FFT,NTT学习笔记

快速傅里叶变换,可以将多项式相乘的时间复杂度从最简单的O(n^2)优化到O(nlgn),详细过程参考算法导论.

FFT的流程大致是:

1):构造多项式,复杂度O(n)

2):求两个多项式的DFT,复杂度O(nlgn)

3):构造多项式乘积的点值表达式,复杂度O(n)

4):求点值表达式的IDFT,复杂度O(nlgn).

下面是两道最简单的习题:

HDU 1402:点击打开链接

求两个大数乘积.

因为一个大数可以看成是一个多项式,每一位上的值都表示对应次数下的系数,因此可以用FFT加速.

本体的一个坑点就是

len = l1+l2-1;

这句代码,可能是精度问题在len更加高位的地方出现了非0值.

#include <bits/stdc++.h>using namespace std;#define pi acos (-1)#define maxn 200010struct plex {    double x, y;    plex (double _x = 0.0, double _y = 0.0) : x (_x), y (_y) {}    plex operator + (const plex &a) const {        return plex (x+a.x, y+a.y);    }    plex operator - (const plex &a) const {        return plex (x-a.x, y-a.y);    }    plex operator * (const plex &a) const {        return plex (x*a.x-y*a.y, x*a.y+y*a.x);    }};void change (plex *y, int len) {    int i, j, k;    for(i = 1, j = len / 2; i < len - 1; i++) {        if (i < j) swap(y[i], y[j]);        k = len / 2;        while (j >= k) {            j -= k;            k /= 2;        }        if (j < k) j += k;    }}void fft(plex y[],int len,int on){    change(y,len);    for(int h = 2; h <= len; h <<= 1)    {        plex wn(cos(-on*2*pi/h),sin(-on*2*pi/h));        for(int j = 0;j < len;j+=h)        {            plex w(1,0);            for(int k = j;k < j+h/2;k++)            {                plex u = y[k];                plex t = w*y[k+h/2];                y[k] = u+t;                y[k+h/2] = u-t;                w = w*wn;            }        }    }    if(on == -1)        for(int i = 0;i < len;i++)            y[i].x /= len;}char a[maxn], b[maxn];plex x1[maxn], x2[maxn];int ans[maxn];int main () {    while (scanf ("%s%s", a, b) == 2) {        int len = 2, l1 = strlen (a), l2 = strlen (b);        while (len < l1*2 || len < l2*2)            len <<= 1;        for (int i = 0; i < l1; i++) {            x1[i] = plex (a[l1-1-i]-'0', 0);        }        for (int i = l1; i < len; i++)            x1[i] = plex (0, 0);        for (int i = 0; i < l2; i++) {            x2[i] = plex (b[l2-1-i]-'0', 0);        }        for (int i = l2; i < len; i++)            x2[i] = plex (0, 0);        fft (x1, len, 1);        fft (x2, len, 1);        for (int i = 0; i < len; i++)            x1[i] = x1[i]*x2[i];        fft (x1, len, -1);        for (int i = 0; i < len; i++) {            ans[i] = (int)(x1[i].x+0.5);        }        for (int i = 0; i < len; i++) {            if (ans[i] >= 10) {                ans[i+1] += ans[i]/10;                ans[i] %= 10;            }        }        len = l1+l2-1;        while (ans[len] <= 0 && len > 0)            len--;        for (int i = len; i >= 0; i--) {            printf ("%d", ans[i]);        }        printf ("\n");    }    return 0;}

HDU 4609:点击打开链接

题意是给出n个长度,任取3个求能组成三角形的概率.

首先记录下每个长度的数量,然后用FFT加速求出任取两个长度下的情况,这里面有重复:

首先减去两次都取同一根的情况,减完之后的结果都/2.

最后只需要所有的情况减去不能组成三角形的情况,将最初的长度序列排序后从小到大枚举下标,假设这条边是最长边,那么如果所有两条边长度小于等于这条边的情况就应该减去,这里用前缀和统计下就好了.

#include <bits/stdc++.h>using namespace std;#define pi acos (-1)#define maxn 611111struct plex {    double x, y;    plex (double _x = 0.0, double _y = 0.0) : x (_x), y (_y) {}    plex operator + (const plex &a) const {        return plex (x+a.x, y+a.y);    }    plex operator - (const plex &a) const {        return plex (x-a.x, y-a.y);    }    plex operator * (const plex &a) const {        return plex (x*a.x-y*a.y, x*a.y+y*a.x);    }};void change (plex y[], int len) {    if (len == 1)        return ;    plex a1[len/2], a2[len/2];    for (int i = 0; i < len; i += 2) {        a1[i/2] = y[i];        a2[i/2] = y[i+1];    }    change (a1, len>>1);    change (a2, len>>1);    for (int i = 0; i < len/2; i++) {        y[i] = a1[i];        y[i+len/2] = a2[i];    }    return  ;}void fft(plex y[],int len,int on){    change(y,len);    for(int h = 2; h <= len; h <<= 1)    {        plex wn(cos(on*2*pi/h),sin(on*2*pi/h));        for(int j = 0;j < len;j+=h)        {            plex w(1,0);            for(int k = j;k < j+h/2;k++)            {                plex u = y[k];                plex t = w*y[k+h/2];                y[k] = u+t;                y[k+h/2] = u-t;                w = w*wn;            }        }    }    if(on == -1)        for(int i = 0;i < len;i++)            y[i].x /= len;}long long num[maxn], sum[maxn];int a[maxn];plex x[maxn];long long n;int main () {    //freopen ("in.txt", "r", stdin);    int t;    scanf ("%d", &t);    while (t--) {        scanf ("%lld", &n);        long long Max = 0;        memset (num, 0, sizeof num);        for (int i = 1; i <= n; i++) {            scanf ("%d", &a[i]);            num[a[i]]++;            Max = max (Max, (long long)a[i]);        }        Max++;        int len = 2;        while (len < Max*2)            len <<= 1;        for (int i = 0; i < len; i++) {            x[i] = plex (num[i], 0);        }        fft (x, len, 1);        for (int i = 0; i < len; i++) {            x[i] = x[i]*x[i];        }        fft (x, len, -1);        for (int i = 0; i < len; i++) {            num[i] = (long long) (x[i].x+0.5);        }        for (int i = 1; i <= n; i++) {//两次取同一个            num[a[i]+a[i]]--;        }        for (int i = 0; i < len; i++) {//重复计算            num[i] /= 2;        }        sum[0] = 0;        for (int i = 1; i < len; i++) {            sum[i] = sum[i-1]+num[i];        }        sort (a+1, a+1+n);        long long tot = n*(n-1)*(n-2)/6, ans = tot;        for (int i = 3; i <= n; i++) {            ans -= sum[a[i]];        }        printf ("%.7f\n", ans*1.0/tot);    }    return 0;}

但是FFT有一个很致命的弱点就是会产生精度误差，在换成long double都不行的时候就需要用到NTT。

NTT就是用数论域中的原根代替FFT中的单位负根，其他的代码完全相同。

求原根的代码：

#include <cstdio>#include <cmath>#include <algorithm>#include <iostream>#include <vector>#include <cstring>using namespace std;int P;const int NUM = 32170;int prime[NUM/4];bool f[NUM];int pNum = 0;void getPrime () {//线性筛选素数    for (int i = 2; i < NUM; ++ i) {        if (!f[i]) {            f[i] = 1;            prime[pNum++] = i;        }        for (int j = 0; j < pNum && i*prime[j] < NUM; ++ j) {            f[i*prime[j]] = 1;            if (i%prime[j] == 0) {                break;            }        }    }}long long getProduct(int a,int b,int P) {//快速求次幂mod    long long ans = 1;    long long tmp = a;    while (b)    {        if (b&1)        {            ans = ans*tmp%P;        }        tmp = tmp*tmp%P;        b>>=1;    }    return ans;}bool judge (int num) {//求num的所有的质因子    int elem[1000];    int elemNum = 0;    int k = P - 1;    for (int i = 0; i < pNum; ++ i) {        bool flag = false;        while (!(k%prime[i])) {            flag = true;            k /= prime[i];        }        if (flag) {            elem[elemNum ++] = prime[i];        }        if (k == 1) {            break;        }        if (k/prime[i]<prime[i]) {            elem[elemNum ++] = prime[i];            break;        }    }    bool flag = true;    for (int i = 0; i < elemNum; ++ i) {        if (getProduct (num, (P-1)/elem[i], P) == 1) {            flag = false;            break;        }    }    return flag;}int main()  {            getPrime();      while (cin >> P)      {          for (int i = 2;;++i)          {              if (judge(i))              {                  cout << i<< endl;                  break;              }          }      }      return 0;  }  

HDU 1402：

随便选一个不太大的模数和他的原根就好了。

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <map>#include <vector>#include <stack>using namespace std;#define mod 1004535809LL#define G 3#define maxn 400005long long qpow (long long a, long long b) {      long long ret=1;      while (b) {          if (b&1) ret = (ret*a)%mod;          a = (a*a)%mod;          b >>= 1;      }      return ret;  }  void change (long long y[], int len) {    for(int i = 1, j = len / 2; i < len - 1; i++) {        if(i < j) swap(y[i], y[j]);        int k = len / 2;        while(j >= k) {            j -= k;            k /= 2;        }        if(j < k) j += k;    }}void ntt(long long y[], int len, int on) {    change (y, len);    for(int h = 2; h <= len; h <<= 1) {        long long wn = qpow(G, (mod-1)/h);        if(on == -1) wn = qpow(wn, mod-2);        for(int j = 0; j < len; j += h) {            long long w = 1;            for(int k = j; k < j + h / 2; k++) {                long long u = y[k];                long long t = w * y[k + h / 2] % mod;                y[k] = (u + t) % mod;                y[k+h/2] = (u - t + mod) % mod;                w = w * wn % mod;            }        }    }    if(on == -1) {        long long t = qpow (len, mod - 2);        for(int i = 0; i < len; i++)            y[i] = y[i] * t % mod;    }}char a[maxn], b[maxn];long long x1[maxn], x2[maxn];long long ans[maxn];int main () {    while (scanf ("%s%s", a, b) == 2) {        int len = 2, l1 = strlen (a), l2 = strlen (b);        while (len < l1*2 || len < l2*2)            len <<= 1;        //cout << len << endl;        for (int i = 0; i < l1; i++) {            x1[i] = a[l1-1-i]-'0';        }        for (int i = l1; i < len; i++)            x1[i] = 0;        for (int i = 0; i < l2; i++) {            x2[i] = b[l2-1-i]-'0';        }        for (int i = l2; i < len; i++)            x2[i] = 0;        ntt(x1, len, 1);        ntt(x2, len, 1);        for (int i = 0; i < len; i++)            x1[i] = x1[i]*x2[i]%mod;        ntt(x1, len, -1);        for (int i = 0; i < len; i++) {            ans[i] = x1[i];        }        for (int i = 0; i < len; i++) {            if (ans[i] >= 10) {                ans[i+1] += ans[i]/10;                ans[i] %= 10;            }        }        len = l1+l2-1;        while (ans[len] <= 0 && len > 0)            len--;        for (int i = len; i >= 0; i--) {            printf ("%lld", ans[i]);        }        printf ("\n");    }    return 0;}