C.Cypher Decypher

题意

找 [ i , j ] [i,j] [i,j]区间中有多少个质数

思路

数据范围为 1 ≤ i ≤ j ≤ 1 0 6 1\le i\le j\le 10^6 1≤i≤j≤106
直接素数筛+前缀和即可

代码

/** @Author: Icey_dying* @Date: 2021-10-02 15:02:33* @LastEditors: Icey_dying* @LastEditTime: 2021-10-02 15:09:02* @FilePath: \Icey_dying\competition\2021\2021.10\2021.10.2\C.cpp*/
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1;
int prime[N], b[N];
int cnt = 0, max1 = 1e6;
int init()
{memset(b, 0, sizeof(b));b[0] = b[1] = 1;for (int i = 2; i <= max1; i++) {if (!b[i])prime[++cnt] = i;for (int j = 1; j <= cnt && prime[j] * i <= max1; j++) {b[prime[j] * i] = 1;if (i % prime[j] == 0)break;}}return 0;
}
int a[1000005];
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int t;init();memset(a, 0, sizeof(a));for (int i = 1; i <= max1; i++)a[i] = a[i - 1] + !(b[i]);cin >> t;int i, j;while (t--) {cin >> i >> j;cout << a[j] - a[i - 1] << endl;}return 0;
}

E.Escape Room

题意

地图上有物体，墙，人和出口，现在问你坐标上是否有东西，如果有则输出相应的，如果没有输出如何最快到达出口的方向

思路

从出口BFS即可，注意题目中上左下右有先后顺序，找完之后记得看看有没有更优先的方向

代码

/** @Author: Icey_dying* @Date: 2021-10-04 10:58:31* @LastEditors: Icey_dying* @LastEditTime: 2021-10-04 11:11:05* @FilePath: \Icey_dying\competition\2021\2021.10\2021.10.2\E.cpp*/
#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
char a[N][N], s[N];
int n, m, f[N][N], v[N][N];
int dx[] = { 0, 0, -1, 0, 1 }, dy[] = { 0, 1, 0, -1, 0 };
queue<int> qx, qy;
int x, y, q;
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> n >> m;for (int i = 0; i < N; i++)for (int j = 0; j < N; j++)a[i][j] = '#';for (int i = 1; i <= n; i++) {cin >> a[i] + 1;a[i][m + 1] = '#';}memset(f, 0x3f, sizeof(f));for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {if (a[i][j] == 'E') {f[i][j] = 0;qx.push(i);qy.push(j);}}}while (qx.size()) {x = qx.front();qx.pop();y = qy.front();qy.pop();for (int i = 1; i <= 4; i++) {if (a[x + dx[i]][y + dy[i]] == 'X' || a[x + dx[i]][y + dy[i]] == '#')continue;if (f[x + dx[i]][y + dy[i]] > f[x][y] + 1) {f[x + dx[i]][y + dy[i]] = f[x][y] + 1;v[x + dx[i]][y + dy[i]] = i;qx.push(x + dx[i]);qy.push(y + dy[i]);}}}for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) {for (int k = 1; k <= 4; k++) {if (f[i][j] + 1 == f[i + dx[k]][j + dy[k]] && v[i + dx[k]][j + dy[k]] > k) {v[i + dx[k]][j + dy[k]] = k;}}}}cin >> q;while (q--) {cin >> x >> y;if (a[x][y] == '#')cout << "W\n";else if (a[x][y] == 'X')cout << "X\n";else if (a[x][y] == 'E')cout << "E\n";else if (f[x][y] == f[0][0])cout << "?\n";else {if (v[x][y] == 1)cout << "L\n";else if (v[x][y] == 2)cout << "D\n";else if (v[x][y] == 3)cout << "R\n";elsecout << "U\n";}}return 0;
}

F.Fixing Subtitles

题意

字幕有延迟，先给出原本的字幕时间和字幕本身和延迟时间，让你输出字幕的正确时间和字幕

思路

这道题只有输入输出麻烦，时间的转换是比较简单的，慢慢调试即可

代码

/** @Author: Icey_dying* @Date: 2021-10-02 15:31:31* @LastEditors: Icey_dying* @LastEditTime: 2021-10-02 16:04:50* @FilePath: \Icey_dying\competition\2021\2021.10\2021.10.2\F.cpp*/
#include <bits/stdc++.h>
using namespace std;
int n, x, y;
void f(int& h, int& m, int& s, int& ms)
{ms += y;s += x + ms / 1000;ms %= 1000;m += s / 60;s %= 60;h += m / 60;m %= 60;
}
int h, m, s, ms, now;
string st;
int main()
{scanf("%d %d.%d", &n, &x, &y);for (int i = 1; i <= n; i++) {scanf("%d", &now);printf("%d\n", now);scanf("%d:%d:%d,%d", &h, &m, &s, &ms);f(h, m, s, ms);printf("%02d:%02d:%02d,%03d", h, m, s, ms);scanf(" --> %d:%d:%d,%d", &h, &m, &s, &ms);//cout << ms << endl;f(h, m, s, ms);printf(" --> %02d:%02d:%02d,%03d\n", h, m, s, ms);// getline(cin, st);getchar();while (getline(cin, st)) {if (st.size() == 0) {cout << "\n";break;}cout << st << "\n";}}return 0;
}

J.Just Send the Email

题意

有一棵树（？），找出每个结点到叶子结点的最小距离（叶子结点本身的距离为1），求和，与总结点数相比

思路

用逆向思维，从叶子结点进行BFS即可

代码

/** @Author: Icey_dying* @Date: 2021-10-04 13:10:13* @LastEditors: Icey_dying* @LastEditTime: 2021-10-04 13:28:44* @FilePath: \Icey_dying\competition\2021\2021.10\2021.10.2\J.cpp*/
#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int N = 1e5 + 5;
long long quickmod(long long a, long long b)
{long long ret = 1;while (b) {if (b & 1)ret = ret * a % mod;a = a * a % mod;b /= 2;}return ret;
}
vector<int> v[N];
queue<int> q;
int n;
bool vis[N];
int depth[N];
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);memset(vis, false, sizeof(vis));cin >> n;for (int i = 2, x; i <= n; i++) {cin >> x;vis[x] = true;v[x].push_back(i);v[i].push_back(x);}memset(depth, 0, sizeof(depth));int cnt = n;for (int i = 1; i <= n; i++)if (!vis[i]) {cnt--;q.push(i);depth[i] = 1;}for (int i = 1; i <= n; i++)vis[i] = !vis[i];while (!q.empty()) {int ans = q.front();// cout << ans << endl;q.pop();for (auto i : v[ans]) {if (!vis[i]) {cnt--;q.push(i);vis[i] = true;depth[i] = depth[ans] + 1;}}if (!cnt)break;}// for (int i = 1; i <= n; i++)//     cout << depth[i] << " \n"[i == n];long long wow = 0;for (int i = 1; i <= n; i++)wow += depth[i];// cout << wow << endl;long long hh = n;int ggcd = __gcd(wow, hh);wow /= ggcd;hh /= ggcd;cout << wow % mod * quickmod(hh, mod - 2) % mod << endl;return 0;
}

K.Kids at the Party

题意

Jaime有5个朋友，现有一些糖果，Jaime想要和朋友一起聚会的前提是他和朋友拥有相同的糖果，且聚会至少有2人，问聚会上可以有几个小朋友（包括Jaime）

思路

直接求糖果可以被几整除即可（e.g. 2,3,4,5,6）

代码

/** @Author: Icey_dying* @Date: 2021-10-02 15:15:01* @LastEditors: Icey_dying* @LastEditTime: 2021-10-02 15:24:37* @FilePath: \Icey_dying\competition\2021\2021.10\2021.10.2\K.cpp*/
#include <bits/stdc++.h>
using namespace std;
const int N = 505;
char a[N];
int t;
int b[10];
int main()
{cin >> t;while (t--) {cin >> a;int ans = 0;memset(b, 0, sizeof(b));if ((a[strlen(a) - 1] - '0') % 2 == 0) {b[2] = 1;ans = 0;if (strlen(a) > 1)ans = (a[strlen(a) - 2] - '0') * 10 + a[strlen(a) - 1] - '0';if (strlen(a) == 1 && (a[strlen(a) - 1] - '0') % 4 == 0)b[4] = 1;if (strlen(a) > 1 && ans % 4 == 0)b[4] = 1;}if ((a[strlen(a) - 1] - '0') % 5 == 0)b[5] = 1;ans = 0;for (int i = 0; i < strlen(a); i++) {ans += a[i] - '0';}if (ans % 3 == 0) {b[3] = 1;if (b[2])b[6] = 1;}ans = 0;for (int i = 2; i <= 6; i++)ans += b[i];if (ans) {int cnt = 0;for (int i = 2; i <= 6; i++) {if (b[i]) {cnt++;cout << i << " \n"[cnt == ans];}}} elsecout << "-1\n";}return 0;
}

L.Leonel and the powers of two

题意

问 2 k 2^k 2k的表达式
规则如下：
2 ( k = 1 ) 2(k=1) 2(k=1)
2 ∗ 2 k − 1 2*2^{k-1} 2∗2k−1(k为奇数)
( 2 k 2 ) 2 (2^{\frac{k}{2}})^2 (22k)2(k为偶数)

思路

直接写一个函数，递归调用即可

代码

/** @Author: Icey_dying* @Date: 2021-10-02 14:56:53* @LastEditors: Icey_dying* @LastEditTime: 2021-10-02 15:00:09* @FilePath: \Icey_dying\competition\2021\2021.10\2021.10.2\L.cpp*/
#include <bits/stdc++.h>
using namespace std;
string f(long long n)
{if (n == 1)return "2";string s;if (n % 2)return "(2*" + f(n - 1) + ")";elsereturn "(" + f(n / 2) + ")^2";
}
int t;
long long n;
int main()
{cin >> t;while (t--) {cin >> n;cout << f(n) << endl;}return 0;
}

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