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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#94129#6137. Sub-cycle GraphCSU2023#AC ✓137ms7240kbC++142.2kb2023-04-05 14:12:262023-04-05 14:12:28

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-04-05 14:12:28]
  • 评测
  • 测评结果:AC
  • 用时:137ms
  • 内存:7240kb
  • [2023-04-05 14:12:26]
  • 提交

answer

#include <bits/stdc++.h>

template <class T>
inline void read(T &res)
{
	char ch; bool flag = false; res = 0;
	while (ch = getchar(), !isdigit(ch) && ch != '-');
	ch == '-' ? flag = true : res = ch ^ 48;
	while (ch = getchar(), isdigit(ch))
		res = res * 10 + ch - 48;
	flag ? res = -res : 0; 
}

template <class T>
inline void put(T x)
{
	if (x > 9)
		put(x / 10);
	putchar(x % 10 + 48);
}

template <class T>
inline void CkMin(T &x, T y) {x > y ? x = y : 0;}

typedef long long ll;
const int N = 2e5 + 5;
const int lim = 2e5;
const int mod = 1e9 + 7;
int T_data, n; ll m;
int a[N], b[N], ex[N], iex[N], fac[N], ifac[N]; 

inline int quick_pow(int x, int k)
{
	int res = 1;
	while (k)
	{
		if (k & 1)
			res = 1ll * res * x % mod;
		x = 1ll * x * x % mod;
		k >>= 1;
	}
	return res;
}

inline int C(int n, int m)
{
	return 1ll * fac[n] * ifac[n - m] % mod * ifac[m] % mod;
}

inline void add(int &x, int y)
{
	x += y;
	x >= mod ? x -= mod : 0;
}

inline void dec(int &x, int y)
{
	x -= y;
	x < 0 ? x += mod : 0;
}

int main()
{
	fac[0] = ex[0] = 1;
	for (int i = 1; i <= lim; ++i)
		add(ex[i] = ex[i - 1], ex[i - 1]);
	iex[lim] = quick_pow(ex[lim], mod - 2);
	for (int i = lim; i >= 1; --i)
		add(iex[i - 1] = iex[i], iex[i]);
	for (int i = 1; i <= lim; ++i)
		fac[i] = 1ll * i * fac[i - 1] % mod;
	ifac[lim] = quick_pow(fac[lim], mod - 2);
	for (int i = lim; i >= 1; --i)
		ifac[i - 1] = 1ll * ifac[i] * i % mod;
	
	read(T_data);
	while (T_data--)
	{
		read(n); read(m);
		if (m > n)
			puts("0");
		else if (m == n)
		{
			int ans = 1;
			for (int i = 1; i < n; ++i)
				ans = 1ll * ans * i % mod;
			ans = 1ll * ans * iex[1] % mod;
			put(ans), putchar('\n');
		}
		else 
		{
			int k = n - m;
			for (int i = 0; i <= k; ++i)
			{
				b[i] = 1ll * ex[k - i] * C(k, i) % mod;
				if ((i & 1) && b[i])
					b[i] = mod - b[i];
			}
			for (int i = k; i <= n; ++i)
				a[i] = b[i - k];
			int ans = 0;
			for (int i = k; i <= n; ++i)
				ans = (1ll * C(n - i + k - 1, k - 1) * a[i] + ans) % mod;
			ans = 1ll * fac[n] * ifac[k] % mod * iex[k] % mod * ans % mod; 
			put(ans), putchar('\n');
			for (int i = 0; i <= n; ++i)
				a[i] = b[i] = 0;
		}
	}
	
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 6ms
memory: 6440kb

input:

3
4 2
4 3
5 3

output:

15
12
90

result:

ok 3 number(s): "15 12 90"

Test #2:

score: 0
Accepted
time: 137ms
memory: 7240kb

input:

17446
3 0
3 1
3 2
3 3
4 0
4 1
4 2
4 3
4 4
5 0
5 1
5 2
5 3
5 4
5 5
6 0
6 1
6 2
6 3
6 4
6 5
6 6
7 0
7 1
7 2
7 3
7 4
7 5
7 6
7 7
8 0
8 1
8 2
8 3
8 4
8 5
8 6
8 7
8 8
9 0
9 1
9 2
9 3
9 4
9 5
9 6
9 7
9 8
9 9
10 0
10 1
10 2
10 3
10 4
10 5
10 6
10 7
10 8
10 9
10 10
11 0
11 1
11 2
11 3
11 4
11 5
11 6
11 7
11...

output:

1
3
3
1
1
6
15
12
3
1
10
45
90
60
12
1
15
105
375
630
360
60
1
21
210
1155
3465
5040
2520
360
1
28
378
2940
13545
35280
45360
20160
2520
1
36
630
6552
42525
170100
393120
453600
181440
20160
1
45
990
13230
114345
643545
2286900
4762800
4989600
1814400
181440
1
55
1485
24750
273735
2047815
10239075
3...

result:

ok 17446 numbers