FFT多项式乘法模板（HDU1402）

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这几天学了下FFT，入门推荐算法导论第三十章，话不多说，上板子。

递归版：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#pragma comment(linker, "/STACK:102400000,102400000")
#include<unordered_map>
#include<unordered_set>
#include<algorithm>
#include<iostream>
#include<fstream>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iomanip>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cctype>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define pt(a) cout<<a<<endl
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define pi acos(-1.0)
typedef pair<int,int> PII;
typedef complex<double> cd;
const ll mod = 1e9+7;
const int N = 2e5+10;

ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};

char a[N],b[N];
cd A[N],B[N];
int s[N];

void fft(cd *x, int n, int type) {
    if(n == 1) return ;
    cd l[n>>1], r[n>>1];
    for (int i = 0; i < n; i += 2) {
        l[i>>1] = x[i];
        r[i>>1] = x[i+1];
    }
    fft(l, n>>1, type);
    fft(r, n>>1, type);
    cd wn(cos(2*pi/n), sin(type*2*pi/n)), w(1, 0), t;
    for(int i = 0; i < (n>>1); i++, w *= wn) {
        t = w*r[i];
        x[i] = l[i] + t;
        x[i+(n>>1)] = l[i] - t;
    }
}

int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);

    while(cin>>a>>b) {
        int n=strlen(a),m=strlen(b);
        mst(A,0),mst(B,0),mst(s,0);
        for(int i=n-1;i>=0;i--) A[n-1-i]=a[i]-'0';
        for(int i=m-1;i>=0;i--) B[m-1-i]=b[i]-'0';
        m += n;
        for(n=1;n<m;n<<=1);
        fft(A,n,1);
        fft(B,n,1);
        for(int i=0;i<n;i++) A[i]*=B[i];
        fft(A,n,-1);
        for(int i=0;i<m;i++) {
            int t=(int)(A[i].real()/n+0.5);
            t += s[i];
            s[i] = t%10;
            s[i+1] += t/10;
        }
        int i;
        for(i=m;i>=1;i--) if(s[i]) break;
        for(;i>=0;i--) cout<<s[i];
        cout<<endl;
    }
    return 0;
}

非递归版：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#pragma comment(linker, "/STACK:102400000,102400000")
#include<unordered_map>
#include<unordered_set>
#include<algorithm>
#include<iostream>
#include<fstream>
#include<complex>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iomanip>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cctype>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define pt(a) cout<<a<<endl
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define pi acos(-1.0)
typedef pair<int,int> PII;
typedef complex<double> cd;
const ll mod = 1e9+7;
const int N = 2e5+10;

ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};

char a[N],b[N];
cd A[N],B[N];
int s[N],R[N];

void fft(cd *x,int n,int type) {
    for (int i = 0; i < n; i++) if(i < R[i]) swap(x[i], x[R[i]]);
    for (int i = 1; i < n; i <<= 1) {
        cd wn(cos(pi/i), type*sin(pi/i));
        for (int j = 0; j < n; j += i<<1) {
            cd w(1, 0);
            for (int k = 0; k < i; k++, w*=wn) {
                cd X = x[j+k], Y = w*x[j+k+i];
                x[j+k] = X+Y;
                x[j+k+i] = X-Y;
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    while(cin>>a>>b) {
        int n=strlen(a),m=strlen(b),L=0;
        mst(A,0),mst(B,0),mst(s,0),mst(R,0);
        for(int i=n-1;i>=0;i--) A[n-1-i]=a[i]-'0';
        for(int i=m-1;i>=0;i--) B[m-1-i]=b[i]-'0';
        m += n;
        for(n=1;n<m;n<<=1) L++;
        for (int i = 0; i < n; i++) {///二进制反转 蝴蝶操作
            R[i] = (R[i>>1]>>1) | ((i&1) << L-1);
        }
        fft(A,n,1);
        fft(B,n,1);
        for(int i=0;i<n;i++) A[i]*=B[i];
        fft(A,n,-1);
        for(int i=0;i<m;i++) {
            int t=(int)(A[i].real()/n+0.5);
            t += s[i];
            s[i] = t%10;
            s[i+1] += t/10;
        }
        int i;
        for(i=m;i>=1;i--) if(s[i]) break;
        for(;i>=0;i--) cout<<s[i];
        cout<<endl;
    }
    return 0;
}