AtCoder Beginner Contest 254
C - K Swap
题意：给定一个整数序列，长度为n。每个数aia_iai能与ai+ka_{i + k}ai+k交换位置，是否能将该序列变换为升序。
思路：该序列可分为k个子序列：
- a0,a0+k,a0+2k,...,a_0,a_{0+k}, a_{0+2k}, ... ,a0,a0+k,a0+2k,...,
- a1,a1+k,a1+2k,...,a_1,a_{1+k}, a_{1+2k}, ... ,a1,a1+k,a1+2k,...,
…- ak−1,ak−1+k,al−1+2k,...,a_k-1,a_{k-1+k}, a_{l-1+2k}, ... ,ak−1,ak−1+k,al−1+2k,...,
将这k个序列内部排序，放回原序列，判断原序列是否为升序。
Code：
// Problem: C - K Swap
// Contest: AtCoder - AtCoder Beginner Contest 254
// URL: https://atcoder.jp/contests/abc254/tasks/abc254_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)/*
author: A Fei
*/
#include <bits/stdc++.h>#define x first
#define y second
#define pi acos(-1)
#define endl '\n'
#define ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mem(x, a) memset(x, a, sizeof x)
#define pb(x) push_back(x)
#define rep(i, l, r) for(int i = l; i <= (r); ++ i)
#define per(i, r, l) for(int i = r; i >= (l); -- i)using namespace std;
typedef long long LL;
typedef double DB;
typedef pair<int, int> PII;
typedef pair<int, double> PID;
typedef pair<double, double> PDD;
template <typename T> void inline read(T &x) {int f = 1; x = 0; char s = getchar();while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();x *= f;
}int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
const int INF = 0x3f3f3f3f;
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
const int N = 2e5 + 10;
int n, k;int main()
{ios;cin >> n >> k;vector<int> a(n);rep(i, 0, n - 1) cin >> a[i];vector<int> c(n);rep(i, 0, k - 1){vector<int> b;for(int j = i; j < n; j += k) b.pb(a[j]);sort(b.begin(), b.end());for(int j = 0; j < b.size(); j ++) c[i + j * k] = b[j];}// for(auto x : c) cout << x << ' ';// cout << endl;if(is_sorted(c.begin(), c.end())) cout << "Yes\n";else cout << "No\n"; return 0;
}D - Together Square
题意：找出有多少对儿：i,ji, ji,j，使得i∗ji * ji∗j为平方数。（1<=i,j<=n1 <= i, j <= n1<=i,j<=n）
思路：
可以考虑每个数都可分解质因子变为：
设N分解质因子为：N=p1a1∗p2a2∗⋯∗pnan其中p1,p2,⋯,pn为N的质因子则N的约数个数为：f(N)=(a1+1)∗(a2+1)∗⋯∗(an+1)设N分解质因子为： \\N = p_1^{a_1} * p_2^{a_2} *\cdots*p_n^{a_n}\\其中p_1,p_2,\cdots,p_n为N的质因子\\ 则N的约数个数为：f(N) = (a_1 + 1) * (a_2 + 1) * \cdots* (a_n + 1) 设N分解质因子为：N=p1a1∗p2a2∗⋯∗pnan其中p1,p2,⋯,pn为N的质因子则N的约数个数为：f(N)=(a1+1)∗(a2+1)∗⋯∗(an+1)
将该数中奇数个的质因子的积（记为XiX_iXi)拿出来计数（记为CiC_iCi）。例：
6：21∗312^1 * 3^121∗31 --> 2∗3=62 * 3 = 62∗3=6
24:23∗312^3 * 3^123∗31 --> 2∗3=62 * 3 = 62∗3=6
6∗24=144=1226 * 24 = 144 = 12 ^ 26∗24=144=122
所以当i,ji, ji,j的XXX相同时，说明i∗ji * ji∗j的质因子个数都是偶数，即i∗ji * ji∗j 为平方数。答案为：C12+C22+...+Ci2C_1^2 + C_2^2 + ... + C_i^2C12+C22+...+Ci2。
Code:
// Problem: D - Together Square
// Contest: AtCoder - AtCoder Beginner Contest 254
// URL: https://atcoder.jp/contests/abc254/tasks/abc254_d
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)/*
author: A Fei
*/
#include <bits/stdc++.h>#define x first
#define y second
#define pi acos(-1)
#define endl '\n'
#define ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mem(x, a) memset(x, a, sizeof x)
#define pb(x) push_back(x)
#define rep(i, l, r) for(int i = l; i <= (r); ++ i)
#define per(i, r, l) for(int i = r; i >= (l); -- i)using namespace std;
typedef long long LL;
typedef double DB;
typedef pair<int, int> PII;
typedef pair<int, double> PID;
typedef pair<double, double> PDD;
template <typename T> void inline read(T &x) {int f = 1; x = 0; char s = getchar();while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();x *= f;
}int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
const int INF = 0x3f3f3f3f;
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~int fun(int n)
{int res = 1;// i <= n / i -- why?:// 最多只会存在一个质因子 > sqrt(n), 两个的话 相乘 > n.// 当我们一直除最后n!=1,此时n就是那个大于sqrt(n)的质因子for(int i = 2; i <= n / i; i ++)if(n % i == 0){int cnt = 0;while(n % i == 0) n /= i, cnt ++;if(cnt % 2) res *= i; }if(n != 1) res *= n;return res;
}int main()
{ios;int n;cin >> n;map<int, int> mp;rep(i, 1, n){int res = fun(i);mp[res] ++;}LL ans = 0;for(auto x : mp) ans += x.y * x.y;cout << ans << endl;return 0;
}E - Small d and k
题意：给定一个无向图，QQQ次询问。
每次询问：一个点走KKK次，能到达所有边的权值和
思路: 简单BFS or DFS
错因: 数组开小了…
至少开：2(双向边)∗M=3∗N2(双向边) * M = 3*N2(双向边)∗M=3∗N。
Code：
// Problem: E - Small d and k
// Contest: AtCoder - AtCoder Beginner Contest 254
// URL: https://atcoder.jp/contests/abc254/tasks/abc254_e
// Memory Limit: 1024 MB
// Time Limit: 3500 ms
//
// Powered by CP Editor (https://cpeditor.org)/*
author: A Fei
*/
#include <bits/stdc++.h>#define x first
#define y second
#define pi acos(-1)
#define endl '\n'
#define ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mem(x, a) memset(x, a, sizeof x)
#define pb(x) push_back(x)
#define rep(i, l, r) for(int i = l; i <= (r); ++ i)
#define per(i, r, l) for(int i = r; i >= (l); -- i)using namespace std;
typedef long long LL;
typedef double DB;
typedef pair<int, int> PII;
typedef pair<int, double> PID;
typedef pair<double, double> PDD;
template <typename T> void inline read(T &x) {int f = 1; x = 0; char s = getchar();while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();x *= f;
}int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
const int INF = 0x3f3f3f3f;
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
const int N = 6e5 + 10, M = 2 * N;
int h[N], ne[M], e[M], idx;
bool st[N];
int n, m;void add(int a, int b)
{ne[idx] = h[a], e[idx] = b, h[a] = idx ++;
}LL bfs(int x, int k)
{mem(st, false);queue<PII> q;q.push({x, 0});LL sum = 0;st[x] = true;while(q.size()){PII t = q.front();q.pop();int x = t.x, y = t.y;// depth ++;if(y <= k) sum += x;for(int i = h[x]; ~i; i = ne[i]){int j = e[i];if(!st[j] && y + 1 <= k) q.push({j, y + 1}), st[j] = true;}}return sum;
}int main()
{ios;mem(h, -1);cin >> n >> m;while(m --){int a, b;cin >> a >> b;add(a, b), add(b, a);}int q, x, k;cin >> q;while(q --){cin >> x >> k;// cout << x << k << endl;cout << bfs(x, k) << endl;}return 0;
}F - Rectangle GCD
题意：给定两个序列：
A1,A2,A3,...,AnA_1,A_2,A_3, ... , A_nA1,A2,A3,...,An
B1,B2,B3,...,BnB_1,B_2,B_3, ... , B_nB1,B2,B3,...,Bn
给两个坐标：(a,b),(c,d)(a, b),(c, d)(a,b),(c,d)，分别为矩阵左上角和右下角（矩阵：Ci,j=Ai+Bj矩阵：C_{i,j} = A_i + B_j矩阵：Ci,j=Ai+Bj）,询问该范围内矩阵的GCDGCDGCD。
思路：
首先一个序列：GCD（a1,a2,...,an）GCD（a_1, a_2,...,a_n）GCD（a1,a2,...,an） = GCD（a1,a2−a1,...,an−an−1）GCD（a_1, a_2 - a_1,...,a_n - a_{n-1}）GCD（a1,a2−a1,...,an−an−1）
即：原序列的最大公约数 = 差分序列的最大公约数。考虑矩阵任一行的数值：
Ai+Bj,Ai+Bj+1,Ai+Bj+2,Ai+Bj+3A_i+B_j,A_i+B_{j+1},A_i+B_{j+2},A_i+B_{j+3}Ai+Bj,Ai+Bj+1,Ai+Bj+2,Ai+Bj+3
做差分变为：
Ai+Bj,Bj+1−Bj,Bj+2−Bj+1,Bj+3−Bj+2A_i+B_j,B_{j+1} - B_{j},B_{j+2} - B_{j+1},B_{j+3} - B_{j + 2}Ai+Bj,Bj+1−Bj,Bj+2−Bj+1,Bj+3−Bj+2
可以发现我们消去了：AiA_iAi
对于查询的矩阵：
黄色部分的GCDGCDGCD为:B的差分序列[b+1,d]B的差分序列[b+1,d]B的差分序列[b+1,d]的gcd1gcd_1gcd1.
左边蓝色的GCDGCDGCD同理为：A的差分序列[a+1,c]A的差分序列[a+1,c]A的差分序列[a+1,c]的gcd2gcd_2gcd2.
综上矩阵的gcdgcdgcd为：GCD(gcd1,gcd2,Aa+Bb−空白快)GCD(gcd_1,gcd_2,A_a+B_b-空白快)GCD(gcd1,gcd2,Aa+Bb−空白快)故可以用两个线段树维护两个序列的差分序列的gcdgcdgcd。
Code:
// Problem: F - Rectangle GCD
// Contest: AtCoder - AtCoder Beginner Contest 254
// URL: https://atcoder.jp/contests/abc254/tasks/abc254_f
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)/*
author: A Fei
*/
#include <bits/stdc++.h>#define x first
#define y second
#define pi acos(-1)
#define endl '\n'
#define ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mem(x, a) memset(x, a, sizeof x)
#define pb(x) push_back(x)
#define rep(i, l, r) for(int i = l; i <= (r); ++ i)
#define per(i, r, l) for(int i = r; i >= (l); -- i)using namespace std;
typedef long long LL;
typedef double DB;
typedef pair<int, int> PII;
typedef pair<int, double> PID;
typedef pair<double, double> PDD;
template <typename T> void inline read(T &x) {int f = 1; x = 0; char s = getchar();while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();x *= f;
}int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
const int INF = 0x3f3f3f3f;
//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
const int N = 2e5 + 10;struct Node
{int l, r;int gcdd;
}tr_a[4 * N], tr_b[4 * N];int n, Q;
int a[N], b[N], da[N], db[N];int gcd(int a, int b)
{return b ? gcd(b, a % b) : a;
}void pushup(Node tr[], int u)
{int l = u << 1, r = u << 1 | 1;tr[u].gcdd = gcd(tr[l].gcdd, tr[r].gcdd);
}void build(Node tr[], int t[], int u, int l, int r)
{tr[u] = {l, r};if(l == r){tr[u] = {l, l, t[l]};// cout << l << ' ' << r << ' ' << t[l] << endl;return ;}int mid = l + r >> 1;build(tr, t, u << 1, l, mid), build(tr, t, u << 1 | 1, mid + 1, r);pushup(tr, u);
}Node query(Node tr[], int u, int l, int r)
{if(l > r) return {0, 0, 0};if(l <= tr[u].l && tr[u].r <= r) return tr[u];int mid = tr[u].l + tr[u].r >> 1;if(r <= mid) return query(tr, u << 1, l, r);else if(l > mid) return query(tr, u << 1 | 1, l, r);else {Node left, right, res;left = query(tr, u << 1, l, r);right = query(tr, u << 1 | 1, l, r);res.gcdd = gcd(left.gcdd, right.gcdd);return res;}
}int main()
{ios;cin >> n >> Q;rep(i, 1, n) cin >> a[i];rep(i, 1, n) cin >> b[i];rep(i, 1, n) da[i] = a[i] - a[i - 1];rep(i, 1, n) db[i] = b[i] - b[i - 1];// rep(i, 1, n) cout << da[i] << ' ';// cout << endl;// rep(i, 1, n) cout << db[i] << ' ';// cout << endl;build(tr_a, da, 1, 1, n);build(tr_b, db, 1, 1, n);while(Q --){int r1, r2, c1, c2;cin >> r1 >> r2 >> c1 >> c2;int g1 = query(tr_b, 1, c1 + 1, c2).gcdd;int g2 = gcd(a[r1] + b[c1], query(tr_a, 1, r1 + 1, r2).gcdd);cout << abs(gcd(g1, g2)) << endl;// cout << a[r1] + b[c1] << ' ' << g1 << ' ' << g2 << endl;}return 0;
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