表达 
        FFT(b,1);
        for (int i = 0; i < l; ++i)//点值做乘法
            a[i] *= b[i];
        FFT(a,-1);//逆DFT
        for (int i = 0; i < l; ++i)
            ans[i] = (int)(a[i].real() + 0.5);
        ans[l] = 0;                             // error-prone :'l' -> '1'
        for (int i = 0; i < l; ++i)
        {
            ans[i+1] += ans[i] / 10;
            ans[i] %= 10;
        }
        int p = l;
        for(; p && !ans[p]; --p);
        for(; ~p; putchar(ans[p--] + '0'));
        puts(""); 
    }
    return 0;
}

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