drawAgirl的博客

n=int(input())for i in range(n): data=list(map(int,input().split())) if data[0]+data[1]>data[2]: print('Case #%d: true'%(i+1)) else: print('Case #%d: false'%(i+1))上面是python的，下面是c++的，c同样...

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <stdlib.h>#include <vector>using namespace std;struct student{ char...

记录每个位置之前和之后的最大最小值。#include <cstdio>#include <algorithm>#include <vector>using namespace std;vector<int> v;int main(){ int n; scanf("%d",&n); for (int ...

#include <stdio.h>int hash_table[130] = {0};int main(){// printf("hello\n"); char s1[90]; char s2[90]; gets(s1); gets(s2); for (int i=0; i<strlen(s2); i++) { ...

我是用哈希表做的。有几个细节要注意下的，a的范围。#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <stdio.h>#include <cstring>#include <stack&g

就是基础的avl操作#include <cstdio>#include <iostream>#include <cmath>#include <algorithm>#include <vector>#include <queue>using namespace std;typedef struct n...

leetcode上有一题和这个基本一样，感觉这题有点水。#include <cstdio>#include <iostream>#include <cmath>#include <algorithm>#include <vector>#include <queue>using namespace std;...

这个和甲级的一道题重复了。今天真的不适合写代码吧，一个下午就ac了3道题，头很大。这道题有点像leetcode朋友圈，应该可以用并查集做，我用了map的方法，具体看代码吧。#include <iostream>#include <vector>#include <algorithm>#include <cstdio>#inclu...

这题可以用双向队列，很容易就解决！具体看代码#include <iostream>#include <vector>#include <algorithm>#include <cstdio>#include <string>#include <queue>#include <unordered_se...

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <stdlib.h>#include <vector>using namespace std;double first[3];double ...

用了哈希表的思想，但最后一个测试点一直没过，为什么啊啊啊啊啊#include <cstdio>#include <stdio.h>#include <iostream>#include <algorithm>#include <cstring>#include <queue>using namespace s...

#include <cstdio>#include <iostream>#include <cstring>int cards[55];void init(){ for (int i=1; i<55; i++) cards[i] = i;}char shape[] = {'S','H','C','D','J'};...

#include <iostream>#include <cstdio>#include <string>#include <algorithm>using namespace std;bool cmp(int a,int b){ return a>b;}void to_array(int n,int num[]){...

我是先复原了二叉树，再对其进行遍历。这里有个技巧，%*c可以读取换行符，然后存储到一个结构数组里，方便后面构造二叉树，这里读取的时候要左右颠倒，因为题目就是要翻转二叉树，干脆读取的时候就翻转了。后面就很基础了。#include <cstdio>#include <iostream>#include <vector>#include <...

用了两种方法，一个是复原了二叉树，一个是直接根据前中序求后序。#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <stdio.h>#include <cstring>#include <s...

#include <iostream>#include <cstdio>#include <cmath>#include <vector>#include <algorithm>using namespace std;const int MAXN = 10010;vector<int> v;int ar...

静态数组，不多说了#include <iostream>#include <cstdio>#include <cmath>#include <vector>#include <algorithm>#include <queue>using namespace std;const int MAXN = 1...

调了好久，最后一个点一直超时，借鉴了下别人的。#include <iostream>#include <cstdio>#include <cmath>#include <vector>#include <algorithm>#include <queue>#include <set>#incl

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <stdio.h>#include <cstring>#include <stack>#include <queue&a