【ACM-ICPC 2018 沈阳赛区网络预赛】不太敢自称官方的出题人题解

A. Gudako and Ritsuka

链接

by Yuki & Asm.Def

期望难度：Hard-

考虑从后往前进行博弈动态规划，在这一过程中维护所有的先手必胜区间。区间不妨采用左开右闭，方便转移。

考虑一次转移，如果当前Servant的后一个位置属于对手，则当前Servant的必胜区间可以通过将后一个Servant的每个必败区间的左端点+1、右端点+x得到；如果后一个位置属于自己，则可以通过将后一个Servant的必胜区间做同样的操作得到。不妨分别对必胜区间左右端点维护一个偏移量，需要从对手进行转移时只需修改偏移量后交换左右端点的集合，然后再在左端点的集合里插入一个0即可。

需要注意的是，这样得到的必胜区间会有重叠，可能导致对下一个对手的必胜区间的统计出错。考虑到每次转移时所有的同类区间的长度的变化量都相同，可以分别用两个优先队列维护这两类区间，每次转移后暴力合并重叠的区间即可。

复杂度\(O((A+B) \log(A+B))\)

//Asm.Def
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 100005;
int A, B, M;
LL N, x;
bool p[maxn], beg_A;

void init()
{
	scanf("%lld%lld%d%d", &N, &x, &A, &B);
	assert(A+B <= 100000);
	assert(A+B >= N);
	assert(A >= 1 && B >= 1);
	M = A+B;
	for(int i = 1;i <= M;++i) p[i] = false;
	int a;
	beg_A = false;
	for(int i = 0;i < A;++i)
	{
		scanf("%d", &a);
		p[a] = 1;
		if(a == 1) beg_A = true;
	}
}

int a[maxn], len;

struct Seg
{
	int l, r;
	LL len;
};
bool operator < (const Seg &a, const Seg &b)
{
	return a.len > b.len;
}

void work()
{
	len = 0;
	int cnt = 0;
	for(int i = 1;i <= M;++i)
	{
		++cnt;
		if(i == M || p[i+1] != p[i])
		{
			a[len++] = cnt;
			cnt = 0;
		}
	}
	bool ans = false;

	static LL loc[maxn];
	static int nxt[maxn], lst[maxn];
	static bool sel[maxn];
	priority_queue<Seg> Q[2];

	bool L = 0, R = 1;
	LL offs[2] = {0};

	loc[len] = 0;
	nxt[len] = lst[len] = len;//用链表记录当前区间端点

	sel[len] = false;
	for(int i = len-1;i >= 0;--i)
	{
		//区间整体右移
		offs[L] += a[i] * x;
		offs[R] += a[i];
		L ^= 1, R ^= 1;
		
		//插入新增的左端点
		loc[i] = -offs[L];
		nxt[i] = nxt[len];
		lst[nxt[i]] = i;
		lst[i] = len;
		nxt[len] = i;
		sel[i] = true;
		Q[L].push( (Seg){i, nxt[i], loc[nxt[i]]-loc[i]} );

		//合并区间
		LL t = offs[R] - offs[L];//R(k)+offs[R] < L(k+1)+offs[L]
		Seg tmp;
		while(!Q[R].empty() && ((tmp = Q[R].top()).len <= t || !sel[tmp.l] || !sel[tmp.r]) )
		{
			Q[R].pop();
			if(sel[tmp.l] && sel[tmp.r])//合并一对跨立的"(](]",
			{
				sel[tmp.l] = sel[tmp.r] = false;
				nxt[lst[tmp.l]] = nxt[tmp.r];
				lst[nxt[tmp.r]] = lst[tmp.l];
				Q[L].push( (Seg){lst[tmp.l], nxt[tmp.r], loc[nxt[tmp.r]]-loc[lst[tmp.l]]} );
			}
			//否则为已被合并的区间，无需处理
		}

	}

	int it = nxt[len];
	ans = false;
	while(it != len && loc[it]+offs[L] < N)
	{
		if(nxt[it] == N || loc[nxt[it]]+offs[R] >= N)
		{
			ans = true;
			break;
		}
		it = nxt[it];
		if(it != len) it = nxt[it];
		if(it == 0) break;
	}
	puts((ans ^ beg_A) ? "Ritsuka" : "Gudako");
}

int main()
{
	clock_t beg = clock();
	int T;
	scanf("%d", &T);
	while(T--)
	{
		init();
		work();
	}
	//printf("%.3f sec\n", double(clock()-beg) / CLOCKS_PER_SEC);
	return 0;
}

B. Call of Accepted

链接

期望难度：Medium

表达式求值问题可以将中缀表达式转换为后缀表达式，其中转换步骤使用调度场算法 与后缀表达式的求值 均在维基百科中有详细的介绍。

根据d运算的描述，\(x\ {\rm d}\ y\)本质上是一个定义在整数集的幂集上的运算，即\(对于任意 S_1, S_2 \in \mathbb{2^Z}且 \min(S_1) \geq 0 且 \min(S_2)\geq 1\)，定义\(S_1\ {\rm d}\ S_2 = \{x\in \mathbb{Z} \mid \exists a\in S_1, \exists b\in S_2, a\leq x \leq ab\}\). 则对于任意一次有定义的\({\rm d}\)运算，都有$\min(S_1\ {\rm d}\ S_2) = \min(S_1) $, \(\max(S_1\ {\rm d}\ S_2)=\max(S_1)\cdot \max(S_2)\). 其中\(\min(S)\)和\(\max(S)\)分别表示集合\(S\)中的最小值和最大值。

再将其余三种运算扩展到\(\mathbb{2^Z}\)上，可得

\(\min(S_1 + S_2)=\min(S_1)+\min(S_2), \max(S_1+S_2)=\max(S_1)+\max(S_2)\)

\(\min(S_1-S_2) = \min(S_1)-\max(S_2), \max(S_1-S_2) = \max(S_1) - \min(S_2)\)

\(\min(S_1*S_2)=\min\{\min(S_1)*\min(S_2),\ \min(S_1)*\max(S_2),\ \max(S_1) * \min(S_2),\ \max(S_1)*\max(S_2)\}\)

\(\max(S_1*S_2)=\max\{\min(S_1)*\min(S_2),\ \min(S_1)*\max(S_2),\ \max(S_1) * \min(S_2),\ \max(S_1)*\max(S_2)\}\)

由于我们只关注最大和最小的结果，所以可以直接用二元组\(<\min(S), \max(S)>\)来表示一个子表达式的运算结果，按照上述扩展定义进行运算即可。

p.s.虽然从实际意义来看d运算不满足结合律，但如果只考虑二元组\(<\min(S), \max(S)>\)的话，有\(<\min(S_1\ {\rm d}\ S_2\ {\rm d}\ \cdots\ {\rm d}\ S_n),\ \max(S_1\ {\rm d}\ S_2\ {\rm d}\ \cdots\ {\rm d}\ S_n)>\ =\ <\min(S_1),\ \max(S_1) \max(S_2) \cdots \max(S_n)>\)，是无需考虑\({\rm d}\)运算的结合顺序的。

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

struct Interval
{
	int L, R;
	Interval(){}
	Interval(int a, int b) : L(a), R(b) {}
	void Print(){printf("[%d,%d]\n", L, R);}
};
Interval operator + (const Interval &a, const Interval &b)
{
	return Interval(a.L + b.L, a.R + b.R);
}
Interval operator - (const Interval &a, const Interval &b)
{
	return Interval(a.L - b.R, a.R - b.L);
}
Interval operator * (const Interval &a, const Interval &b)
{
	int mn = min(min(a.L * b.L, a.L * b.R), min(a.R * b.L, a.R * b.R));
	int mx = max(max(a.L * b.L, a.L * b.R), max(a.R * b.L, a.R * b.R));
	return Interval(mn, mx);
}
Interval f(const Interval &a, const Interval &b)
{
	assert(a.L >= 0 && b.L >= 1);
	return Interval(a.L, a.R * b.R);
}
int RPNLen, RPNNumLen;
Interval RPNNum[105];
char s[105], RPN[105];

inline int precedence(char ope) {
	if (ope == '+') return 1;
	if (ope == '-') return 1;
	if (ope == '*') return 2;
	if (ope == '/') return 2;
	if (ope == 'd') return 3;
	return 0;
}

void expressionToRPN() {
	int stackLen = 0, x = 0;
	char stack[105];
	RPNLen = 0;
	RPNNumLen = 0;
	for (int i = 0; s[i] != '\0'; i++) {
		if (s[i] >= '0' && s[i] <= '9') {
			if (i == 0 || s[i-1] < '0' || s[i-1] > '9') {
				RPNLen++;
				RPN[RPNLen] = 'N';
				x = s[i] - '0';
			} else {
				x = x * 10 - '0' + s[i];
			}
			if(s[i+1] < '0' || s[i+1] > '9')
				RPNNum[++RPNNumLen] = Interval(x, x);
		} else if (s[i] == '(') {
			stackLen++;
			stack[stackLen] = s[i];
		} else if (s[i] == ')') {
			while (stack[stackLen] != '(') {
				RPNLen++;
				RPN[RPNLen] = stack[stackLen];
				stackLen--;
			}
			stackLen--;
		} else {
			while (stackLen > 0 && precedence(s[i]) <= precedence(stack[stackLen])) {
				RPNLen++;
				RPN[RPNLen] = stack[stackLen];
				stackLen--;
			}
			stackLen++;
			stack[stackLen] = s[i];
		}
	}
	while (stackLen > 0) {
		RPNLen++;
		RPN[RPNLen] = stack[stackLen];
		stackLen--;
	}
}

Interval CalcRPN() {
	int RPNNumCnt = 0, stackLen = 0;
	Interval stack[105];
	for (int i = 1; i <= RPNLen; i++) {
		if (RPN[i] == 'N') {
			RPNNumCnt++;
			stackLen++;
			stack[stackLen] = RPNNum[RPNNumCnt];
		} else {
			Interval b = stack[stackLen];
			stackLen--;
			Interval a = stack[stackLen];
			stackLen--;
			Interval result;
			//printf("%c\n", RPN[i]);
			//a.Print(), b.Print();
			if(RPN[i] == '+') result = a + b;
			if(RPN[i] == '-') result = a - b;
			if(RPN[i] == '*') result = a * b;
			if(RPN[i] == 'd') result = f(a, b);
			//result.Print();
			stackLen++;
			stack[stackLen] = result;
		}
	}
	return stack[stackLen];
}

int main() {
	while (scanf("%s", s) == 1) {
		expressionToRPN();
		Interval ans = CalcRPN();
		printf("%d %d\n", ans.L, ans.R);
	}
	return 0;
}

附对拍用的std.py

# Haizs
import re
class Interval:
    def __init__(self, l, r):
        self.l = l
        self.r = r

    def __str__(self):
        return "%d %d" % (self.l, self.r)

    def __add__(self, b):
        return Interval(self.l + b.l, self.r + b.r)

    def __sub__(self, b):
        return Interval(self.l - b.r, self.r - b.l)

    def __mul__(self, b):
        mn = min(min(self.l*b.l, self.r*b.r), min(self.l*b.r, self.r*b.l))
        mx = max(max(self.l*b.l, self.r*b.r), max(self.l*b.r, self.r*b.l))
        return Interval(mn, mx)

    def __pow__(self, b):
        return Interval(self.l, self.r * b.r)

a = input()
while a:
    a = a.replace("d", "**")
    b = re.sub(r"(\d+)", r"Interval(\1,\1)", a)
    # print(b)
    print(eval(b))
    try:
        a = input()
    except:
        break

by catsworld & Asm.Def

C. Convex Hull

链接

期望难度：Medium

Solution 1

不考虑外层循环的情况，那么答案显然是：

\[ans = \sum_{i=1}^{\sqrt{x}} \mu(i) * \frac{1}{6}(\frac{x}{i^2}+1) * (\frac{x}{i^2}) * (2(\frac{x}{i^2}))+1) * i^4 \]

在加了外层循环的情况下，考虑计算\(\mu(i) * i^4\)的系数
令\(sum(i)=\sum_{j=1}^i j^2\)
对于每一个\(\mu(i) * i^4\)，它在全部的答案中出现次数为\(n-i^2+1\)次，可以推出系数为\(\sum_{j=i^2}^n sum(\frac{j}{i^2})\)
令\(Max=\frac{n}{i^2}\)
考虑sum括号中的取值，可以发现，一定有\(i*i个1，i*i个2...i*i个Max-1，(n-i*i+11-(Max-1)*i*i)个Max\)
所以，最终的系数为

\[\begin{aligned} &i*i*(\sum_{j=1}^{Max-1}sum(j)) + (n-i*i+1-(Max-1)*i*i)*sum(Max)\\ = &Max*Max*(Max+1)*(Max-1)/12+(n-i*i+1-(Max-1)*i*i)*sum(Max) \end{aligned} \]

考虑到模数很大，计算过程中需要用类似于分治乘法的思路或int128。

By WYJ2015

Solution2

答案可转化为

\[\sum_{i=1}^n gay(i) \cdot (n+1-i) \mod p \]

在\(\sum\limits_{i=1}^{n} gay(i) (n+1-i)\)中，\(i^2 \cdot (n+1-i)\)被计入答案当且仅当i不含有平方因子。不妨考虑对所有i的因子进行容斥，即

\[\begin{aligned} Ans &= \sum_{x=1}^{[\sqrt{n}]} \mu(x) \sum_{k=1}^{[\frac{n}{x^2}]} (k x^2)^2 * (n+1-k x^2) \mod p\\ &= \sum_{x=1}^{[\sqrt{n}]} \mu(x) \left( (n+1)x^4 \sum_{k=1}^{[\frac{n}{x^2}]} k^2 - x^6 \sum_{k=1}^{[\frac{n}{x^2}]} k^3 \right) \mod p\\ \end{aligned} \]

其中，平方和与立方和为

\[\begin{aligned} \sum_{i=1}^n i^2 &= \frac{n(n+1)(2n+1)}{6}\\ \sum_{i=1}^n i^3 &= \left(\frac{n(n+1)}{2}\right)^2 \end{aligned} \]

直接枚举x后代入计算即可。

在计算\(x \times y \mod p\)时，由于p的最大值为\(10^{11}\)，可能会超过long long的表示范围，可以将x拆成\((a\cdot 2^{20} + b)\)，将y拆成\((c\cdot 2^{20} + d)\)，将每次乘法的运算数范围限制在\(2^{20}\)以内，可以直接用((((((a * c) << 20) + (a * d + b * c)) % mod) << 20) + b * d) % mod计算出结果。

时间复杂度\(O(\sqrt{N})\)

by Asm.Def

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
typedef long long LL;

int mu[maxn], P[maxn], pcnt;
bool not_p[maxn];
LL N, mod;

const LL lb = (1LL << 20) - 1;

inline LL mult(LL x, LL y, LL mod)
{
	LL a = x >> 20, b = x & lb;
	LL c = y >> 20, d = y & lb;
	return ( ( ( ( ( (a * c) << 20) + (a * d + b * c) ) % mod) << 20) + b * d) % mod;
}

inline LL S2(LL N)
{
	return mult(mult(N, N+1, 6*mod), (2*N+1), 6*mod) / 6;
	//return (__int128(N) * (N+1) * (2*N+1) / 6) % mod;
}

inline LL S3(LL N)
{
	LL t = mult(N, N+1, mod<<1) >> 1;
	//LL t = (__int128(N) * (N+1) / 2) % mod;
	return mult(t, t, mod);
}

void init()
{
	mu[1] = 1;
	for(int i = 2;i < maxn;++i)
	{
		if(!not_p[i]) P[pcnt++] = i, mu[i] = -1;
		for(int j = 0;j < pcnt;++j)
		{
			if(i * P[j] >= maxn) break;
			not_p[i * P[j]] = true;
			if(i % P[j] == 0)
			{
				mu[i * P[j]] = 0;
				break;
			}
			mu[i * P[j]] = -mu[i];
		}
	}
}

void work()
{
	LL ans = 0;
	for(int i = 1;LL(i) * i <= N;++i) if(mu[i])
	{
		LL t = LL(i) * i, t2 = mult(t, t, mod);
		ans = (ans + mu[i] * mult( mult(t2,N+1,mod), S2(N/t), mod) ) % mod;
		ans = (ans - mu[i] * mult(mult(t2,t,mod), S3(N/t), mod)) % mod;
	}
	printf("%lld\n", (ans + mod) % mod);
}

int main()
{
	init();

	while(~scanf("%lld%lld", &N, &mod))
	{
		work();
	}
	return 0;
}

D. Made In Heaven

链接

by XLC

期望难度：Easy

K短路模板题。由于数据均为随机生成，直接预处理出每个点到终点的最短路后A*搜索即可。

E. The Cake Is A Lie

链接

by Haizs

期望难度：Medium

二分答案，那么每次check相当于是给定一个半径的圆，然后问这个圆最多覆盖多少个点，我们可以枚举一个点，然后再枚举每个与他距离<=2r的点，就可以求出所有的相交弧，离散化之后，求出覆盖最多次的弧，就是答案了。复杂度\(O(n^2\log(n)\cdot \log(30000))\)。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 605;
const double eps = 1e-6;
int i, j, t, n, m, k, z, y, x;
double l, r, mid;
int R;
int cmp(double x)
{
    if (fabs(x) < eps) return 0;
    if (x > 0) return 1;
    return -1;
}
struct point
{
    double x, y;
    point() {}
    point(double _x, double _y)
        : x(_x), y(_y) {}
    void input()
    {
        scanf("%lf%lf", &x, &y);
    }
    friend point operator+(const point &a, const point &b)
    {
        return point(a.x + b.x, a.y + b.y);
    }
    friend point operator-(const point &a, const point &b)
    {
        return point(a.x - b.x, a.y - b.y);
    }
    double norm()
    {
        return sqrt(x * x + y * y);
    }
    double angl()
    {
        return atan2(y, x);
    }
} poi[maxn];
double dis[maxn][maxn], ang[maxn][maxn];
pair<double, int> pdi[maxn];
#define mp(a, b) make_pair(a, b)
bool check(double r)
{
    int i, j, t, ans = 1, k;
    double d;
    for (i = 1; i <= n; i++)
    {
        t = 0;
        for (j = 1; j <= n; j++)
            if (i != j)
            {
                if (cmp(dis[i][j] - 2.0 * r) > 0) continue;
                d = acos(dis[i][j] / (2.0 * r));
                pdi[++t] = mp(ang[i][j] - d, 1);
                pdi[++t] = mp(ang[i][j] + d, -1);
            }
        sort(pdi + 1, pdi + t + 1);
        k = 1;
        for (j = 1; j <= t; j++)
        {
            k += pdi[j].second;
            ans = max(ans, k);
        }
    }
    return ans >= m;
}
int main()
{
    // cout << time(NULL) << endl;
    // freopen("tcial.in", "r", stdin);
    // freopen("tcial.out", "w", stdout);
    int T, I;
    scanf("%d", &T);
    for (I = 1; I <= T; I++)
    {
        scanf("%d%d", &n, &m);
        for (i = 1; i <= n; i++) poi[i].input();
        scanf("%d", &R);
        if (m > n)
        {
            printf("The cake is a lie.\n");
            continue;
        }
        for (i = 1; i <= n; i++)
            for (j = 1; j <= n; j++)
                dis[i][j] = (poi[j] - poi[i]).norm(), ang[i][j] = (poi[j] - poi[i]).angl();
        l = R;
        r = 30000;
        while (r - l > eps)
        {
            mid = (l + r) / 2;
            if (check(mid - R))
                r = mid;
            else
                l = mid;
        }
        printf("%.6f\n", r);
    }
    // cout << time(NULL) << endl;
    return 0;
}

F. Fantastic Graph

https://nanti.jisuanke.com/t/31446](https://nanti.jisuanke.com/t/31446)

by Infi

期望难度：Easy

添加源点s,汇点t。
对于原图的边，定义流量为[0,1]，s对于N个点都连边，流量为[L,R],M个点对t都连边，流量为[L,R]。那么就变成了有源汇上下界可行流问题。根据相关方法建图即可。

G. Spare Tire

链接

by ZGH

期望难度：Easy+

观察递推方程，不难看出通项公式的形式：

\[a_n = k p^n + an^2 + bn + c \]

代入后解得\(p=1, a=1, b=1, c=-k\)
即\(a_n = n^2 + n\)

则答案为

\[\begin{aligned} &\sum_{i=1}^{n} [\gcd(i, m)=1] (i^2 + i)\\ =&\sum_{d|m \land d \leq n} \mu(d) \cdot \sum_{t=1}^{[\frac{n}{d}]} ((td)^2 + td)\\ =&\sum_{d|m \land d \leq n} \mu(d) \cdot \left( d^2 \cdot \sum_{t=1}^{[\frac{n}{d}]} t^2 + d\cdot \sum_{t=1}^{[\frac{n}{d}]} t \right) \end{aligned} \]

对m分解质因数后dfs枚举所有满足条件且\(\mu(d)\)不为0的d，然后用求和公式计算后半部分的贡献。

H. Hamming Weight

链接

by Asm.Def

期望难度：Hard

将N表示为

\[N = \sum_{i=0}^{n-1} A_i 2^i,\ A_i \in \{0,1\} \]

由于位与运算每一位是独立的，不妨对每一位单独考虑它对答案的贡献：

\[Ans(N) = \sum_{i=0}^{n-1}\left[A_i \cdot (1 +\sum_{j=0}^{i-1}A_j\cdot 2^j )+ 2^i \cdot \sum_{j=i+1}^{n-1} A_j\cdot 2^{j-i-1} \right]^2 \]

如果直接用FFT计算每次平方，时间复杂度为\(O(n^2 \log n)\) ，难以接受。

考虑在N的某个区间\([L, R)\)上定义答案，并将答案写成多项式的形式，即

\[Ans_{[L,R)}(x)=\sum_{i=L}^{R-1} \left[A_i \cdot (1 +\sum_{j=L}^{i-1}A_j\cdot x^{j-L} )+ x^{i-L} \cdot \sum_{j=i+1}^{R-1} A_j\cdot x^{j-i-1} \right]^2 \]

其中有一项常数项，不方便合并，因此先将答案拆开：

\[\begin{aligned} &Ans_{[L,R)}(x)\\ =&\sum_{i=L}^{R-1} \left[A_i^2 + 2A_i(A_i \cdot \sum_{j=L}^{i-1}A_j\cdot x^{j-L} + x^{i-L} \cdot \sum_{j=i+1}^{R-1} A_j\cdot x^{j-i-1}) \\+ (A_i \cdot \sum_{j=L}^{i-1}A_j\cdot x^{j-L} + x^{i-L} \cdot \sum_{j=i+1}^{R-1} A_j\cdot x^{j-i-1})^2 \right] \end{aligned} \]

考虑到\(A_i\)的取值范围为0或1，即\(A_i^2 = A_i\)，可将答案转化为

\[\begin{aligned} &Ans_{[L,R)}(x)\\ =&\sum_{i=L}^{R-1} A_i + 2 \sum_{i=L}^{R-1} A_i \cdot \left( \sum_{j=L}^{i-1}A_j\cdot x^{j-L} + \sum_{j=i+1}^{R-1} A_j\cdot x^{j-L-1} \right) + \sum_{i=L}^{R-1} \left(A_i\sum_{j=L}^{i-1}A_j\cdot x^{j-L} + \sum_{j=i+1}^{R-1} A_j\cdot x^{j-L-1} \right)^2 \end{aligned} \]

令

\[\begin{array}{ccl} &S(i)&=&\sum\limits_{j=i}^{n-1} A_i\\ &F_{[L,R)}(x)&=&\sum\limits_{i=L}^{R-1}A_i x^i\\ &Ans1_{[L,R)}(x)&=&\sum\limits_{i=L}^{R-1} A_i \cdot \left( \sum\limits_{j=L}^{i-1}A_j\cdot x^{j-L} + \sum\limits_{j=i+1}^{R-1} A_j\cdot x^{j-L-1} \right)\\ &Ans2_{[L,R)}(x)&= &\sum\limits_{i=L}^{R-1} \left(A_i\sum\limits_{j=L}^{i-1}A_j\cdot x^{j-L} + \sum\limits_{j=i+1}^{R-1} A_j\cdot x^{j-L-1} \right)^2 \end{array} \]

则\(Ans_{[L,R)}(x) = S(L)-S(R) + 2Ans1_{[L,R)}(x) + Ans2_{[L,R)}(x)\)。

考虑如何合并两个相邻区间\([L, mid)\)、\([mid,R)\)的答案。

\[\begin{aligned} Ans1_{[L,R)}(x)=&\sum_{i=L}^{R-1} A_i \cdot \left( \sum_{j=L}^{i-1}A_j\cdot x^{j-L} + \sum_{j=i+1}^{R-1} A_j\cdot x^{j-L-1} \right)\\ =&\sum_{i=L}^{mid-1} A_i \cdot \left( \sum_{j=L}^{i-1}A_j\cdot x^{j-L} + \sum_{j=i+1}^{mid-1} A_j\cdot x^{j-L-1} + \sum_{j=mid}^{R-1} A_j \cdot x^{j-L-1} \right)\\ &+ \sum_{i=mid}^{R-1} A_i \cdot \left( \sum_{j=mid}^{i-1}A_j\cdot x^{j-L} + \sum_{j=i+1}^{R-1} A_j\cdot x^{j-L-1} + \sum_{j=L}^{mid-1} x^{j-L} \right)\\ =&Ans1_{[L,mid)}(x) + Ans1_{[mid,R)}(x)\cdot x^{mid-L}\\ &+ \left(\sum_{i=L}^{mid-1} A_i \right) \cdot \left( \sum_{j=mid}^{R-1} A_j \cdot x^{j-L-1} \right)\cdot + \left(\sum_{i=mid}^{R-1} A_i \right) \cdot \left( \sum_{j=L}^{mid-1} A_j \cdot x^{j-L} \right)\\ =&Ans1_{[L,mid)}(x) + Ans1_{[mid,R)}(x)\cdot x^{mid-L} \\&+\sum_{j=mid}^{R-1}(S(L)-S(mid))\cdot A_j x^{j-L-1}+\sum_{j=L}^{mid-1}(S(mid)-S(R))\cdot A_j x^{j-L} \end{aligned} \]

\[\begin{aligned} Ans2_{[L,R)}(x)=&\sum_{i=L}^{R-1} \left(A_i\sum_{j=L}^{i-1}A_j\cdot x^{j-L} + \sum_{j=i+1}^{R-1} A_j\cdot x^{j-L-1} \right)^2\\ =&\sum_{i=L}^{mid-1} \left(A_i\sum_{j=L}^{i-1}A_j\cdot x^{j-L} + \sum_{j=i+1}^{mid-1} A_j\cdot x^{j-L-1} + \sum_{j=mid}^{R-1}A_j\cdot x^{j-L-1} \right)^2 \\&+ \sum_{i=mid}^{R-1} \left(A_i\sum_{j=mid}^{i-1}A_j\cdot x^{j-L} + \sum_{j=i+1}^{R-1} A_j\cdot x^{j-L-1} + A_i\sum_{j=L}^{mid-1} A_j\cdot x^{j-L} \right)^2\\ =&Ans2_{[L,mid)}(x) + Ans2_{[mid, R)}(x) \cdot x^{2(mid-L)}\\ &+2\sum_{i=L}^{mid-1} \left(A_i\sum_{j=L}^{i-1}A_j\cdot x^{j-L} + \sum_{j=i+1}^{mid-1}A_j\cdot x^{j-L-1} \right)\cdot F_{[mid,R)}(x) x^{mid-L-1} \\ &+2\sum_{i=mid}^{R-1} A_i \left( \sum_{j=mid}^{i-1}A_j\cdot x^{j-mid} + \sum_{j=i+1}^{R-1}A_j\cdot x^{j-mid-1} \right) \cdot x^{mid-L} \cdot F_{[L,mid)}(x)\\ &+F_{[mid,R)}(x)^2\cdot (mid-L) \cdot x^{2(mid-L-1)}+F_{[L,mid)}(x)^2\cdot [S(mid) - S(R)] \end{aligned} \]

化简到这里，就可以通过两次长度为\((mid-L)\)和\((R-mid)\)的FFT运算和若干次多项式加法、数乘和移位，由\([L,mid)\)和\([mid,R)\)在\(O((R-L)\log(R-L))\)的时间内求出\(Ans1_{[L,R)}(x), Ans2_{[L,R)}(x)\)。由于待求的相当于x=2时的点值，所以每次用FFT进行乘法后都可以对结果进行一次进位，从而确保在相乘的过程中系数不会溢出。

对\(Ans_{[0,n)}(2)​\)分治求解，总复杂度\(O(n\log^2 n)​\).

//Asm.Def
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005, mod = 998244353, g = 3, maxk = 1 << 19;
typedef long long LL;
typedef vector<int> Poly;
int A[maxn], N, Sum[maxn];

void Print(const Poly &ans)
{
	for(int i = ans.size()-1;i >= 0;--i) printf("%d ", ans[i]);
	puts("");
}

int powmod(int a, int n)
{
	int ans = 1;
	while(n)
	{
		if(n & 1) ans = (LL) ans * a % mod;
		a = (LL) a * a % mod;
		n >>= 1;
	}
	return ans;
}

void Carry(Poly &x)
{
	int C = 0, c;
	while(x.size() > 1)
	{
		if(x.back() == 0) x.pop_back();
		else break;
	}

	for(int i = 0;i < x.size() || C;++i)
	{
		if(i < x.size())
		{
			x[i] += C;
			C = x[i] >> 1;
			x[i] = x[i] & 1;
		}
		else
		{
			x.push_back(C & 1);
			C >>= 1;
		}
	}
}

//Poly operator * (const Poly &a, const Poly &b)
//{
//	Poly ans(a.size() + b.size() - 1);
//	for(int i = 0;i < (int) a.size();++i)
//		for(int j = 0;j < (int) b.size();++j)
//			ans[i+j] += a[i] * b[j];
//	Carry(ans);
//	return ans;
//}

void Shuff(int A[], int n)//n为位数
{
	static int B[maxk];
	int N = 1 << n, j, mx = 1 << (n-1);
	for(int i = 0, it = 0;i < N;++i)
	{
		B[i] = A[it];
		j = mx;
		while(it & j)
		{
			it ^= j;
			j >>= 1;
		}
		it ^= j;
	}
	for(int i = 0;i < N;++i) A[i] = B[i];
}

void DFT(Poly &A, int n, int a)//n为位数
{
	static int B[maxk], pw[20];
	int N = 1 << n;
	pw[n-1] = a;
	for(int i = n-1;i;--i) pw[i-1] = (LL) pw[i] * pw[i] % mod;
	A.resize(N);
	for(int i = 0;i < N;++i) B[i] = A[i];
	Shuff(B, n);
	for(int i = 0;i < n;++i)
	{
		int d = (1 << i), x0 = pw[i];
		for(int j = 0;j < N;j += (d<<1))
		{
			for(int k = j, x = 1;k < j+d;++k)
			{
				int t = (LL) x * B[k+d] % mod;
				B[k+d] = (B[k] + mod - t) % mod;
				B[k] = (B[k] + t) % mod;
				x = (LL) x * x0 % mod;
			}
		}
	}
	for(int i = 0;i < N;++i) A[i] = B[i];
	//Print(A);
}

Poly operator * (const Poly &x, const Poly &y)
{
	static Poly A, B;
	A = x, B = y;
	while(A.size() > 1)
	{
		if(A.back() == 0) A.pop_back();
		else break;
	}
	while(B.size() > 1)
	{
		if(B.back() == 0) B.pop_back();
		else break;
	}
	int n = 0;
	while((1<<n) < A.size() + B.size()) ++n;

	int a = powmod(g, (mod-1) >> n);
	DFT(A, n, a);
	DFT(B, n, a);

	for(int i = 0;i < (1 << n);++i) A[i] = (LL) A[i] * B[i] % mod;
	DFT(A, n, powmod(a, mod-2));
	int t = powmod((1 << n), mod-2);
	for(int i = 0;i < (1 << n);++i) A[i] = (LL) A[i] * t % mod;

	//Print(A);
	Carry(A);
	return A;
}

Poly operator * (const Poly &a, int x)
{
	Poly ans(a.size());
	for(int i = 0;i < (int) a.size();++i)
		ans[i] = a[i] * x;
	Carry(ans);
	return ans;
}

Poly operator + (const Poly &a, const Poly &b)
{
	Poly ans(max(a.size(), b.size()));
	for(int i = 0;i < (int) ans.size();++i)
	{
		ans[i] = 0;
		if(i < (int) a.size()) ans[i] += a[i];
		if(i < (int) b.size()) ans[i] += b[i];
	}
	return ans;
}

Poly operator << (const Poly &x, int n)
{
	Poly ans(x.size() + n, 0);
	for(int i = 0;i < x.size();++i) ans[n+i] = x[i];
	return ans;
}

//void Multi(const Poly &a, const Poly &b, Poly &ans)
//{
//	ans.resize(a.size() + b.size() - 1);
//	for(int i = 0;i < (int) a.size();++i)
//		for(int j = 0;j < (int) b.size();++j)
//			ans[i+j] += a[i] * b[j];
//}

void init()
{
	for(int i = 1;i <= N;++i) scanf("%1d", &A[N-i]);
	Sum[N] = 0;
	for(int i = N-1;i >= 0;--i)
		Sum[i] = Sum[i+1] + A[i];
}


void Solve(int L, int R, Poly &Ans1, Poly &Ans2)
{
	static Poly SL, SR, AL, AR;
	if(R - L == 1)
	{
		Ans1.resize(1);Ans1[0] = 0;
		Ans2.resize(1);Ans2[0] = 0;
		return;
	}
	int mid = (L + R) >> 1;
	Poly Ans1L, Ans1R, Ans2L, Ans2R;
	Solve(L, mid, Ans1L, Ans2L);
	Solve(mid, R, Ans1R, Ans2R);
	//printf("Solve (%d,%d)\n", L, R);
	//
	//Print(Ans1L);
	//Print(Ans1R);
	//Print(Ans2L);
	//Print(Ans2R);
	//puts("");

	AL.clear();
	AL.resize(mid-L);
	for(int i = L;i < mid;++i) AL[i-L] = A[i];

	AR.clear();
	AR.resize(R-mid);
	for(int i = mid;i < R;++i) AR[i-mid] = A[i];

	SL.clear();
	SL.resize(mid-L);
	for(int i = L;i < mid-1;++i)
		SL[i-L] = A[i] * (Sum[i+1]-Sum[mid]) + A[i+1] * (i+1-L);

	SR.clear();
	SR.resize(R-mid);
	for(int i = mid;i < R-1;++i)
		SR[i-mid] = A[i] * (Sum[i+1]-Sum[R]) + A[i+1] * (Sum[mid]-Sum[i+1]);

	//puts("");
	//Print(AL);
	//Print(AR);
	//Print(SL);
	//Print(SR);
	
	Ans2 = Ans2L + ((AR * SL) << (mid-L)) + ((AR * AR * (mid - L)) << (2 * (mid-L-1))) + (Ans2R << (2 * (mid-L))) + ((AL * SR) << (mid-L+1)) + AL * AL * (Sum[mid]-Sum[R]);
	Carry(Ans2);

	Ans1 = Ans1L + (Ans1R << (mid-L));
	if((int) Ans1.size() < (R-L-1))
		Ans1.resize(R-L-1);
	for(int i = L;i < mid;++i)
		Ans1[i-L] += A[i] * (Sum[mid] - Sum[R]);
	for(int i = mid;i < R;++i)
		Ans1[i-L-1] += A[i] * (Sum[L] - Sum[mid]);
	Carry(Ans1);
	//Print(Ans1);
	//Print(Ans2);
}

void work()
{
	Poly ans1, ans2;
	Solve(0, N, ans1, ans2);
	//Print(ans1), Print(ans2);
	ans2 = ans2 + (ans1 << 1);
	ans2[0] += (Sum[0] - Sum[N]);
	Carry(ans2);
	int sum = 0;
	for(int i = ans2.size()-1;i >= 0;--i) printf("%d", ans2[i]);
	puts("");
	//for(int i = 0;i < (int) ans.size();++i)
	//	sum += ans[i];
	//printf("%d\n", sum % 1000000007);
}

int main()
{
	time_t beg = clock();
	while(~scanf("%d", &N))
	{
		init();
		work();
	}
	//printf("%.2f sec", double(clock() - beg) / CLOCKS_PER_SEC);
	return 0;
}

I. Lattice's basics in digital electronics

链接

by Joker

期望难度：Easy

签到题。直接根据题意模拟即可，可以采用map来减少编码难度。

J. Ka Chang

链接

期望难度：Medium-

按每一层的结点个数分类讨论，设阈值为\(T\)。

当第\(L\)层的结点个数\(Size_L <T\)时，每次\(1\ L\ X\)操作只需枚举这一层的所有结点，维护它们对每个结点的答案产生的贡献即可。

当第\(L\)层的结点个数\(Size_L >= T\)时，这样的层不超过\(\frac{N}{T}\)个，对于操作1可以直接对每个这样的层维护增加了多少point，对于每次询问直接枚举一遍即可。

对于第一种情况，可以用树状数组维护dfs序列上的区间和，时间复杂度\(O(Q(T\cdot \log N))\);
对于第二种情况，时间复杂度\(O(Q\cdot \frac{N}{T})\)

则总时间复杂度为\(O(Q \cdot (T \log N + \frac{N}{T}))\)，取\(T=\sqrt{\frac{N}{\log N}}\) 最优。

时间复杂度\(O(Q\cdot\sqrt{N\log N})\) 。

by bird_14

K. Supreme Number

链接

by morejarphone

期望难度：Easy-

考虑到答案中任意一位都必须是1或质数，可知答案只可能由1、2、3、5、7构成。由于任意两个不为1的数字构成的两位数一定可以被11整除，所以答案中除1外的数字只能出现一次；1最多出现2次，因为111可以被3整除；而2、5、7三者一定不会有两者同时出现。因此满足条件的整数不会超过四位，全部预处理出来即可。