【HDU1402】A * B Problem Plus（FFT）

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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

学习及模板

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;}

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