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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#269757#7754. Rolling For DaysFISHER_WA 1ms5936kbC++202.4kb2023-11-29 22:44:002023-11-29 22:44:00

Judging History

你现在查看的是最新测评结果

  • [2023-11-29 22:44:00]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:5936kb
  • [2023-11-29 22:44:00]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
inline int Mod(int x) { return x + ((x >> 31) & mod); }
inline int add(int x, int y) { return Mod(x + y - mod); }
inline int sub(int x, int y) { return Mod(x - y); }
inline int mul(int x, int y) { return 1ll * x * y % mod; }
inline int sqr(int x) { return 1ll * x * x % mod; }
int qpow(int a, int b) {
	if (!b) return 1;
	int t = qpow(a, b >> 1);
	t = mul(t, t);
	if (b & 1) return mul(t, a);
	return t;
}
const int maxn = 1000, maxm = 12;
int fac[maxn + 5], ifac[maxn + 5];
void init(int n) {
	fac[0] = 1;
	for (int i = 1; i <= n; i++) fac[i] = mul(fac[i - 1], i);
	ifac[n] = qpow(fac[n], mod - 2);
	for (int i = n - 1; ~i; i--) ifac[i] = mul(ifac[i + 1], i + 1);
}
inline int C(int m, int n) { return mul(fac[m], mul(ifac[m - n], ifac[n])); }
int a[maxm + 5], b[maxm + 5];
int sa[1 << maxm], sb[1 << maxm];
int f[maxn + 5][1 << maxm], g[maxn + 5][1 << maxm];
int sf[maxn + 5][1 << maxm], sg[maxn + 5][1 << maxm], sg2[maxn + 5][1 << maxm];
int p[maxn + 5][maxn + 5];
inline void upd(int& x, int y) { x = add(x, y); }
int main() {
	int n, m, N = 0;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++) scanf("%d", &a[i]);
	for (int i = 0; i < m; i++) scanf("%d", &b[i]), N += b[i];
	int S = 1 << m;
	for (int s = 0; s < S; s++) {
		sa[s] = n;
		for (int i = 0; i < m; i++)
			if ((s >> i) & 1) sa[s] -= a[i], sb[s] += b[i];
	}
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++) p[i][j] = add(p[i][j - 1], mul(n - j + 1, qpow(i - j + 1, mod - 2))); 
	init(n);
	sg[0][0] = ifac[n];
	for (int i = 1; i <= N; i++)
		for (int s = 0; s < S; s++) {
			if (i < sb[s]) continue;
			for (int x = 0; x < m; x++)
				if ((s >> x) & 1) {
					int fs = s ^ (1 << x);
					upd(f[i][s], mul(add(sub(mul(sg[i - 1][fs], p[sa[fs] + sb[fs]][i]), sg2[i - 1][fs]), sf[i - 1][fs]), mul(C(i - 1 - sb[fs], b[x] - 1), fac[sa[fs] - i + sb[fs]])));
					upd(g[i][s], mul(sg[i - 1][fs], mul(C(i - 1 - sb[fs], b[x] - 1), fac[sa[fs] - i + sb[fs]])));
				}
			sf[i][s] = add(sf[i - 1][s], mul(f[i][s], ifac[sa[s] - i + sb[s]]));
			sg[i][s] = add(sg[i - 1][s], mul(g[i][s], ifac[sa[s] - i + sb[s]]));
			sg2[i][s] = add(sg2[i - 1][s], mul(g[i][s], mul(ifac[sa[s] - i + sb[s]], p[sa[s] + sb[s]][i])));
		}
	int ans = f[N][S - 1];
	for (int i = 0; i < m; i++) ans = mul(ans, mul(fac[a[i]], ifac[a[i] - b[i]]));
	printf("%d", ans);
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 5628kb

input:

2 2
1 1
1 1

output:

2

result:

ok answer is '2'

Test #2:

score: 0
Accepted
time: 1ms
memory: 5936kb

input:

4 2
2 2
2 1

output:

582309210

result:

ok answer is '582309210'

Test #3:

score: -100
Wrong Answer
time: 0ms
memory: 5628kb

input:

5 5
1 1 1 1 1
0 0 0 0 1

output:

0

result:

wrong answer expected '5', found '0'