关键在于清空字符数组和使用scanf进行输入 #include <stdio.h> #include <string.h> #include <fstream> #include <iostream> using namespace std; int main() { int num; while (scanf("%d", &num) != EOF) { char earlest[20]; char lastest[20]; ch…
没什么难的,简单模拟题 #include <iostream> using namespace std; int main() { int num; cin>>num; int cost = 0; int curFloor = 0; while (num--) { int floor; cin>>floor; int tmp = floor - curFloor; cost += tmp > 0 ? 6 * tmp : -4 * tmp; cost += 5; c…
利用广度优先搜索,找出每层的叶子节点的个数. #include <iostream> #include <vector> #include <queue> #include <fstream> using namespace std; vector<vector<int>> tree; vector<int> ans; void BFS(int s) { queue<pair<int, int>>…
简单模拟题,注意读懂题意就行 #include <iostream> #include <queue> using namespace std; #define CUSTOMER_MAX 1000+1 #define INF 0x6fffffff #ifndef LOCAL // #define LOCAL #endif LOCAL int n; // number of windows <=20 int m ;// queue capacity <=10 int k;…
转换进制&&逆序可以在一起进行,有一点技巧,不要用十进制数来表示低进制,容易溢出. #include <iostream> #include <vector> using namespace std; bool isPrime(int n) { if(n < 2) return false; if(n == 2) return true; if(n % 2 == 0) return false; for(int i = 3; i < n; i += 2)…
简单模拟题,遍历一遍即可.考察输入输出. #include <iostream> #include <string> #include <stdio.h> #include <iomanip> using namespace std; #define N 3 int main() { char res[3]={'W','T','L'}; char max_res[N]; int i,j; float tmp,sum=1,odd; for(i=0;i<N…
这题给定了一个图,我用DFS的思想,来求出在图中去掉某个点后还剩几个相互独立的区域(连通子图). 在DFS中,每遇到一个未访问的点,则对他进行深搜,把它能访问到的所有点标记为已访问.一共进行了多少次这样的搜索, 就是我们要求的独立区域的个数. #include <iostream> #include <fstream> #include <memory.h> using namespace std; const int maxNum = 1001; bool visit…
PAT (Advanced Level) Practice 1006 Sign In and Sign Out (25 分) 凌宸1642 题目描述: At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of s…
At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked a…
简单题.构造出二叉搜索树,然后check一下. #include<stdio.h> #include<algorithm> using namespace std; +; struct Node { int left; int right; int val; } s[maxn]; int n; int a[maxn]; int ans[maxn],tot; int h[maxn]; int k; void dfs(int x) { ans[k++]=s[x].val; ) dfs(…