Manthan, Codefest 16

暴力 A - Ebony and Ivory

import java.util.*;
import java.io.*;

public class Main   {
    public static void main(String[] args)  {
        Scanner cin = new Scanner (new BufferedInputStream (System.in));
        int a = cin.nextInt ();
        int b = cin.nextInt ();
        int c = cin.nextInt ();
        for (int i=0; a*i<=c; ++i)  {
            int d = c - a * i;
            if (d % b == 0) {
                System.out.println ("Yes");
                return ;
            }
        }
        System.out.println ("No");
    }
}

数学 B - A Trivial Problem

题意:问n!的后缀0的个数为m个的n的范围.

分析:出现0的一定是2*5产生的,而2的数字有很多,所以找到最小的数字之前5的总个数为m的.二分来找.

#include <bits/stdc++.h>

int number(int x)   {
    int ret = 0;
    while (x)   {
        x /= 5;
        ret += x;
    }
    return ret;
}

int main()  {
    int m;  scanf ("%d", &m);
    int left = 1, right = (int) 1e9;
    while (left < right)   {
        int mid = left + right >> 1;
        if (number (mid) < m)   left = mid + 1;
        else    right = mid;
    }
    std::vector<int> ans;
    for (;;)    {
        if (number (left) == m) ans.push_back (left);
        else    break;
        left++;
    }
    int sz = ans.size ();
    printf ("%d\n", sz);
    for (int i=0; i<sz; ++i)    {
        if (i > 0)  putchar (' ');
        printf ("%d", ans[i]);
    }
    puts ("");

    return 0;
}

Trie + DP C - Spy Syndrome 2

题意:有一句话被变成全小写并且删掉空格并且翻转单词,然后给出可能的单词.问原来可能的这句话.

分析:首先把单词插入到字典树上,这里为了节约内存把所有单词并在一起.结点保存了该单词在单词串的位置以便输出.然后文本串倒过来在字典树上DP搜索,最后正的输出,那么可以找到可行的一句话.

#include <bits/stdc++.h>

const int N = 1e4 + 5;
const int M = 1e6 + 1e5;
const int NODE = M;
char text[N], words[M];
int ch[NODE][26], val[NODE], pos[NODE];
int n, m, sz;
int nex[N], wl[N];

int idx(char c) {
    return tolower (c) - 'a';
}
void trie_init()    {
    memset (ch[0], 0, sizeof (ch[0]));
    sz = 1;
}
void trie_insert(char *str, int end, int id, int p)  {
    int u = 0;
    for (int c, i=0; i<end; ++i)   {
        c = idx (str[i]);
        if (!ch[u][c])  {
            memset (ch[sz], 0, sizeof (ch[sz]));
            val[sz] = 0;    pos[sz] = 0;
            ch[u][c] = sz++;
        }
        u = ch[u][c];
    }
    val[u] = id;    pos[id] = p;
}
void trie_query()   {
    memset (nex, -1, sizeof (nex));
    memset (wl, 0, sizeof (wl));
    nex[n] = 0;
    for (int i=n; i>0; --i)    {
        if (nex[i] == -1)   continue;
        int u = 0;
        for (int c, j=i-1; j>=0; --j)  {
            c = idx (text[j]);
            if (!ch[u][c])  break;
            u = ch[u][c];
            if (val[u] > 0) {
                wl[j] = pos[val[u]];
                nex[j] = i;
            }
        }
    }
}

int main()  {
    scanf ("%d", &n);
    scanf ("%s", text);
    scanf ("%d", &m);
    trie_init ();
    for (int L=0, i=1; i<=m; ++i) {
        scanf ("%s", words + L);
        int len = strlen (words + L);
        trie_insert (words + L, len, i, L);
        L += len + 1;
    }
    trie_query ();
    int now = 0;
    while (now < n)   {
        if (now > 0)    putchar (' ');
        printf ("%s", words + wl[now]);
        now = nex[now];
    }
    puts ("");

    return 0;
}

DFS + 二分 D - Fibonacci-ish

题意:在n个数找出一组数字满足fn = fn-1 + fn-2, 问最大长度.

分析:n的范围小,可以考虑n^2枚举两个起点,因为要考虑到个数的问题,这里我选择一种方便的写法:首先不考虑个数,只预处理两个数能否到下一个数字.然后考虑个数,类似DFS的vis功能,深搜时-1,回溯时+1

#include <bits/stdc++.h>

const int N = 1e3 + 5;
const int MOD = 1e9 + 7;
int a[N], A[N];
int nex[N][N];
int cnt[N];
int ans;

void DFS(int i, int j, int step)    {
    if (step > ans) ans = step;
    int k = nex[i][j];
    if (k == -1)    return ;
    else if (cnt[k] > 0)    {
        --cnt[k];
        DFS (j, k, step + 1);
        ++cnt[k];
    }
}

int main()  {
    int n;  scanf ("%d", &n);
    for (int i=0; i<n; ++i) {
        scanf ("%d", &a[i]);    A[i] = a[i];
    }
    std::sort (A, A+n);
    int m = std::unique (A, A+n) - A;
    for (int i=0; i<n; ++i) {
        a[i] = std::lower_bound (A, A+m, a[i]) - A;
        cnt[a[i]]++;
    }
    for (int i=0; i<m; ++i) {
        for (int j=0; j<m; ++j) {
            int k = std::lower_bound (A, A+m, A[i] + A[j]) - A;
            if (k >= m || A[i] + A[j] != A[k])  nex[i][j] = -1;
            else    nex[i][j] = k;
        }
    }
    ans = 2;
    for (int i=0; i<m; ++i) {
        --cnt[i];
        for (int j=0; j<m; ++j) {
            if (cnt[j] <= 0)    continue;
            --cnt[j];
            DFS (i, j, 2);
            ++cnt[j];
        }
        ++cnt[i];
    }
    printf ("%d\n", ans);

    return 0;
}

二分查找 + RMQ + 组合数学 E - Startup Funding

题意:对于每一个li = i,找到一个ri,使得最大.从n个结果中选择k个,最小值的期望.

分析:第一个问题,考虑前缀max (vk)是递增的,考虑前缀min(ck)是递减的,两者取min那么是单峰函数,二分查找.第二个问题,首先对结果排序,假设最小值为ans[i],那么选中它当最小值的概率是C(n-i, k-1) / C (n, k).p * ans[i]求和就是期望.发现公式可以递推.

#include <bits/stdc++.h>

const int N = 1e6 + 5;
int mx[N][21], mn[N][21];
int best[N];
int n, k;

void build_max()	{
	for (int j=1; (1<<j)<=n; ++j)	{
		for (int i=1; i+(1<<j)-1<=n; ++i)	{
			mx[i][j] = std::max (mx[i][j-1], mx[i+(1<<(j-1))][j-1]);
		}
	}
}
int query_max(int l, int r)	{
	int k = 0;	while (1<<(k+1) <= r-l+1)	++k;
	return std::max (mx[l][k], mx[r-(1<<k)+1][k]);
}

void build_min()	{
	for (int j=1; (1<<j)<=n; ++j)	{
		for (int i=1; i+(1<<j)-1<=n; ++i)	{
			mn[i][j] = std::min (mn[i][j-1], mn[i+(1<<(j-1))][j-1]);
		}
	}
}
int query_min(int l, int r)	{
	int k = 0;	while (1<<(k+1) <= r-l+1)	++k;
	return std::min (mn[l][k], mn[r-(1<<k)+1][k]);
}

int p(int l, int r)	{
	if (l > r || l < 1 || r > n)	return 0;
	return std::min (100 * query_max (l, r), query_min (l, r));
}

int main()	{
	scanf ("%d%d", &n, &k);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &mx[i][0]);
	}
	build_max ();
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &mn[i][0]);
	}
	build_min ();
	for (int i=1; i<=n; ++i)	{
		int low = i, high = n;
		while (low + 1 < high)	{
			int mid = low + high >> 1;
			int v1 = 100 * query_max (i, mid);
			int v2 = query_min (i, mid);
			if (v1 < v2)	low = mid;
			else	high = mid;
		}
		best[i-1] = std::max (p (i, low), p (i, high));
	}
	std::sort (best, best+n);
	double prob = 1.0 * k / n;
	double ans = prob * best[0];
	for (int i=1; i<=n-k; ++i)	{
		prob = prob * (n - i - k + 1) / (n - i);
		ans += prob * best[i];
	}
	printf ("%.12f\n", ans);

	return 0;
}

树形DP F - The Chocolate Spree

题意:树上选择两条不相交的路径,且两条路径权值和最大.

分析:因为权值>0, 所以起点或终点一定在叶子结点上,第一次DFS,得到best[u]:u结点的子树下得到最大权值和(一条),以及down[u]:从结点u出发到叶子节点选择一条路的最大权值和.第二次DFS扫描每一个结点,从儿子中选择一个,它子树best[v1]作为一条路径,还有一条从前缀i以及后缀i+1中选择,更新最大值就是答案.

#include <bits/stdc++.h>

typedef long long ll;
const int N = 1e5 + 5;
std::vector<int> edge[N];
int a[N];
ll best[N], down[N];
ll ans;
int n;

void DFS(int u, int fa)	{
	std::vector<ll> downs;
	for (auto v: edge[u])	{
		if (v == fa)	continue;
		DFS (v, u);
		best[u] = std::max (best[u], best[v]);
		downs.push_back (down[v]);
	}
	ll mx1 = 0, mx2 = 0;
	for (auto d: downs)	{
		if (d > mx1)	{
			mx2 = mx1;	mx1 = d;
		}
		else if (d > mx2)	{
			mx2 = d;
		}
	}
	best[u] = std::max (best[u], mx1 + mx2 + a[u]);
	down[u] = mx1 + a[u];
	ans = std::max (ans, best[u]);
}

void DFS2(int u, int fa, ll up)	{
	up += a[u];
	std::vector<int> children;
	for (auto v: edge[u])	{
		if (v == fa)	continue;
		children.push_back (v);
	}
	int sz = children.size ();
	if (sz == 0)	return ;
	std::vector<ll> prebest (sz + 1), sufbest (sz + 1);		//前缀(1~i-1)最优的一条路径
	prebest[0] = 0;
	for (int i=0; i<sz; ++i)	{
		prebest[i+1] = std::max (prebest[i], best[children[i]]);
	}
	sufbest[sz] = 0;
	for (int i=sz-1; i>=0; --i)	{							//后缀(i+1~sz-1)最优的一条路径
		sufbest[i] = std::max (sufbest[i+1], best[children[i]]);
	}
	std::vector<ll> predown (sz + 1), predown2 (sz + 1);	//前缀两条到叶子节点最优的路径
	predown[0] = predown2[0] = 0;
	for (int i=0; i<sz; ++i)	{
		predown[i+1] = predown[i];
		predown2[i+1] = predown2[i];
        ll x = down[children[i]];
        if (x > predown[i+1])	{
			predown2[i+1] = predown[i+1];
			predown[i+1] = x;
        }
        else if (x > predown2[i+1])	{
			predown2[i+1] = x;
        }
	}
	std::vector<ll> sufdown (sz + 1), sufdown2 (sz + 1);	//后缀两条到叶子节点最优的路径
	sufdown[sz] = sufdown2[sz] = 0;
	for (int i=sz-1; i>=0; --i)	{
		sufdown[i] = sufdown[i+1];
		sufdown2[i] = sufdown2[i+1];
        ll x = down[children[i]];
        if (x > sufdown[i])	{
			sufdown2[i] = sufdown[i];
			sufdown[i] = x;
        }
        else if (x > sufdown2[i])	{
			sufdown2[i] = x;
        }
	}
	for (int i=0; i<sz; ++i)	{
		ll cur = std::max (prebest[i], sufbest[i+1]);
		cur = std::max (cur, up + std::max (predown[i], sufdown[i+1]));
		cur = std::max (cur, a[u] + predown[i] + sufdown[i+1]);
		cur = std::max (cur, a[u] + predown[i] + predown2[i]);
		cur = std::max (cur, a[u] + sufdown[i+1] + sufdown2[i+1]);
		cur += best[children[i]];
		ans = std::max (ans, cur);
	}
	for (int i=0; i<sz; ++i)	{
		int v = children[i];
		ll new_up = up;
		new_up = std::max (new_up, a[u] + std::max (predown[i], sufdown[i+1]));
		DFS2 (v, u, new_up);
	}
}

int main()	{
	scanf ("%d", &n);
	for (int i=1; i<=n; ++i)	scanf ("%d", a+i);
	for (int u, v, i=1; i<n; ++i)	{
		scanf ("%d%d", &u, &v);
		edge[u].push_back (v);
		edge[v].push_back (u);
	}
	DFS (1, 0);
	DFS2 (1, 0, 0);
	printf ("%I64d\n", ans);

	return 0;
}

DFS序 + 线段树 + bitset G - Yash And Trees　　

题意:两种操作; 1.v的子树的所有结点权值+x 2. 询问v子树%m后是素数的个数

分析:1操作想到线段树的成段更新,树变成线段用DFS序,每个结点有它'统治"的范围(子树). 然而后者统计用普通数组很难实现.用到了bitset这个容器,里面可以表示m位的01,本题表示一个结点子树所拥有的数值(%m),最后只要&primes就是素数个数.那么如何实现+x呢,因为每一位表示数值,往前一位表示+1,那么<<x, 还有可能移位超出去了,还要| >>(m - x).

#include <bits/stdc++.h>

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
const int N = 1e5 + 5;
std::bitset<1000> tree[N<<2], primes, ret;
std::vector<int> edge[N];
int lazy[N<<2];
int a[N], id[N], fl[N], fr[N];
int n, m, q, tot;

void add(int &x, int y)  {
    x += y;
    if (x >= m) x %= m;
}

void push_up(int o) {
    tree[o] = tree[o<<1] | tree[o<<1|1];
}
void rotate(int o, int x)   {
    add (lazy[o], x);
    tree[o] = (tree[o] << x) | (tree[o] >> (m - x));
}
void push_down(int o)   {
    if (lazy[o] != 0)   {
        rotate (o << 1, lazy[o]);
        rotate (o << 1 | 1, lazy[o]);
        lazy[o] = 0;
    }
}
void build(int l, int r, int o) {
    if (l == r) {
        tree[o].set (a[id[l]]%m);   return ;
    }
    int mid = l + r >> 1;
    build (lson);   build (rson);
    push_up (o);
}
void updata(int ql, int qr, int x, int l, int r, int o) {
    if (ql <= l && r <= qr) {
        rotate (o, x); return ;
    }
    push_down (o);
    int mid = l + r >> 1;
    if (ql <= mid)  updata (ql, qr, x, lson);
    if (qr > mid)   updata (ql, qr, x, rson);
    push_up (o);
}
void query(int ql, int qr, int l, int r, int o) {
    if (ql <= l && r <= qr) {
        ret |= tree[o]; return ;
    }
    push_down (o);
    int mid = l + r >> 1;
    if (ql <= mid)  query (ql, qr, lson);
    if (qr > mid)   query (ql, qr, rson);
}

void DFS(int u, int fa) {
    id[fl[u]=++tot] = u;
    for (auto v: edge[u])  {
        if (v != fa)    DFS (v, u);
    }
    fr[u] = tot;
}

bool is_prime(int x)    {
    if (x == 2 || x == 3)   return true;
    if (x % 6 != 1 && x % 6 != 5)   return false;
    for (int i=5; i*i<=x; i+=6) {
        if (x % i == 0 || x % (i + 2) == 0) return false;
    }
    return true;
}

int main()  {
    scanf ("%d%d", &n, &m);
    for (int i=1; i<=n; ++i) {
        scanf ("%d", a+i);
    }
    for (int u, v, i=0; i<n-1; ++i) {
        scanf ("%d%d", &u, &v);
        edge[u].push_back (v);
        edge[v].push_back (u);
    }
    tot = 0;
    DFS (1, 0);
    for (int i=2; i<m; ++i) {
        if (is_prime (i))   primes.set (i);
    }
    build (1, n, 1);
    scanf ("%d", &q);
    while (q--) {
        int op, v, x;   scanf ("%d%d", &op, &v);
        if (op == 1)    {
            scanf ("%d", &x);
            x %= m;
            updata (fl[v], fr[v], x, 1, n, 1);
        }
        else    {
            ret.reset ();
            query (fl[v], fr[v], 1, n, 1);
            ret &= primes;
            printf ("%d\n", (int) ret.count ());
        }
    }

    return 0;
}

暴力 || 莫队+线段树 H - Fibonacci-ish II

题意:q次询问,每次对l和r的范围内的数字去重,然后升序排序,计算fib[j] * a[j]的和．

分析：目前只会暴力的思路: 先排序, 然后每一个数原先对应的询问区间内累加，Ｏ(ｎｑ)复杂度险过

#include <bits/stdc++.h>

const int N = 3e4 + 5;
std::pair<int, int> a[N];
int fib[N];
int ql[N], qr[N], last[N], step[N];
int ans[N];

int main()	{
	int n, m;	scanf ("%d%d", &n, &m);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", &a[i].first);
		a[i].second = i;
	}
	std::sort (a+1, a+1+n);
	fib[0] = 1;	fib[1] = 1;
	for (int i=2; i<=n; ++i)	fib[i] = (fib[i-2] + fib[i-1]) % m;
	int q;	scanf ("%d", &q);
	for (int i=0; i<q; ++i)	{
		scanf ("%d%d", ql+i, qr+i);
		last[i] = -1;
	}
	for (int i=1; i<=n; ++i)	{
		int v = a[i].first % m;
		for (int j=0; j<q; ++j)	{
			if (a[i].second < ql[j] || a[i].second > qr[j])	continue;
			if (a[i].first == last[j])	continue;
			ans[j] = (ans[j] + v * fib[step[j]++]) % m;
			last[j] = a[i].first;
		}
	}
	for (int i=0; i<q; ++i)	printf ("%d\n", ans[i]);

	return 0;
}