代码

#include<cstdio>
#include<algorithm>
#include<ctype.h>
#include<string.h>
#include<math.h>

using namespace std;
#define ll long long
#define travel(i,x) for(int i=h[x];i;i=pre[i])

inline char read() {
	static const int IN_LEN = 1000000;
	static char buf[IN_LEN], *s, *t;
	return (s == t ? t = (s = buf) + fread(buf, 1, IN_LEN, stdin), (s == t ? -1 : *s++) : *s++);
}
template<class T>
inline void read(T &x) {
	static bool iosig;
	static char c;
	for (iosig = false, c = read(); !isdigit(c); c = read()) {
		if (c == '-') iosig = true;
		if (c == -1) return;
	}
	for (x = 0; isdigit(c); c = read()) x = ((x + (x << 2)) << 1) + (c ^ '0');
	if (iosig) x = -x;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *ooh = obuf;
inline void print(char c) {
	if (ooh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), ooh = obuf;
	*ooh++ = c;
}
template<class T>
inline void print(T x) {
	static int buf[30], cnt;
	if (x == 0) print('0');
	else {
		if (x < 0) print('-'), x = -x;
		for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
		while (cnt) print((char)buf[cnt--]);
	}
}
inline void flush() { fwrite(obuf, 1, ooh - obuf, stdout); }
const double pi = acos(-1);
const int N = 1<<18;
int n, m, p;
struct cp{
	double a, b;
	inline cp operator +(const cp &rhs)const{ return (cp){a+rhs.a, b+rhs.b};}
	inline cp operator -(const cp &rhs)const{ return (cp){a-rhs.a, b-rhs.b};}
	inline cp operator *(const cp &rhs)const{ return (cp){a*rhs.a-b*rhs.b, a*rhs.b+b*rhs.a};}
	inline cp operator *(const double rhs)const{ return (cp){a*rhs, b*rhs};}
}a[N], b[N];
inline void FFT(cp *f, int g){
	for(int i=0, j=0; i<p; ++i){
		if(i>j) swap(f[i], f[j]);
		for(int k=p>>1; (j^=k)<k; k>>=1);
	}
	for(int i=2; i<=p; i<<=1){
		int tmp=i>>1;
		register cp w0=(cp){cos(2*pi/i), g*sin(2*pi/i)};
		for(int j=0; j<p; j+=i){
			register cp w=(cp){1, 0};
			for(int k=j; k<j+tmp; ++k){
				register cp t=f[k+tmp]*w;
				f[k+tmp]=f[k]-t;
				f[k]=f[k]+t;
				w=w*w0;
			}
		}
	}
}
int main() {
	read(n), read(m), ++n, ++m;
	for(int i=0; i<n; ++i){static int tmp; read(tmp); a[i].a=tmp;}
	for(int i=0; i<m; ++i){static int tmp; read(tmp); b[i].a=tmp;}
	for(p=1; p<n+m-1; p<<=1);
	FFT(a, 1), FFT(b, 1);
	for(int i=0; i<p; ++i) a[i]=a[i]*b[i];
	FFT(a, -1);
	for(int i=0; i<n+m-1; ++i) print((int)(a[i].a/p+0.5)), print(' ');
	return flush(), 0;
}

评测结果