Wannafly Winter Camp Day5 Div1 E题 Fast Kronecker Transform 转化为NTT或FFT

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Catalog

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目录

• Catalog
  • Problem：传送门
• Solution：
  • AC_Code：
  • Problem Description：

Problem：传送门

 原题目描述在最下面。
 对给定的式子算解。
 (0leq kleq n+m,c_k=(sum_{i+j=k}i imes j imes sigma_{a_i,b_j}) mod;998244353)，其中(当且仅当a=b时，sigma_{a_i,b_j}=1。)

Solution：

 我们发现只有当(a_i)和(b_j)相等时才会对答案造成贡献。

 一共(2e^5)个数字，我们枚举每一个数字算贡献，同时分情况讨论：当这个
数字出现次数小于阀值(T)时，我们(O(n^2))暴力算；当出现次数大于(T)时，我们用(FFTorNTT)计算。

 听说这题有人(FFT)丢精度就(wa)了，我就干脆用(NTT)写了，刚好这个模数也是费马质数嘛，接下来看怎么把上述奇怪的卷积转化成一个可以用(FFT)和(NTT)计算的卷积。

 我们把(a)数组和(b)数组中数字(x)所有出现的位置提出来放到新数组里面去，比如(x)出现在(a)的(1,3,4)位置，也出现在(b)的(2,4,5,6)位置。

 当(k)等于(5)时，数字(x)产生的贡献(1*4,2*3)
 当(k)等于(6)时，数字(x)产生的贡献(1*5)

 我们知道一般的卷积式子是长这样的：(C_k=sum_{i+j=k}A_i imes B_j)

 哇哦，感觉这个式子和那个贡献好相似啊。

 我们就构造这样两个数组(A_{a_i}=a_i,B_{b_j}=b_j)可以得到：
(A[]={0,1,0,3,4},B[]={0,0,2,0,4,5,6})

 我们对这两个数组求他们多项式乘法的结果，然后将结果的贡献累计到答案数组中就行啦。

 然后当阀值(T)为(10000)时，复杂度就可以承受了。

AC_Code：

感谢日天的板子
这题复杂度的分析dls讲题时分析的很清楚啦，一位群友的笔记截图在下面

#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
namespace lh {
#define o2(x) (x)*(x)
    using namespace std;
    typedef long long LL;
    typedef unsigned long long uLL;
    typedef pair<int, LL> pii;
}

using namespace lh;
const int MX = 2e5 + 5;
//const int P = (479 << 21) + 1;
const int P = 998244353;
const int MOD = 998244353;
const int G = 3;
const int NUM = 20;
struct my_NTT {
    LL wn[NUM];
    LL a[MX << 1], b[MX << 1];
    LL pow (LL a, LL x, LL mod) {
        LL ans = 1;
        a %= mod;
        while (x) {
            if (x & 1) ans = ans * a % mod;
            x >>= 1;
            a = a * a % mod;
        }
        return ans;
    }
    //在程序的开头就要放
    void init() {
        for (int i = 0; i < NUM; i++) {
            int t = 1 << i;
            wn[i] = pow (G, (P - 1) / t, P);
        }
    }
    void Rader (LL F[], int len) {
        int j = len >> 1;
        for (int i = 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;
        }
    }
    void NTT (LL F[], int len, int t) {
        Rader (F, len);
        int id = 0;
        for (int h = 2; h <= len; h <<= 1) {
            id++;
            for (int j = 0; j < len; j += h) {
                LL E = 1;
                for (int k = j; k < j + h / 2; k++) {
                    LL u = F[k];
                    LL v = E * F[k + h / 2] % P;
                    F[k] = (u + v) % P;
                    F[k + h / 2] = (u - v + P) % P;
                    E = E * wn[id] % P;
                }
            }
        }
        if (t == -1) {
            for (int i = 1; i < len / 2; i++) swap (F[i], F[len - i]);
            LL inv = pow (len, P - 2, P);
            for (int i = 0; i < len; i++) F[i] = F[i] * inv % P;
        }
    }
    void Conv (LL a[], LL b[], int len) {
        NTT (a, len, 1);
        NTT (b, len, 1);
        for (int i = 0; i < len; i++) a[i] = a[i] * b[i] % P;
        NTT (a, len, -1);
    }
    void gao (LL A[], LL B[], int n, int m, LL ans[]) {//0~n-1
        int len = 1;
        while (len < n + m) len <<= 1;
        for (int i = 0; i < n; i++) a[i] = A[i];
        for (int i = 0; i < m; i++) b[i] = B[i];
        for (int i = n; i < len; i++) a[i] = 0;
        for (int i = m; i < len; i++) b[i] = 0;
        Conv (a, b, len);
        for (int i = 0; i < len; i++) ans[i] = (ans[i]+a[i])%MOD;
    }
}ntt;
const int MXN = 2e5 + 5;
int n, m;
int ar[MXN], br[MXN];
LL A[MXN], B[MXN];
std::vector<int> all[MXN], bll[MXN];
LL ans[MXN];
void solve1(int id) {
    for(int i = 0; i < all[id].size(); ++i) {
        for(int j = 0; j < bll[id].size(); ++j) {
            ans[all[id][i]+bll[id][j]] += (LL)all[id][i] * bll[id][j];
            ans[all[id][i]+bll[id][j]] %= MOD;
        }
    }
}
void solve2(int id) {
    for(int i = 0; i <= n+m; ++i) A[i] = B[i] = 0;
    for(int i = 0; i < all[id].size(); ++i) A[all[id][i]] = all[id][i];
    for(int i = 0; i < bll[id].size(); ++i) B[bll[id][i]] = bll[id][i];
    ntt.gao(A, B, all[id].back()+1, bll[id].back()+1, ans);
}
int main(int argc, char const *argv[]) {
    scanf("%d%d", &n, &m); ++n, ++m;
    ntt.init();
    std::vector<int> vs;
    for(int i = 0; i < n; ++i) scanf("%d", &ar[i]), vs.push_back(ar[i]);
    for(int i = 0; i < m; ++i) scanf("%d", &br[i]), vs.push_back(br[i]);
    sort(vs.begin(), vs.end());
    vs.erase(unique(vs.begin(), vs.end()), vs.end());
    for(int i = 0, tmp; i < n; ++i) {
        tmp = lower_bound(vs.begin(), vs.end(), ar[i]) - vs.begin();
        all[tmp].push_back(i);
    }
    for(int i = 0, tmp; i < m; ++i) {
        tmp = lower_bound(vs.begin(), vs.end(), br[i]) - vs.begin();
        bll[tmp].push_back(i);
    }
    for(int i = 0; i < vs.size(); ++i) {
        if(all[i].size() + bll[i].size() <= 10000) solve1(i);
        else solve2(i);
    }
    for(int i = 0; i <= n + m-2; ++i) printf(i!=n+m-2?"%lld ":"%lld
", ans[i]);
    return 0;
}

Problem Description：

果然，我的$FFTwa了$ ```cpp #include #define fi first #define se second #define pb push_back namespace lh { #define o2(x) (x)*(x) using namespace std; typedef long long LL; typedef unsigned long long uLL; typedef pair pii; }

using namespace lh;
const int MX = 2e5 + 5;
//const int P = (479 << 21) + 1;
const int P = 998244353;
const int MOD = 998244353;
const int G = 3;
const int NUM = 20;

struct my_FFT {
const double pi = acos (-1.0);
static const int MM = MX * 4;
int len, res[MM], mx; //开大4倍
struct cpx {
double r, i;
cpx (double r = 0, double i = 0) : r (r), i (i) {};
cpx operator+ (const cpx &b) {return cpx (r + b.r, i + b.i);}
cpx operator- (const cpx &b) {return cpx (r - b.r, i - b.i);}
cpx operator* (const cpx &b) {return cpx (rb.r - ib.i,i*b.r + r * b.i);}
} va[MM], vb[MM];
void rader (cpx F[], int len) { //len = 2^M,reverse F[i] with F[j] j为i二进制反转
int j = len >> 1;
for (int i = 1; i < len - 1; ++i) {
if (i < j) swap (F[i], F[j]); // reverse
int k = len >> 1;
while (j >= k) j -= k, k >>= 1;
if (j < k) j += k;
}
}
void FFT (cpx F[], int len, int t) {
rader (F, len);
for (int h = 2; h <= len; h <<= 1) {
cpx wn (cos (-t * 2 * pi / h), sin (-t * 2 * pi / h) );
for (int j = 0; j < len; j += h) {
cpx E (1, 0); //旋转因子
for (int k = j; k < j + h / 2; ++k) {
cpx u = F[k];
cpx v = E * F[k + h / 2];
F[k] = u + v;
F[k + h / 2] = u - v;
E = E * wn;
}
}
}
if (t == -1) //IDFT
for (int i = 0; i < len; ++i) F[i].r /= len;
}
void Conv (cpx a[], cpx b[], int len) { //求卷积
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);
}
void gao (LL a[], LL b[], int n, int m, LL ans[]) {
len = 1;
mx = n + m;
while (len <= mx) len <<= 1; //mx为卷积后最大下标
for (int i = 0; i < len; i++) va[i].r =va[i].i = vb[i].r = vb[i].i = 0;
for (int i = 0; i < n; i++) va[i].r = a[i];//根据题目要求改写
for (int i = 0; i < m; i++) vb[i].r = b[i];//根据题目要求改写
Conv (va, vb, len);
for (int i = 0; i < len; ++i) ans[i] = (ans[i]+(LL)(va[i].r + 0.5))%MOD;
}

}fft;
const int MXN = 2e5 + 5;
int n, m;
int ar[MXN], br[MXN];
LL A[MXN], B[MXN];
std::vector all[MXN], bll[MXN];
LL ans[MXN*10];
void solve1(int id) {
for(int i = 0; i < all[id].size(); ++i) {
for(int j = 0; j < bll[id].size(); ++j) {
ans[all[id][i]+bll[id][j]] += (LL)all[id][i] * bll[id][j];
ans[all[id][i]+bll[id][j]] %= MOD;
}
}
}
void solve2(int id) {
for(int i = 0; i <= n+m; ++i) A[i] = B[i] = 0;
for(int i = 0; i < all[id].size(); ++i) A[all[id][i]] = all[id][i];
for(int i = 0; i < bll[id].size(); ++i) B[bll[id][i]] = bll[id][i];
fft.gao(A, B, all[id].back()+1, bll[id].back()+1, ans);
}
int main(int argc, char const *argv[]) {
scanf("%d%d", &n, &m); ++n, ++m;
std::vector vs;
for(int i = 0; i < n; ++i) scanf("%d", &ar[i]), vs.push_back(ar[i]);
for(int i = 0; i < m; ++i) scanf("%d", &br[i]), vs.push_back(br[i]);
sort(vs.begin(), vs.end());
vs.erase(unique(vs.begin(), vs.end()), vs.end());
for(int i = 0, tmp; i < n; ++i) {
tmp = lower_bound(vs.begin(), vs.end(), ar[i]) - vs.begin();
all[tmp].push_back(i);
}
for(int i = 0, tmp; i < m; ++i) {
tmp = lower_bound(vs.begin(), vs.end(), br[i]) - vs.begin();
bll[tmp].push_back(i);
}
for(int i = 0; i < vs.size(); ++i) {
if(all[i].size() + bll[i].size() <= 10000) solve1(i);
else solve2(i);
}
for(int i = 0; i <= n + m-2; ++i) printf(i!=n+m-2?"%lld ":"%lld ", ans[i]);
return 0;
}