【HDU1402】A * B Problem Plus（FFT）_a * b problem plus杭电-优快云博客

本文链接：https://blog.youkuaiyun.com/oranges_c/article/details/54783465

本文详细介绍了快速傅立叶变换（FFT）算法及其在多项式乘法中的应用。通过具体的代码实现展示了如何利用FFT进行高效的高精度计算，并提供了一个完整的Java实现示例。

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记录一个菜逼的成长。。

题目链接
用java,普通的高精度也能过。
但这里记下用FFT的方法

学习资料里解释的很好。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const double PI = acos(-1);
const int INF = 0x3f3f3f3f;
struct complex
{
    double r,i;
    complex(double _r = 0.0,double _i = 0.0)
    {
        r = _r; i = _i;
    }
    complex operator +(const complex &b)
    {
        return complex(r+b.r,i+b.i);
    }
    complex operator -(const complex &b)
    {
        return complex(r-b.r,i-b.i);
    }
    complex operator *(const complex &b)
    {
        return complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
};
///雷德算法--倒位序
void Rader(complex F[],int len)
{
    for( int i = 1,j = len >> 1; i < len-1; i++ ){
        if(i < j)swap(F[i],F[j]);
        int k = len >> 1;
        while(j >= k){
            j -= k;
            k >>= 1;
        }
        if(j < k)j += k;
    }
}
/**
 * 做FFT
 * len必须为2^k形式，
 * on==1时是DFT，on==-1时是IDFT
**/
void fft(complex F[],int len,int dft)
{
    Rader(F,len);
    for( int h = 2; h <= len; h <<= 1 ){
        complex wn(cos(-dft*2*PI/h),sin(-dft*2*PI/h));
        for( int j = 0; j < len; j += h ){
            complex w(1,0);
            for( int k = j; k < j + (h >> 1); k++ ){
                complex u = F[k];
                complex t = w*F[k+(h>>1)];
                F[k] = u + t;
                F[k+(h>>1)] = u - t;
                w = w * wn;
            }
        }
    }
    if(dft == -1)
        for( int i = 0; i < len; i++ )
            F[i].r /= len;
}
void Conv(complex a[],complex b[],int len)
{
    fft(a,len,1);
    fft(b,len,1);
    for( int i = 0; i < len; i++ )
        a[i] = a[i] * b[i];
    fft(a,len,-1);
}
const int maxn = 200000 + 10;
char str1[maxn],str2[maxn];
int sum[maxn];
complex a[maxn],b[maxn];
int main()
{
    while(~scanf("%s%s",str1,str2)){
        int len1 = strlen(str1);
        int len2 = strlen(str2);
        int len = 1;
        while(len < len1*2 || len < len2*2)len <<= 1;
        for( int i = 0; i < len1; i++ ){
            a[i] = complex(str1[len1-i-1]-'0',0);
        }
        for( int i = len1; i < len; i++ ){
            a[i] = complex(0,0);
        }
        for( int i = 0; i < len2; i++ ){
            b[i] = complex(str2[len2-i-1]-'0',0);
        }
        for( int i = len2; i < len; i++ ){
            b[i] = complex(0,0);
        }

        Conv(a,b,len);

        ///处理精度
        for( int i = 0; i < len; i++ )
            sum[i] = (int)(a[i].r + 0.5);
        ///处理进位
        for( int i = 0; i < len; i++ ){
            sum[i+1] += sum[i] / 10;
            sum[i] %= 10;
        }
        len = len1 + len2;
        while(len > 0 && sum[len] <= 0)len--;
        for( ;len >= 0; len-- )
            printf("%d",sum[len]);
        puts("");
    }
    return 0;
}