PAT2019

19.9.8周日下午13:30-16:45

发挥不好。回来把题目重新做一遍。

第一题

7-1 Forever

题目：

“Forever number” is a positive integer A with K digits, satisfying the following constrains:

the sum of all the digits of A is m;

the sum of all the digits of A+1 is n; and

the greatest common divisor of m and n is a prime number which is greater than 2.

Now you are supposed to find these forever numbers.

Input Specification:

Each input file contains one test case. For each test case, the first line contains a positive integer N (≤5). Then N lines follow, each gives a pair of K (3<K<10) and m (1<m<90), of which the meanings are given in the problem description.

Output Specification:

For each pair of K and m, first print in a line Case X, where X is the case index (starts from 1). Then print n and A in the following line. The numbers must be separated by a space. If the solution is not unique, output in the ascending order of n. If still not unique, output in the ascending order of A. If there is no solution, output No Solution.

Sample Input:

2

6 45

7 80

Sample Output:

Case 1

10 189999

10 279999

10 369999

10 459999

10 549999

10 639999

10 729999

10 819999

10 909999

Case 2

No Solution

这题坑最多。

AC代码

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>

#include <algorithm>

#include <cmath>

#include <string>

#include <set>

#include <vector>

#include <map>

using namespace std;

bool isprime(int a)

{

if (a <= 2) return false;

int sqr = sqrt(1.0*a);

for (int i = 2; i <= sqr; i++) if (a%i == 0) return false;

return true;

}

int get_sum(int x) {

int summ = 0;

while (x) {

summ += x % 10;

x /= 10;

}

return summ;

}

int gcd(int a, int b)

{

if (b == 0) return a;

return gcd(b, a%b);

}

int main()

{

int N, k, m, n, t;

cin >> N;

for (int i = 1; i <= N; i++)

{

scanf("%d%d", &k, &m);

int minn=pow(10.0,k-1), maxn = pow(10.0, k), addition, zf = 1;

printf("Case %dn", i);

for (t = 1; t < k; ++t) {

n = m + 1 - t * 9;

if (!isprime(gcd(m, n))) continue;

minn += pow(10.0, t) - 1 ;

addition = pow(10.0, t);

for (int j = minn, cnt = 0; j < maxn; j += addition) {

if (cnt % 10 == 9) { cnt++; continue; }

if (get_sum(j) == m) { zf = 0; printf("%d %dn", n, j); }

cnt++;

}

}

if (zf) { printf("No Solution"); }

}

}

第二题

7-2 Merging Linked Lists

题目：

Given two singly linked lists L1=a1→a2→⋯→an−1→an and L2=b1→b2→⋯→bm−1→b​m.If n≥2m, you are supposed to reverse and merge the shorter one into the longer one to obtain a list like a1→a2→b​m→a3→a4→bm−1⋯. For example, given one list being 6→7 and the other one 1→2→3→4→5, you must output 1→2→7→3→4→6→5.

Input Specification:

Each input file contains one test case. For each case, the first line contains the two addresses of the first nodes of L1and L2, plus a positive N (≤10^​5) which is the total number of nodes given. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Data Next

where Address is the position of the node, Data is a positive integer no more than 10^​5

​​ , and Next is the position of the next node. It is guaranteed that no list is empty, and the longer list is at least twice as long as the shorter one.

Output Specification:

For each case, output in order the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 01000 7

02233 2 34891

00100 6 00001

34891 3 10086

01000 1 02233

00033 5 -1

10086 4 00033

00001 7 -1

Sample Output:

01000 1 02233

02233 2 00001

00001 7 34891

34891 3 10086

10086 4 00100

00100 6 00033

00033 5 -1

签到题难度，AC代码

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>

#include <algorithm>

#include <cmath>

#include <string>

#include <set>

#include <vector>

#include <map>

using namespace std;

struct Node {

int address, data, next;

} node[100001];

int main()

{

int head1, head2, n, address, data, next;

scanf("%d%d%d", &head1, &head2, &n);

for (int i = 0; i < n; i++) {

scanf("%d%d%d", &address, &data, &next);

node[address] = { address, data, next };

}

vector<Node> list1, list2, result;

for (int p = head1; p != -1; p = node[p].next)

list1.push_back(node[p]);

for (int p = head2; p != -1; p = node[p].next)

list2.push_back(node[p]);

if (list1.size() > list2.size()) {

int j = list2.size() - 1;

for (int i = 0; i < list1.size(); i = i + 2) {

result.push_back(list1[i]);

if (i + 1 < list1.size()) result.push_back(list1[i + 1]);

if (j >= 0) result.push_back(list2[j--]);

}

}

else {

int j = list1.size() - 1;

for (int i = 0; i < list2.size(); i = i + 2) {

result.push_back(list2[i]);

if (i + 1 < list2.size()) result.push_back(list2[i + 1]);

if (j >= 0) result.push_back(list1[j--]);

}

}

for (int i = 0; i + 1 < result.size(); i++)

printf("%05d %d %05dn", result[i].address, result[i].data, result[i + 1].address);

printf("%05d %d -1n", result[result.size() - 1].address, result[result.size() - 1].data);

return 0;

}

第三题

7-3 Postfix Expression

题目：
Given a syntax tree (binary), you are supposed to output the corresponding postfix expression, with parentheses reflecting the precedences of the operators.

Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 20) which is the total number of nodes in the syntax tree. Then N lines follow, each gives the information of a node (the i-th line corresponds to the i-th node) in the format:

data left_child right_child
where data is a string of no more than 10 characters, left_child and right_child are the indices of this node’s left and right children, respectively. The nodes are indexed from 1 to N. The NULL link is represented by −1. The figures 1 and 2 correspond to the samples 1 and 2, respectively.

Output Specification:
For each case, print in a line the postfix expression, with parentheses reflecting the precedences of the operators.There must be no space between any symbols.

Sample Input 1:
8

8 7
a -1 -1
4 1
2 5
b -1 -1
d -1 -1
-1 6
c -1 -1
Sample Output 1:
(((a)(b)+)(©(-(d))))

Sample Input 2:
8
2.35 -1 -1

6 1
-1 4
% 7 8
2 3
a -1 -1
str -1 -1
871 -1 -1
Sample Output 2:
(((a)(2.35)*)(-((str)(871)%))+)
 

这题不难，就是最后一个测试点死活过不去。

不知道问题出在哪。出来讨论发现很多人也是卡在最后一个点。

有些人不知道怎么就过了。

我一开始用的方案是重新建一颗标准树。不过好像不必这么麻烦。

下面这个方案是别人的思路。

细节在于，他只对左孩空右孩非空时，先输出根，再访问右孩。

而我在考场写的特判条件是，【根值=='-' 且 左右孩仅有1个非空 】时，先输出根，若左孩非空，访问左，若右孩非空访问右。

博主提到可能存在正负号±，不能只考虑负号。晕了，原地取正是什么操作？

别人的AC代码

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>

#include <algorithm>

#include <cmath>

#include <string>

#include <set>

#include <vector>

#include <map>

using namespace std;

struct Node {

string data;

int lchild, rchild;

} node[21];

bool occured[21] = { false };

void postorder(int root) {

cout << "(";

if (node[root].lchild == -1 && node[root].rchild != -1) {

cout << node[root].data;

postorder(node[root].rchild);

cout << ")";

}

else {

if (node[root].lchild != -1) postorder(node[root].lchild);

if (node[root].rchild != -1) postorder(node[root].rchild);

cout << node[root].data << ")";

}

}

int main()

{

int n, root;

string data;

scanf("%d", &n);

for (int i = 1; i < n + 1; i++) {

cin >> node[i].data >> node[i].lchild >> node[i].rchild;

if (node[i].lchild != -1) occured[node[i].lchild] = true;

if (node[i].rchild != -1) occured[node[i].rchild] = true;

}

for (root = 1; root < n + 1 && occured[root]; root++);

postorder(root);

return 0;

}

第四题

7-4 Dijkstra Sequence

题目：

Dijkstra’s algorithm is one of the very famous greedy algorithms. It is used for solving the single source shortest path problem which gives the shortest paths from one particular source vertex to all the other vertices of the given graph. It was conceived by computer scientist Edsger W. Dijkstra in 1956 and published three years later.

In this algorithm, a set contains vertices included in shortest path tree is maintained. During each step, we find one vertex which is not yet included and has a minimum distance from the source, and collect it into the set. Hence step by step an ordered sequence of vertices, let’s call it Dijkstra sequence, is generated by Dijkstra’s algorithm.

On the other hand, for a given graph, there could be more than one Dijkstra sequence. For example, both { 5, 1, 3, 4, 2 } and { 5, 3, 1, 2, 4 } are Dijkstra sequences for the graph, where 5 is the source. Your job is to check whether a given sequence is Dijkstra sequence or not.

Input Specification:

Each input file contains one test case. For each case, the first line contains two positive integers Nv(≤10^3) and Ne(≤10 ^5), which are the total numbers of vertices and edges, respectively. Hence the vertices are numbered from 1 to Nv. Then Ne lines follow, each describes an edge by giving the indices of the vertices at the two ends, followed by a positive integer weight (≤100) of the edge. It is guaranteed that the given graph is connected.

Finally the number of queries, K, is given as a positive integer no larger than 100, followed by K lines of sequences, each contains a permutationof the Nv vertices. It is assumed that the first vertex is the source for each sequence.

All the inputs in a line are separated by a space.

Output Specification:

For each of the K sequences, print in a line Yes if it is a Dijkstra sequence, or No if not.

Sample Input:

5 7

1 2 2

1 5 1

2 3 1

2 4 1

2 5 2

3 5 1

3 4 1

4

5 1 3 4 2

5 3 1 2 4

2 3 4 5 1

3 2 1 5 4

Sample Output:

Yes

Yes

Yes

No

谁能想到最后一题居然是最简单的题。

而第一个20分题居然是坑最多的题。

如果顺序颠倒过来做就好了。

AC代码

#define _CRT_SECURE_NO_WARNINGS

#include <iostream>

#include <algorithm>

#include <cmath>

#include <string>

#include <set>

#include <vector>

#include <map>

using namespace std;

const int INF = 0x3fffffff;

int n, G[1001][1001], d[1001], query[1001];

bool Dijkstra(int root){

fill(d, d+1001, INF);

bool vis[1001] = {false};

d[root] = 0;

for (int i = 0; i < n; i++){

int u, min = INF;

for (int j = 1; j < n + 1; j++)

if (d[j] < min && !vis[j])

min = d[j];

if (d[query[i]] == min) u = query[i];

else return false;

vis[u] = true;

for (int j = 1; j < n + 1; j++)

if (G[u][j] && !vis[j] && d[j] > d[u] + G[u][j])

d[j] = d[u] + G[u][j];

}

return true;

}

int main()

{

int m, u, v, distance, k;

scanf("%d%d", &n, &m);

for (int i = 0; i < m; i++){

scanf("%d%d%d", &u, &v, &distance);

G[u][v] = G[v][u] = distance;

}

scanf("%d",&k);

for (int i = 0; i < k; i++){

for (int j = 0; j < n; j++) scanf("%d", &query[j]);

bool isD = Dijkstra(query[0]);

printf("%sn", isD ? "Yes" : "No");

}

return 0;

}

参考：//https://blog.csdn.net/weixin_44742521/article/details/100752614

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