LG P1919

\(\text{FFT}\)

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#define re register
using namespace std;

const int N = 2e6 + 1e5;
int c[N], rev[N];
char s[N];

const double Pi = acos(-1.0);
struct complex{
	double x, y;
	inline complex(double xx = 0, double yy = 0){x = xx, y = yy;}
	inline complex operator + (const complex &b) const {return complex(x + b.x, y + b.y);}
	inline complex operator - (const complex &b) const {return complex(x - b.x, y - b.y);}
	inline complex operator * (const complex &b) const {return complex(x * b.x - y * b.y, x * b.y + y * b.x);}
}a[N], b[N];

inline void FFT(complex *a, int lim, int inv)
{
	for(re int i = 0; i < lim; i++)
	if (i < rev[i]) swap(a[i], a[rev[i]]);
	for(re int mid = 1; mid < lim; mid <<= 1)
	{
		complex I = complex(cos(Pi / mid), inv * sin(Pi / mid));
		for(re int i = 0; i < lim; i += mid << 1)
		{
			complex W = complex(1, 0);
			for(re int j = 0; j < mid; j++, W = W * I)
			{
				complex x = a[i + j], y = W * a[i + j + mid];
				a[i + j] = x + y, a[i + j + mid] = x - y;
			}
		}
	}
}

int main()
{
	scanf("%s", s);
	int n = strlen(s);
	for(re int i = 0; i < n; i++) a[i].x = s[n - i - 1] ^ 48;
	scanf("%s", s);
	int m = strlen(s);
	for(re int i = 0; i < m; i++) b[i].x = s[m - i - 1] ^ 48;
	
	int limit = 1;
	while (limit < n + m - 1) limit <<= 1;
	int bit = 0;
	while ((1 << bit) < limit) ++bit;
	for(re int i = 0; i < limit; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
	
	FFT(a, limit, 1), FFT(b, limit, 1);
	for(re int i = 0; i < limit; i++) a[i] = a[i] * b[i];
	FFT(a, limit, -1);
	for(re int i = 0; i < limit; i++) c[i] = int(a[i].x / limit + 0.5);
	for(re int i = 0; i < limit; i++)
	{
		if (c[i] >= 10) c[i + 1] += c[i] / 10, limit = ((i + 1) == limit ? limit + 1 : limit);
		c[i] %= 10;
	}
	while (limit && !c[limit]) --limit;
	for(re int i = limit; i >= 0; i--) printf("%d", c[i]);
}

\(\text{NTT}\)

#include <cstdio>
#include <iostream>
#include <cstring>
#define re register
using namespace std;

const int N = 2e6 + 1e5;
const int P = 998244353, g = 3;
int a[N], b[N], rev[N];
char s[N];

inline int fpow(int x, int y)
{
	int res = 1;
	for(; y; y >>= 1)
	{
		if (y & 1) res = 1LL * res * x % P;
		x = 1LL * x * x % P;
	}
	return res;
}

inline void NTT(int *a, int lim, int inv)
{
	if (lim == 1) return;
	for(re int i = 0; i < lim; i++)
	if (i < rev[i]) swap(a[i], a[rev[i]]);
	for(re int mid = 1, I; mid < lim; mid <<= 1)
	{
		I = fpow(g, (P - 1) / (mid << 1));
		if (inv == -1) I = fpow(I, P - 2);
		for(re int i = 0, W; i < lim; i += mid << 1)
		{
			W = 1;
			for(re int j = 0, x, y; j < mid; j++, W = 1LL * W * I % P)
			{
				x = a[i + j], y = 1LL * W * a[i + j + mid] % P;
				a[i + j] = (x + y) % P, a[i + j + mid] = (x - y + P) % P;
			}
		}
	}
}

int main()
{
	scanf("%s", s);
	int n = strlen(s);
	for(re int i = 0; i < n; i++) a[i] = s[n - i - 1] ^ 48;
	scanf("%s", s);
	int m = strlen(s);
	for(re int i = 0; i < m; i++) b[i] = s[m - i - 1] ^ 48;
	
	int limit = 1;
	while (limit < n + m - 1) limit <<= 1;
	int bit = 0;
	while ((1 << bit) < limit) ++bit;
	for(re int i = 0; i < limit; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
	
	NTT(a, limit, 1), NTT(b, limit, 1);
	for(re int i = 0; i < limit; i++) a[i] = 1LL * a[i] * b[i] % P;
	NTT(a, limit, -1);
	int inv = fpow(limit, P - 2);
	for(re int i = 0; i < limit; i++) a[i] = 1LL * a[i] * inv % P;
	for(re int i = 0; i < limit; i++)
	{
		if (a[i] >= 10) a[i + 1] += a[i] / 10, limit = ((i + 1) == limit ? limit + 1 : limit);
		a[i] %= 10;
	}
	while (limit && !a[limit]) --limit;
	for(re int i = limit; i >= 0; i--) printf("%d", a[i]);
}
posted @ 2021-08-16 20:13  leiyuanze  阅读(25)  评论(0编辑  收藏  举报