2019ICPC上海网络赛

D - Counting Sequences I

题意：

给 (n) , (1 le n le 3000)

问满足 (sum_{i=1}^na_i = prod_{i=1}^na_i)

的方案数，对 (1e9+7) 取模

思路：

可以暴力打表，用排列公式

[dfrac {N!} {a_1!a_2!...a_m!} ]

加上发现的一些剪枝，可以在一分钟左右算出全部答案

#include<cstdio>
#include<iostream>
using namespace std;

const int mod = 1e9 + 7;
const int N = 3005;
typedef long long ll;

ll fac[N], inv[N];
int a[N], n;
ll ans;

ll qpow(ll a, ll b)
{
	ll ans = 1;
	while (b)
	{
		if (b & 1) ans = ans * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return ans;
}

void dfs(int pos, int u, int add, int pro)
{
	if (pos > n)
	{
		if (add == pro)
		{
			ll tmp = fac[n], cnt = 1;
			for (int i = 2; i <= n; i++)
				if (a[i] == a[i - 1]) cnt++;
				else
				{
					tmp = tmp * inv[cnt] % mod;
					cnt = 1;
				}
			tmp = tmp * inv[cnt] % mod;
			ans = (ans + tmp) % mod;
		}
		return;
	}
	if (pro > 2*n) return;
	if (u == 1 and (n - pos + 1) + add != pro) return;
	if ((n - pos + 1) + add < pro) return;
	for (int i = u; i >= 1; i--)
	{
		a[pos] = i;
		dfs(pos + 1, i, add + i, pro * i);
		//a[pos] = 0;
	}
}
int main()
{
	//freopen("D:\1.txt", "w", stdout);
	fac[0] = 1;
	for (int i = 1; i < N; i++)
		fac[i] = fac[i - 1] * i % mod;

	inv[N - 1] = qpow(fac[N - 1], mod - 2);//阶乘的逆元
	for (int i = N - 2; i >= 0; i--)
		inv[i] = inv[i + 1] * (i + 1) % mod; 

	
	putchar('{');
	for (n = 2; n <= 3000; n++)
	{	
		ans = 0;
		dfs(1, n, 0, 1);
		printf("%lld", ans);
		if (n != 3000)
			putchar(',');
		else
			putchar('}');
	}

	return 0;
}

F - Rhyme scheme

题目大意：

用字符表示集合的划分

给定 (n) 问第 (k) 小的划分

思路：

如图，所有的叶子节点从左到右就是 $n = 4 $ 的时候的所有集合划分

设状态表示 (f[n][i][j]) 表示的是总层数 (n) , 目前层数 (i) ，且由根节点到它的路径上最大的字母为 (j) 的点的集合 ,集合的属性是该节点下叶子节点的数目

显然当 (i == n) 时，(f[n][i][j]=1) ，即到了最后一层，都是叶子节点。

否则如上图蓝色标注的 (B) 节点，若 最大字母不变则有 (j) 棵相同的子树，否则加上一颗增加了的子树。

(f[n][i][j] = f[n][i+1][j]*j + f[n][i+1][j+1])

然后就可以直接用这些信息去找答案了，类似找 (k) 大这样。

需要注意 (B_{26}) 爆了 $long long $

#include<bits/stdc++.h>
using namespace std;

int T, n;
__int128 k;
__int128 f[30][30][30];
void read(__int128& x) {
	x = 0;
	int f = 1;
	char ch = getchar();
	while (!(ch >= '0' && ch <= '9')) ch = getchar();
	x = x * 10 + ch - '0';
	while ((ch = getchar()) >= '0' && ch <= '9')
		x = x * 10 + ch - '0';
	x *= f;
}

void init() {
	for (int n = 1; n <= 26; n++)
		for (int i = n; i >= 1; i--)
			if (i == n) {
				for (int j = 1; j <= i; j++) f[n][i][j] = 1ll;
			}
			else {
				for (int j = 1; j <= i; j++) f[n][i][j] = f[n][i + 1][j] * j + f[n][i + 1][j + 1];
			}
}

int main() {
	init();
	scanf("%d", &T); int c = 0;
	while (T--) {
		scanf("%d", &n);
		read(k);

		printf("Case #%d: ", ++c);
		int now = 0, j;
		for (int i = 1; i <= n; i++) {
			for (j = 1; j <= now; j++) {
				if (k <= f[n][i][now]){
					break;
				}
				k -= f[n][i][now];
			}
			putchar('A' + j - 1);
			now = max(now, j);
		}
		puts("");
	}
}

C - Triple

题意

给三个数组 (A,B,C) 问有多少个 ((i,j,k)) 使得

(A_i,B_j,C_k) 中较小的两个数的和大于等于最大的数

(1 le T le 100)

(1 le A_i,B_i,C_i,n le 100,000)

There are at most (20) test cases with (N>1000)

思路

用容斥思想，所有不和法的方案就是 较小的两数相加小于第三个数的方案。

处理出 所有的 (A_i + B_j) ，然后对于每一个 (C_k) ,只要加上所有小于 (C_k) 的方案数就可以

这里小数据用暴力，大数据用 多项式乘法

/*
 * @Author: zhl
 * @Date: 2020-11-09 15:23:52
 */
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i = a;i <= b;i++)
#define mem(a,b) memset((a),(b),sizeof(a))

using namespace std;

typedef long long ll;
const double pi = acos(-1.0);
const int N = 6e5 + 10;

struct cp {
	double x, y;
	cp() {}
	cp(double _x, double _y) {
		x = _x; y = _y;
	}
	cp operator + (cp b) {
		return cp(x + b.x, y + b.y);
	}
	cp operator -(cp b) {
		return cp(x - b.x, y - b.y);
	}
	cp operator *(cp b) {
		return cp(x * b.x - y * b.y, x * b.y + y * b.x);
	}
};
int rev[N];
int bit = 0;
int lim;
void FFT(cp* a, int inv) {

	for (int i = 0; i < lim; i++) {
		if (i < rev[i]) {
			swap(a[i], a[rev[i]]);
		}
	}

	for (int mid = 1; mid < lim; mid <<= 1) {
		cp temp(cos(pi / mid), inv * sin(pi / mid));
		for (int i = 0; i < lim; i += mid * 2) {
			cp omega(1, 0);
			for (int j = 0; j < mid; j++, omega = omega * temp) {
				cp x = a[i + j], y = omega * a[i + j + mid];
				a[i + j] = x + y, a[i + j + mid] = x - y;
			}
		}
	}
}


int T, n, A[N], B[N], C[N], tA[N], tB[N], tC[N];
cp x[N], y[N];
ll sum[N];

ll solve_small(int* a, int* b, int* c) {
	for (int i = 0; i <= c[n - 1]; i++)sum[i] = 0;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			sum[a[i] + b[j]]++;
		}
	}
	ll ans = 0;
	for (int i = 1; i <= c[n - 1]; i++)sum[i] += sum[i - 1];
	for (int i = 0; i < n; i++) ans += sum[c[i] - 1];
	return ans;
}

ll solve_big(int* a, int* b, int* c) {

	for (int i = 0; i <= lim; i++)x[i] = cp(a[i], 0), y[i] = cp(b[i], 0);

	FFT(x, 1); FFT(y, 1);
	for (int i = 0; i <= lim; i++)x[i] = x[i] * y[i];
	FFT(x, -1);

	mem(sum, 0);
	ll ans = 0;
	for (int i = 0; i <= c[n - 1]; i++)sum[i] = signed(x[i].x / lim + 0.5);
	for (int i = 1; i <= c[n - 1]; i++)sum[i] += sum[i - 1];
	for (int i = 0; i < n; i++) ans += sum[c[i] - 1];
	return ans;
}

int main() {
	scanf("%d", &T); int c = 0;
	while (T--) {
		scanf("%d", &n);

		mem(tA, 0); mem(tB, 0); mem(tC, 0);
		lim = 1; bit = 0;
		while (lim <= (2 * n))lim <<= 1, bit++;
		mem(rev, 0);
		for (int i = 0; i < lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));

		for (int i = 0; i < n; i++)scanf("%d", A + i), tA[A[i]]++; sort(A, A + n);
		for (int i = 0; i < n; i++)scanf("%d", B + i), tB[B[i]]++; sort(B, B + n);
		for (int i = 0; i < n; i++)scanf("%d", C + i), tC[C[i]]++; sort(C, C + n);

		printf("Case #%d: ", ++c);
		if (n <= 1000) {
			printf("%lld
", 1ll * n * n * n - solve_small(A, B, C) - solve_small(A, C, B) - solve_small(B, C, A));
		}
		else {
			printf("%lld
", 1ll * n * n * n - solve_big(tA, tB, C) - solve_big(tA, tC, B) - solve_big(tB, tC, A));;
		}
	}
}