【51nod1028】【大数乘法 V2】【fft】

题目大意

给出2个大整数A,B，计算A*B的结果。

解题思路

fft，然而会卡精度，使用模运算的fft即可解决问题。

code

#include
#include
#include
#define LD double
#define LL long long
#define min(a,b) ((a
#define max(a,b) ((a>b)?a:b)
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
using namespace std;
int const maxn=1e5,g=3,mo=1004535809;
int n,m,A,B,w[maxn*4+10],a[maxn*4+10],b[maxn*4+10],t[maxn*4+10];
void read(int &num,int *a){
    num=0;int v=0;char ch=getchar();
    for(;(ch<'0')||(ch>'9');ch=getchar());
    for(;(ch>='0')&&(ch<='9');a[num++]=ch-'0',ch=getchar());
    int mx=num/2-1;fo(i,0,mx)swap(a[i],a[num-i-1]);
}
int up(LD x){return int(x)+((int(x)==x)?0:1);}
void DFT(int *a,int tag){
    fo(i,0,n-1){
        int pos=0;
        for(int j=0,ii=i;j<m;pos=(pos<<1)+(ii&1),ii=ii>>1,j++);
        t[pos]=a[i];
    }
    for(int i=2;i<=n;i=i<<1){
        int half=i>>1;
        fo(j,0,half-1){
            int wi=(tag>0)?w[n/i*j]:w[n-n/i*j];
            for(int k=j;kint x=t[k],y=1ll*wi*t[k+half]%mo;
                t[k]=(x+y)%mo;
                t[k+half]=(x-y+mo)%mo;
            }
        }
    }
    fo(i,0,n-1)a[i]=t[i];
}
int Pow(int x,int y){
    int z=1;
    while(y){
        if(y&1)z=1ll*z*x%mo;
        x=1ll*x*x%mo;
        y=y>>1;
    }
    return z;
}
int main(){
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
    read(A,a);read(B,b);
    m=up(log(max(A,B)<<1)/log(2));n=1<<m;
    w[0]=1;w[1]=Pow(g,(mo-1)/n);
    fo(i,2,n)w[i]=1ll*w[i-1]*w[1]%mo;
    DFT(a,1);DFT(b,1);
    fo(i,0,n-1)a[i]=1ll*a[i]*b[i]%mo;
    DFT(a,-1);
    int ni=Pow(n,mo-2);
    fo(i,0,n-1)a[i]=1ll*a[i]*ni%mo;
    fo(i,0,n-1){
        a[i+1]+=a[i]/10;
        a[i]%=10;
    }
    for(;a[n]==0;n--);
    fd(i,n,0)putchar(a[i]+'0');
    return 0;
}