步梯 abc 367

Tasks - AtCoder Beginner Contest 367

A - Shout Everyday (atcoder.jp)

签到

#include <bits/stdc++.h>
typedef long long ll;
const int N = 2;

void solve()
{
    int a,b,c;
    std::cin >> a >> b >> c;
    bool ok = true;
    if(b > c) {
        if(a >= b && a <= 24) ok = false;
        if(a < c) ok = false;
    }
    else {
        if(a >= b && a <= c) ok = false;
    }
    if(ok) std::cout << "Yes\n";
    else std::cout << "No\n";
    return;
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int _ = 1;
    //std::cin >> _ ;
    while(_ --) {
        solve();
    }
    return 0;
}

B - Cut .0 (atcoder.jp)

反转去除前导零

#include <bits/stdc++.h>
typedef long long ll;
const int N = 2e6 +50;

void solve()
{
    std::string s;
    std::cin >> s;
    std::reverse(s.begin(), s.end());
    int n = s.size();
    int cnt = 0;
    //std::cout << s << '\n';
    while(s.size())
    {
        if(s[0] != '0') break;
        if(s[0] == '.') {
            s.erase(0,1);
            break;
        }
        if(s[0] == '0') s.erase(0,1);
    }
    if(s[0] == '.') s.erase(0,1);
    std::reverse(s.begin(), s.end());
    std::cout << s << '\n';
    return;
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int _ = 1;
    //std::cin >> _ ;
    while(_ --) {
        solve();
    }
    return 0;
}

C - Enumerate Sequences (atcoder.jp)

爆搜

#include <bits/stdc++.h>
typedef long long ll;
const int N = 3e5 + 50;
int a[N],R[N];
	int n,k;
	void dfs (int m, int s) {
		if(m > n) {
			if(s % k == 0) {
				for(int i = 1; i <= n; i++) 
					std::cout << a[i] << " ";
				std::cout << '\n';
			}
		}
		for(int i = 1; i <= R[m-1]; i++) {
			a[m] = i;
			dfs(m+1,s + i);
		}
	}

void solve() {

	std::cin >> n >> k;
	for(int i = 0; i < n; i++) std::cin >> R[i];
	dfs(1,0);
	return;
}

int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	int _ = 1;
	//std::cin >> _;
	while(_ --) {
		solve();
	}
	return 0;
}

D - Pedometer (atcoder.jp)

考虑到$s,t$ 在环上的大小关系分为两类

$s < t$ 断环为链 $Sum_{t-1} - Sum_{s-1}$ % m = 0

$(a-b) \mod m = 0$ , $a,b$ 对 $m$ 同余

枚举 $t$ ,令 $s = t - 1$ ,$cnt$ 统计相同余数

最巧妙的地方

for(int t = 2 ; t <= n; t++) {
    cnt[sum[t-2]] ++;// s -> t - 1 ， s - 1 -> t - 2
    res += cnt[sum[t-1]]; // 保证 与t配对的s 均是小于t的
}

$s > t$

$Sum_n - Sum_{s-1} + Sum_{t-1}$ = $suf_s + Sum_{t-1}$

($a + b$ ) % m = 0, $a \mod m + b \mod m = m$

同理枚举 $s$

#include <bits/stdc++.h>
typedef long long ll;
const int N = 2e6 + 50;

void solve()
{
    int n,m;
    std::cin >> n >> m;
    std::vector<int> a(n);
    std::vector<ll> sum(n+1),suf(n+2);
    std::map<int,int> cnt;
    for(int i = 0;i < n;i ++) {
        std::cin >> a[i];
        sum[i+1] = (sum[i] + a[i]) % m;
        //std::cout << sum[i+1] << " ";
    }
    //std::cout << '\n';
    for(int i = n ;i >= 1;i --) {
        suf[i] = (suf[i+1] + a[i-1]) % m;
        //std::cout << suf[i] << " ";
    }
    //std::cout << '\n';
    ll res = 0;
    for(int t = 2 ; t <= n; t++) {
        cnt[sum[t-2]] ++;// s -> t - 1
        res += cnt[sum[t-1]];
    }
    cnt.clear();
    for(int s = 2; s <= n; s++) {
        cnt[sum[s-2]] ++; // t -> s - 1
        res += cnt[(m - suf[s]) % m];// 防止后缀和取模等于0 产生cnt[m] 
    }
    std::cout << res << '\n';
    return;
}
int main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int _ = 1;
    //std::cin >> _ ;
    while(_ --) {
        solve();
    }
    return 0;
}

E - Permute K times (atcoder.jp)

问$A_7$经过次变换值为$A_x$

经过一次 $A_6$，经过两次$A_4$ ，经过三次$A_3$ ， 经过四次$A_6$

可以构建$i \quad x[i]$基环树森林

每次转移就是在转移边

用倍增 $st[i][0] = x[i]$ 表示$i$ 的 $2^{i}$ 转移后点是多少

$k \le 10^{18}$

$st[i][j] = st[st[i][j-1]][j-1]$

$2^{j} = 2^{j-1} + 2^{j-1}$

经过$2^{j-1}$ 次转移的点，再转移$2^{j-1}$ 次

#include <bits/stdc++.h>
typedef long long ll;
const int N = 3e5 + 50;
int st[N][63];
void solve() {
	int n;
	ll k;
	std::cin >> n >> k;
	std::vector<int> X(n),A(n);
	for(int i = 0; i < n; i++) std::cin >> X[i];
	for(int i = 0; i < n; i++) std::cin >> A[i];

	for(int i = 1; i <= n; i++) {
		st[i][0] = X[i-1];
	}
	for(int j = 1; j <= 62; j ++) {
		for(int i = 1; i <= n; i++) {
			st[i][j] = st[st[i][j-1]][j-1];
		}
	}

	for(int i = 1; i <= n; i++) {
		int x = i;
		ll  y = k;
		for(int j = 62; j >= 0; j --) {
			if(y >= (1ll << j)) {
				x = st[x][j];
				y -= (1ll << j);
			}
		}
		std::cout << A[x-1] << " ";
	}
	return;
}

int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	int _ = 1;
	//std::cin >> _;
	while(_ --) {
		solve();
	}
	return 0;
}

F - Rearrange Query (atcoder.jp)

• Problem Statement

  You are given sequences of positive integers of length $N$: $A=(A_1,A_2,\ldots,A_N)$ and $B=(B_1,B_2,\ldots,B_N)$.

  You are given $Q$ queries to process in order. The $i$-th query is explained below.

  • You are given positive integers $l_i,r_i,L_i,R_i$. Print Yes if it is possible to rearrange the subsequence $(A_{l_i},A_{l_i+1},\ldots,A_{r_i})$ to match the subsequence $(B_{L_i},B_{L_i+1},\ldots,B_{R_i})$, and No otherwise.

  哈希

  将每个$i$ 映射到更大的范围，利用前缀和查询是否相等

  #include <bits/stdc++.h>
  typedef long long ll;
  const int N = 2e6 + 50;
  const ll base = 1145141;
  const ll mod = 1e9 + 7;
  void solve()
  {
      int n,q;
      std::cin >> n >> q;
      std::vector<int> a(n),b(n);
      std::vector<ll> p(n+1),ha(n+1,0),hb(n+1,0);
      for(int i = 0; i < n; i++) {
          std::cin >> a[i];
      }
      for(int i = 0; i < n; i++) {
          std::cin >> b[i];
      }
      p[0] = 1;
      for(int i = 1; i <= n; i++) {
          p[i] = p[i-1] * base % mod;
      }
      for(int i = 0; i < n; ++i) {
          ha[i+1] = (ha[i] + p[a[i]]) % mod;
      }
      for(int i = 0; i < n; ++i) {
          hb[i+1] = (hb[i] + p[b[i]]) % mod;
      }
      while(q--) {
          int l,r,L,R;
          std::cin >> l >> r >> L >> R;
          l--;
          L--;
          if((ha[r] - ha[l] + mod) % mod  == (hb[R] - hb[L] + mod) % mod) {
              std::cout << "Yes\n";
          }
          else {
              std::cout << "No\n";
          }
      }
      return;
  }
  int main()
  {
      std::ios::sync_with_stdio(false);
      std::cin.tie(nullptr);
      int _ = 1;
      //std::cin >> _ ;
      while(_ --) {
          solve();
      }
      return 0;
  }