1144 1145 1146 1147模拟练习

1144 The Missing Number (20 分)

Sample Input:

10
5 -25 9 6 1 3 4 2 5 17

Sample Output:

7

Note

1. 找到第一个没有出现的整数

Code

#include<iostream>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<string>
#include<fstream>
#include<map>
using namespace std;
const int maxn = 1e6;
int visited[maxn];
int main(){
#ifdef _DEBUG 
	ifstream cin("data.txt");
#endif
	int  num, i;
	cin >> num;
	for(i = 0; i < num; i++){
		int temp;
		cin >> temp;
		if(temp >= 0 && temp < maxn){
			visited[temp] = 1;
		}
	}
	for(i = 1; i < maxn; i++){
		if(visited[i] == 0 ) break;
	}
	cout << i;



#ifdef _DEBUG 
	cin.close();
#endif
	return 0;
}

1145 Hashing - Average Search Time (25 分)

Sample Input:

4 5 4
10 6 4 15 11
11 4 15 2

Sample Output:

15 cannot be inserted.
2.8

Note

1. 题意输入 tablesize num queries，下一行存入myhash，存不进去的记 cannot be inserted， 最后一行输入查询序列，输出平均查询长度
2. 注意： 查找元素时，如果位置上没有元素则查询结束

Code

#include<iostream>
#include<algorithm>
#include<vector>
#include<fstream>
#include<queue>
#include<string>
#include<cstring>
#include<cmath>
#include<map>
using namespace std;
const int maxn = 1e7;
bool isprime(int n ){
	if (n <= 1 ) return 0;
	for(int i = 2; i <= sqrt(n); i++){
		if(n % i == 0) return 0;
	}
	return 1;
}
int main(){ 
#ifdef _DEBUG
	ifstream cin("data2.txt");
#endif
    int num, i, tablesize, quries;
	cin >> tablesize >> num >> quries;
	while(isprime(tablesize) == 0) tablesize++;
	vector<int> myhash(tablesize, -1);
	for(i = 0; i < num; i++){
		int temp;
		cin >> temp;
		int key = temp % tablesize, cnt = 1;
		while(myhash[key] != -1 && cnt < tablesize){
			key =( temp % tablesize + cnt * cnt) % tablesize;
			cnt++;
		}
		if(cnt < tablesize) myhash[key] = temp;
		else 	if(cnt == tablesize) 
			printf("%d cannot be inserted.\n", temp);
	}
	int finalcnt = 0;
	for(i = 0; i < quries; i++){
		int temp;
		cin >> temp;
		int key = temp % tablesize, cnt = 1;
		finalcnt++;
		while(myhash[key] != temp && cnt <= tablesize && myhash[key] != -1){
			key =( temp % tablesize + cnt * cnt) % tablesize;
			cnt++;
			finalcnt++;
		}
	
	}
	printf("%.1lf", finalcnt * 1.0 / quries);
#ifdef _DEBUG
	 cin.close();
#endif
	return 0;
}

1146 Topological Order (25 分)

Sample Input:

6 8
1 2
1 3
5 2
5 4
2 3
2 6
3 4
6 4
5
1 5 2 3 6 4
5 1 2 6 3 4
5 1 2 3 6 4
5 2 1 6 3 4
1 2 3 4 5 6

Sample Output:

3 4

Note

1. 题意： 输入有向图，问queries个序列是否为拓扑排序。
2. 思路： 记录每个节点的入度indegree（num), 以及每个节点的出边inta[maxn]，检测每个节点输出时入度是否为零，并将该节点的出边节点入度–
3. 敲的过程中又犯了最基础的输入没完break的问题。。。

Code

#include<iostream>
#include<vector>
#include<map>
#include<fstream>
using namespace std;
const int maxn = 1e6;
vector<int> a[maxn];
vector<int> indegree;
int main(){ 
#ifdef _DEBUG
	ifstream cin("data3.txt");
#endif
    int i, j, k, num, edge, queries, cnt = 0, tempa, tempb;
	cin >> num >> edge;
	indegree.resize(num + 1);
	for(i = 0; i < edge; i++){
		cin >> tempa >> tempb;
		indegree[tempb]++;
		a[tempa].push_back(tempb);
	}
	cin >> queries;
	for(i = 0; i < queries; i++){
		vector <int> tempdegree, save(num);
		tempdegree = indegree;
        for(j = 0; j < num; j++)
            cin >> save[j];
		for(j = 0; j < num; j++){
			tempa = save[j];
			if(tempdegree[tempa] != 0 ) {
				if(cnt == 0) cout << i;
				else cout  << " " << i;
				cnt++;
				break;
			}
			for(int k = 0; k < a[tempa].size(); k++){
				tempdegree[a[tempa][k]]--;
			}
		}
	} 
#ifdef _DEBUG
	cin.close();
#endif
	return 0;
}

1147 Heaps (30 分)

Sample Input:

3 8
98 72 86 60 65 12 23 50
8 38 25 58 52 82 70 60
10 28 15 12 34 9 8 56

Sample Output:

Max Heap
50 60 65 72 12 23 86 98
Min Heap
60 58 52 38 82 70 25 8
Not Heap
56 12 34 28 9 8 15 10

Note

1. 题意：判断是否为大顶堆小顶堆，输入的是层次遍历结果，输出后序遍历结果
2. 思路： 判断是否为堆利用了insert的主要逻辑 a[i] > a[i/2],BST的层次序列转后序，利用递归左右子树刚好为 *2 和 *2 + 1
3. 近些年最后的题目码量上都相对变少，主要还是考的思路和简化

Code

#include<iostream>
#include<algorithm>
#include<vector>
#include<fstream>
using namespace std;
void PreTraverse(vector <int> a, int index){   
    if(index >= a.size()) return;
    PreTraverse(a, index * 2 );
    PreTraverse(a, index * 2 +1);
    if(index != 1) cout << a[index] << " ";
	else   cout << a[index] << endl;
}
int main(){ 
#ifdef _DEBUG
	ifstream cin("data4.txt");
#endif
    int treenum, num, i, j, index;
    cin >> treenum >> num;
    while(treenum--){   
        vector <int> a(num + 1);
        int flag = 0;
        for(i = 1; i <= num; i++) cin >> a[i];
        if(a[1] > a[num] ){
            flag = 1;    
            for(j = 2; j <= num; j++) 
                if(a[j] >a[j/2]){
                    flag = 0;
                    break;
                }
        }else{   
            flag = 2;
            for(j = 2; j <= num; j++) 
                if(a[j] < a[j/2]){
                    flag = 0;
                    break;
            }
        }
        if(flag == 0) cout << "Not Heap";
        else if(flag == 1) cout << "Max Heap";
        else cout << "Min Heap";
		cout << endl;
        PreTraverse(a, 1);
    }
#ifdef _DEBUG
	cin.close();
#endif
	return 0;
}

Summary

1. ( temp % tablesize + cnt * cnt) % tablesize与(temp + cnt * cnt ) % tablesize;相等吗，为什么两种写法都可过呢
2. 前中后层次序列之间的转换：