# luoguP3702 [SDOI2017]序列计数

https://www.luogu.org/problemnew/show/P3702

$Ans[i + j - 1]$ 这一项的值包含 $f1[i] * f2[j]$ 的值，而其他项不包含

$Ans[i + j - 2]$ 这一项的值包含 $f1[i - 1] * f2[j - 1]$ 的值，而其他项不包含

1. NTT
#include <bits/stdc++.h>
#define double long double
#define CIOS ios::sync_with_stdio(false);
#define For(i, a, b) for(register int i = a; i <= b; i++)
#define Rof(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> ";
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename T>
f = 0; T fu = 1; char c = getchar();
while(c < '0' || c > '9') {if(c == '-') fu = -1; c = getchar();}
while(c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}
f *= fu;
}

template <typename T>
void print(T x) {
if(x < 0) putchar('-'), x = -x;
if(x < 10) putchar(x + 48);
else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
print(x); putchar(t);
}

const int N = 405, mod = 20170408;
inline int mul(int x, int y) { return (int)(1ll * x * y % (ll)mod); }

int r[N], n, m, p, len = 1, tot = 0;
ll A[N], B[N], C[N], ans[N], a[N], b[N], c[N], d[N];

const int P[3] = {469762049, 998244353, 167772161}, G = 3, Gi[3] = {P[0] / G + 1, P[1] / G + 1, P[2] / G + 1};

ll fpow(ll x, ll y, ll p) {
ll ans = 1;
while(y) {
if(y & 1) ans = ans * x % p;
y >>= 1; x = x * x % p;
}
return ans;
}

void NTT(ll *a, int tp, int t) {
for(register int i = 1; i < len; i++) if(i < r[i]) swap(a[i], a[r[i]]);
for(register int mid = 1; mid < len; mid <<= 1) {
ll wn = fpow(tp == 1 ? G : Gi[t], (P[t] - 1) / (mid << 1), P[t]);
for(register int i = 0; i < len; i += (mid << 1)) {
ll w = 1;
for(register int j = 0; j < mid; j++, w = w * wn % P[t]) {
ll x = a[i + j], y = a[i + j + mid] * w % P[t];
a[i + j] = (x + y) % P[t]; a[i + j + mid] = (x - y + P[t]) % P[t];
}
}
}
if(tp == -1) {
ll inv = fpow(len, P[t] - 2, P[t]);
for(register int i = 0; i < len; i++) a[i] = a[i] * inv % P[t];
}
}

ll inv(ll x, ll y) { return fpow(x, y - 2, y); }

ll times(ll x, ll y, ll p) {
ll ans = 0;
while(y) {
if(y & 1) ans = (ans + x) % p;
y >>= 1; x = (x + x) % p;
}
return ans;
}

void mul(ll *A, ll *B, ll *ans) {
for(register int i = 1; i <= p; i++) A[i - 1] = A[i], B[i - 1] = B[i]; A[p] = B[p] = 0;
len = 1; tot = 0;
while(len <= (p << 1)) len <<= 1, tot++;
for(register int i = 1; i < len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (tot - 1));
memcpy(a, A, sizeof(a)); memcpy(b, A, sizeof(b)); memcpy(c, A, sizeof(c));
memcpy(d, B, sizeof(d)); NTT(a, 1, 0); NTT(d, 1, 0); for(register int i = 0; i < len; i++) a[i] = a[i] * d[i] % P[0]; NTT(a, -1, 0);
memcpy(d, B, sizeof(d)); NTT(b, 1, 1); NTT(d, 1, 1); for(register int i = 0; i < len; i++) b[i] = b[i] * d[i] % P[1]; NTT(b, -1, 1);
memcpy(d, B, sizeof(d)); NTT(c, 1, 2); NTT(d, 1, 2); for(register int i = 0; i < len; i++) c[i] = c[i] * d[i] % P[2]; NTT(c, -1, 2);
for(register int i = 0; i <= (p << 1); i++) {
ll p = 1ll * P[0] * P[1];
ll s = times(a[i], times(P[1], inv(P[1], P[0]), p), p) + times(b[i], times(P[0], inv(P[0], P[1]), p), p); s %= p;
ll k = (c[i] % P[2] - s % P[2] + P[2]) % P[2] * inv(p % P[2], P[2]) % P[2];
ll Ans = (s + (k * P[0] % mod * P[1] % mod)) % mod;
ans[i + 1] = Ans;
}
for(register int i = 1; i <= p; i++) ans[i] = (ans[i] + ans[i + p]) % mod, ans[i + p] = 0;
for(register int i = p; i >= 1; i--) A[i] = A[i - 1], B[i] = B[i - 1]; A[0] = B[0] = 0;
}

int fpow(int x, int y) {
int ans = 1;
while(y) {
if(y & 1) ans = mul(ans, x);
y >>= 1; x = mul(x, x);
}
return ans;
}

int pri[5000000], plen;
bool isp[20000005];

void init_pri(int n, ll *a) {
isp[1] = 1;
for(register int i = 2; i <= n; i++) {
if(!isp[i]) { pri[++plen] = i; }
for(register int j = 1; j <= plen && i * pri[j] <= n; j++) {
isp[i * pri[j]] = 1;
if(i % pri[j] == 0) break;
}
}
for(register int i = 1; i <= p; i++) a[i] = 0;
for(register int i = 1; i <= n; i++) if(isp[i]) ++a[i % p + 1];
}

int main() {
ans[1] = 1;
while(n) {
if(n & 1) {
memcpy(B, ans, sizeof(B));
mul(A, B, ans);
}
n >>= 1;
memcpy(B, A, sizeof(B));
memcpy(C, A, sizeof(C));
mul(B, C, A);
}
int fir = ans[1];
memset(ans, 0, sizeof(ans));
memset(A, 0, sizeof(A));
for(register int i = 1; i <= m; i++) ++A[i % p + 1];
ans[1] = 1; n = t;
while(n) {
if(n & 1) {
memcpy(B, ans, sizeof(B));
mul(A, B, ans);
}
n >>= 1;
memcpy(B, A, sizeof(B));
memcpy(C, A, sizeof(C));
mul(B, C, A);
}
cout << (ans[1] - fir + mod) % mod << endl;
return 0;
}

2.FFT & 暴力卷积（当然 FFT 是错误的）

#include <bits/stdc++.h>
#define CIOS ios::sync_with_stdio(false);
#define For(i, a, b) for(register int i = a; i <= b; i++)
#define Rof(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> ";
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename T>
f = 0; T fu = 1; char c = getchar();
while(c < '0' || c > '9') {if(c == '-') fu = -1; c = getchar();}
while(c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}
f *= fu;
}

template <typename T>
void print(T x) {
if(x < 0) putchar('-'), x = -x;
if(x < 10) putchar(x + 48);
else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
print(x); putchar(t);
}

const int N = 405, P = 20170408;
inline int mul(int x, int y) { return (int)(1ll * x * y % (ll)P); }

/*const double PI = acos(-1.0);

struct cp {
double x, y;
cp (double xx = 0, double yy = 0) {
x = xx; y = yy;
}
}a[N], b[N];

cp operator + (const cp a, const cp b) { return cp(a.x + b.x, a.y + b.y); }
cp operator - (const cp a, const cp b) { return cp(a.x - b.x, a.y - b.y); }
cp operator * (const cp a, const cp b) { return cp(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); }*/

int r[N], n, m, p, len = 1, tot = 0;
int A[N], B[N], C[N], ans[N];

/*void FFT(cp *a, int d) {
for(register int i = 1; i < len; i++) if(i < r[i]) swap(a[i], a[r[i]]);
for(register int mid = 1; mid < len; mid <<= 1) {
cp wn = cp(cos(PI / mid), sin(PI / mid) * d);
for(register int i = 0; i < len; i += (mid << 1)) {
cp w = cp(1.0, 0.0);
for(register int j = 0; j < mid; j++, w = w * wn) {
cp x = a[i + j], y = a[i + j + mid] * w;
a[i + j] = x + y; a[i + j + mid] = x - y;
}
}
}
}*/

/*void mul(int *A, int *B, int *ans) {
len = 1, tot = 0;
while(len <= (p << 1)) len <<= 1, tot++;
for(register int i = 1; i <= len; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (tot - 1));
for(register int i = 1; i <= len; i++) a[i - 1] = cp(A[i], 0), b[i - 1] = cp(B[i], 0);
FFT(a, 1); FFT(b, 1);
for(register int i = 0; i < len; i++) a[i] = a[i] * b[i];
FFT(a, -1);
for(register int i = 1; i <= len; i++) ans[i] = (int)((ll)(a[i - 1].x / (double)len + 0.5) % (ll)P);
for(register int i = 1; i <= p; i++) ans[i] = (ans[i] + ans[i + p]) % P, ans[i + p] = 0;
}*/

/*void mul(int *A, int *B, int *ans) {
for(register int i = 1; i <= p << 1; i++) A[i - 1] = A[i], B[i - 1] = B[i], ans[i] = 0; ans[0] = 0;
for(register int i = 0; i < p; i++) {
for(register int j = 0; j < p; j++) {
ans[i + j] += mul(A[i], B[j]);
if(ans[i + j] >= P) ans[i + j] -= P;
}
}
for(register int i = p << 1; i >= 1; i--) ans[i] = ans[i - 1], A[i] = A[i - 1], B[i] = B[i - 1]; A[0] = B[0] = 0;
for(register int i = 1; i <= p; i++) ans[i] = (ans[i] + ans[i + p]) % P, ans[i + p] = 0;
}*/

void mul(int *A, int *B, int *ans) {
memset(ans, 0, (N + 1) * 4);
for(register int i = 1; i <= p; i++) {
for(register int j = 1; j <= p; j++) {
ans[i + j - 1] += mul(A[i], B[j]);
ans[i + j - 1] %= P;
}
}
for(register int i = 1; i <= p; i++) ans[i] = (ans[i] + ans[i + p]) % P, ans[i + p] = 0;
}

int fpow(int x, int y) {
int ans = 1;
while(y) {
if(y & 1) ans = mul(ans, x);
y >>= 1; x = mul(x, x);
}
return ans;
}

int pri[5000000], plen;
bool isp[20000005];

void init_pri(int n, int *a) {
isp[1] = 1;
for(register int i = 2; i <= n; i++) {
if(!isp[i]) { pri[++plen] = i; }
for(register int j = 1; j <= plen && i * pri[j] <= n; j++) {
isp[i * pri[j]] = 1;
if(i % pri[j] == 0) break;
}
}
for(register int i = 1; i <= p; i++) a[i] = 0;
for(register int i = 1; i <= n; i++) if(isp[i]) ++a[i % p + 1];
}

int main() {
ans[1] = 1;
while(n) {
if(n & 1) {
memcpy(B, ans, sizeof(B));
mul(A, B, ans);
}
n >>= 1;
memcpy(B, A, sizeof(B));
memcpy(C, A, sizeof(C));
mul(B, C, A);
}
int fir = ans[1];
memset(ans, 0, sizeof(ans));
memset(A, 0, sizeof(A));
for(register int i = 1; i <= m; i++) ++A[i % p + 1];
ans[1] = 1; n = t;
while(n) {
if(n & 1) {
memcpy(B, ans, sizeof(B));
mul(A, B, ans);
}
n >>= 1;
memcpy(B, A, sizeof(B));
memcpy(C, A, sizeof(C));
mul(B, C, A);
}
cout << (ans[1] - fir + P) % P << endl;
return 0;
}
posted @ 2018-11-04 17:59 LJC00118 阅读(...) 评论(...) 编辑 收藏