CF1548C The Three Little Pigs

传送


这题的题解太妙了，虽然是dp，但从头到尾没一步是在我意料之内的……


一句话题意：给一个(n)，(q)组询问，每次让求(sum_{i=1}^{n} C_{3i}^x extrm{mod} 10^9+7).（(1 leqslant n leqslant 10^6, 1 leqslant q leqslant 2 * 10^5)）


咋dp的呢？

令(dp[x][m]=sumlimits_{i=0}^{n - 1} C_{3i+m}^x(m = 0,1,2)).那么(ans[x] = dp[x][0]+C_{3n}^x).

而(sumlimits_{m=0}^2dp[x][m]=sumlimits_{i=0}^{n-1}(C_{3i}^x+C_{3i+1}^x+C_{3i+2}^x)=sumlimits_{i=0}^{3n-1}C_i^x).因为这三项相加相当于将所有(iin[1,3n-1])都访问过了。

接下来，根据( extrm{Hockey-Stick Identity})，有(sumlimits_{i=0}^{3n-1}C_i^x = C_{3n}^{x+1}).（其实这一步不知道也行，预处理出来就好了）

于是就有(dp[x][0]+dp[x][1]+dp[x][2] = C_{3n}^{x+1} (1)).

又根据杨辉三角，能得出关系式：
(dp[x][1] = dp[x][0]+dp[x - 1][0] (2))，

(dp[x][2] = dp[x][1] + dp[x - 1][1] (3)).

最后将((1)(2)(3))联立，就能解得递推式

[egin{align*} dp[x][0] &= frac1{3}(C_{3n}^{x+1} - 2dp[x-1][0] - dp[x - 1][1]) \ dp[x][1] &= frac1{3}(C_{3n}^{x+1} + dp[x-1][0] - dp[x - 1][1]) \ dp[x][2] &= frac1{3}(C_{3n}^{x+1} + dp[x-1][0] + 2dp[x - 1][1]) end{align*}]

边界条件：(dp[0][0] = dp[0][1] = dp[0][2] = 0).

时间复杂度(O(n+q)).


太妙了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<queue>
#include<assert.h>
#include<ctime>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
#define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 3e6 + 5;
const ll mod = 1e9 + 7;
In ll read()
{
	ll ans = 0;
	char ch = getchar(), las = ' ';
	while(!isdigit(ch)) las = ch, ch = getchar();
	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
	if(las == '-') ans = -ans;
	return ans;
}
In void write(ll x)
{
	if(x < 0) x = -x, putchar('-');
	if(x >= 10) write(x / 10);
	putchar(x % 10 + '0');
}

In ll ADD(ll a, ll b) {return a + b < mod ? a + b : a + b - mod;}
In ll quickpow(ll a, ll b)
{
	ll ret = 1;
	for(; b; b >>= 1, a = a * a % mod)
		if(b & 1) ret = ret * a % mod;
	return ret;
}

int n, m, Q;

ll f[maxn], inv[maxn], dp[maxn][3], inv3;
In ll C(int n, int m) {return f[n] * inv[m] % mod * inv[n - m] % mod;}
In void init()
{
	inv3 = quickpow(3, mod - 2);
	f[0] = inv[0] = 1;
	for(int i = 1; i <= m; ++i) f[i] = f[i - 1] * i % mod;
	inv[m] = quickpow(f[m], mod - 2);
	for(int i = m - 1; i; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
	dp[0][0] = dp[0][1] = dp[0][2] = n;
	for(int i = 1; i <= m; ++i)
	{
		ll c = C(m, i + 1);
		dp[i][0] = ADD(ADD(c, mod - dp[i - 1][0] * 2 % mod), mod - dp[i - 1][1]) * inv3 % mod;
		dp[i][1] = ADD(ADD(c, dp[i - 1][0]), mod - dp[i - 1][1]) * inv3 % mod;
		dp[i][2] = ADD(ADD(c, dp[i - 1][0]), dp[i - 1][1] * 2 % mod) * inv3 % mod;
	}
}

int main()
{
	n = read(), Q = read(); m = n * 3;
	init();
	for(int i = 1; i <= Q; ++i)
	{
		int x = read();
		write(ADD(dp[x][0], C(m, x))), enter;
	}
	return 0;
}