提交记录 5175


用户 题目 状态 得分 用时 内存 语言 代码长度
HowNegative 1002i. 【模板题】多项式乘法 Accepted 100 29.702 ms 6048 KB C++ 2.04 KB
提交时间 评测时间
2018-08-10 13:12:07 2020-08-01 00:13:30
#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 int N = 1<<18, P = 998244353;
int n, m, p, a[N], b[N], w[N];
inline ll Pow(ll x, register int y=P-2) {
	ll ass=1;
	for(; y; y>>=1, x=x*x%P) if(y&1) ass=ass*x%P;
	return ass;
}
inline int Mod(int x){ return x<P?x:x-P;}
inline void NTT(int *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=1; i<p; i<<=1){
		w[0]=1;
		for(int j=1, w0=(g==1?Pow(3, (P-1)/i/2):Pow(Pow(3, (P-1)/i/2))); j<i; ++j) w[j]=(ll)w[j-1]*w0%P;
		for(int j=0; j<p; j+=i<<1){
			for(int k=j; k<j+i; ++k){
				int t=(ll)f[k+i]*w[k-j]%P;
				f[k+i]=Mod(f[k]-t+P);
				f[k]=Mod(f[k]+t);
			}
		}
	}
}
int main() {
	read(n), read(m), ++n, ++m;
	for(int i=0; i<n; ++i) read(a[i]);
	for(int i=0; i<m; ++i) read(b[i]);
	for(p=1; p<n+m-1; p<<=1);
	NTT(a, 1), NTT(b, 1);
	for(int i=0; i<p; ++i) a[i]=(ll)a[i]*b[i]%P;
	NTT(a, -1);
	for(int i=0, I=Pow(p); i<n+m-1; ++i) print((ll)a[i]*I%P), print(' ');
	return flush(), 0;
}

CompilationN/AN/ACompile OKScore: N/A

Subtask #1 Testcase #18.88 us36 KBAcceptedScore: 0

Subtask #1 Testcase #228.932 ms5 MB + 764 KBAcceptedScore: 100

Subtask #1 Testcase #312.486 ms2 MB + 24 KBAcceptedScore: 0

Subtask #1 Testcase #412.58 ms2 MB + 4 KBAcceptedScore: 0

Subtask #1 Testcase #59.69 us36 KBAcceptedScore: 0

Subtask #1 Testcase #68.83 us36 KBAcceptedScore: 0

Subtask #1 Testcase #78.34 us36 KBAcceptedScore: 0

Subtask #1 Testcase #827.839 ms5 MB + 160 KBAcceptedScore: 0

Subtask #1 Testcase #927.819 ms5 MB + 160 KBAcceptedScore: 0

Subtask #1 Testcase #1026.758 ms4 MB + 580 KBAcceptedScore: 0

Subtask #1 Testcase #1129.702 ms5 MB + 928 KBAcceptedScore: 0

Subtask #1 Testcase #1225.216 ms3 MB + 684 KBAcceptedScore: 0

Subtask #1 Testcase #137.57 us36 KBAcceptedScore: 0


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