PAT – 第32页

L2-004. 这是二叉搜索树吗？-PAT团体程序设计天梯赛GPLT

一棵二叉搜索树可被递归地定义为具有下列性质的二叉树：对于任一结点，

其左子树中所有结点的键值小于该结点的键值；
其右子树中所有结点的键值大于等于该结点的键值；
其左右子树都是二叉搜索树。
所谓二叉搜索树的“镜像”，即将所有结点的左右子树对换位置后所得到的树。

给定一个整数键值序列，现请你编写程序，判断这是否是对一棵二叉搜索树或其镜像进行前序遍历的结果。

输入格式：

输入的第一行给出正整数N（<=1000）。随后一行给出N个整数键值，其间以空格分隔。

输出格式：

如果输入序列是对一棵二叉搜索树或其镜像进行前序遍历的结果，则首先在一行中输出“YES”，然后在下一行输出该树后序遍历的结果。数字间有1个空格，一行的首尾不得有多余空格。若答案是否，则输出“NO”。

输入样例1：
7
8 6 5 7 10 8 11
输出样例1：
YES
5 7 6 8 11 10 8
输入样例2：
7
8 10 11 8 6 7 5
输出样例2：
YES
11 8 10 7 5 6 8
输入样例3：
7
8 6 8 5 10 9 11
输出样例3：
NO

分析：假设它是二叉搜索树，一开始isMirror为FALSE，根据二叉搜索树的性质将已知的前序转换为后序，转换过程中，如果发现最后输出的后序数组长度不为n，那就设isMirror为true，然后清空后序数组，重新再转换一次（根据镜面二叉搜索树的性质），如果依旧转换后数组大小不等于n，就输出no否则输出yes

#include <cstdio>
#include <vector>
using namespace std;
bool isMirror;
vector<int> pre, post;
void getpost(int root, int tail) {
    if(root > tail) return ;
    int i = root + 1, j = tail;
    if(!isMirror) {
        while(i <= tail && pre[root] > pre[i]) i++;
        while(j > root && pre[root] <= pre[j]) j--;
    } else {
        while(i <= tail && pre[root] <= pre[i]) i++;
        while(j > root && pre[root] > pre[j]) j--;
    }
    if(i - j != 1) return ;
    getpost(root + 1, j);
    getpost(i, tail);
    post.push_back(pre[root]);
}
int main() {
    int n;
    scanf("%d", &n);
    pre.resize(n);
    for(int i = 0; i < n; i++)
        scanf("%d", &pre[i]);
    getpost(0, n - 1);
    if(post.size() != n) {
        isMirror = true;
        post.clear();
        getpost(0, n - 1);
    }
    if(post.size() == n) {
        printf("YES\n%d", post[0]);
        for(int i = 1; i < n; i++)
            printf(" %d", post[i]);
    } else {
        printf("NO");
    }
    return 0;
}

#include <cstdio>

#include <vector>

using namespace std;

bool isMirror;

vector<int> pre, post;

void getpost(int root, int tail) {

if(root > tail) return ;

int i = root + 1, j = tail;

if(!isMirror) {

while(i <= tail && pre[root] > pre[i]) i++;

while(j > root && pre[root] <= pre[j]) j--;

} else {

while(i <= tail && pre[root] <= pre[i]) i++;

while(j > root && pre[root] > pre[j]) j--;

}

if(i - j != 1) return ;

getpost(root + 1, j);

getpost(i, tail);

post.push_back(pre[root]);

}

int main() {

int n;

scanf("%d", &n);

pre.resize(n);

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

scanf("%d", &pre[i]);

getpost(0, n - 1);

if(post.size() != n) {

isMirror = true;

post.clear();

getpost(0, n - 1);

}

if(post.size() == n) {

printf("YES\n%d", post[0]);

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

printf(" %d", post[i]);

} else {

printf("NO");

}

return 0;

}

1043. Is It a Binary Search Tree (25)-PAT甲级真题

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
Both the left and right subtrees must also be binary search trees.
If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a line “YES” if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or “NO” if not. Then if the answer is “YES”, print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:
7
8 6 5 7 10 8 11
Sample Output 1:
YES
5 7 6 8 11 10 8
Sample Input 2:
7
8 10 11 8 6 7 5
Sample Output 2:
YES
11 8 10 7 5 6 8
Sample Input 3:
7
8 6 8 5 10 9 11
Sample Output 3:
NO
题目大意：给定一个整数键值序列，现请你编写程序，判断这是否是对一棵二叉搜索树或其镜像进行前序遍历的结果。
分析：假设它是二叉搜索树，一开始isMirror为false，根据二叉搜索树的性质将已知的前序转换为后序，转换过程中，如果发现最后输出的后序数组长度不为n，那就设isMirror为true，然后清空后序数组，重新再转换一次（根据镜面二叉搜索树的性质），如果依旧转换后数组大小不等于n，就输出NO否则输出YES

#include <cstdio>
#include <vector>
using namespace std;
bool isMirror;
vector<int> pre, post;
void getpost(int root, int tail) {
    if(root > tail) return ;
    int i = root + 1, j = tail;
    if(!isMirror) {
        while(i <= tail && pre[root] > pre[i]) i++;
        while(j > root && pre[root] <= pre[j]) j--;
    } else {
        while(i <= tail && pre[root] <= pre[i]) i++;
        while(j > root && pre[root] > pre[j]) j--;
    }
    if(i - j != 1) return ;
    getpost(root + 1, j);
    getpost(i, tail);
    post.push_back(pre[root]);
}
int main() {
    int n;
    scanf("%d", &n);
    pre.resize(n);
    for(int i = 0; i < n; i++)
        scanf("%d", &pre[i]);
    getpost(0, n - 1);
    if(post.size() != n) {
        isMirror = true;
        post.clear();
        getpost(0, n - 1);
    }
    if(post.size() == n) {
        printf("YES\n%d", post[0]);
        for(int i = 1; i < n; i++)
            printf(" %d", post[i]);
    } else {
        printf("NO");
    }
    return 0;
}

#include <cstdio>

#include <vector>

using namespace std;

bool isMirror;

vector<int> pre, post;

void getpost(int root, int tail) {

if(root > tail) return ;

int i = root + 1, j = tail;

if(!isMirror) {

while(i <= tail && pre[root] > pre[i]) i++;

while(j > root && pre[root] <= pre[j]) j--;

} else {

while(i <= tail && pre[root] <= pre[i]) i++;

while(j > root && pre[root] > pre[j]) j--;

}

if(i - j != 1) return ;

getpost(root + 1, j);

getpost(i, tail);

post.push_back(pre[root]);

}

int main() {

int n;

scanf("%d", &n);

pre.resize(n);

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

scanf("%d", &pre[i]);

getpost(0, n - 1);

if(post.size() != n) {

isMirror = true;

post.clear();

getpost(0, n - 1);

}

if(post.size() == n) {

printf("YES\n%d", post[0]);

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

printf(" %d", post[i]);

} else {

printf("NO");

}

return 0;

}

1023. Have Fun with Numbers (20)-PAT甲级真题（大整数运算）

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number consisting exactly the numbers from 1 to 9, only in a different permutation. Check to see the result if we double it again!

Now you are suppose to check if there are more numbers with this property. That is, double a given number with k digits, you are to tell if the resulting number consists of only a permutation of the digits in the original number.

Input Specification:

Each input file contains one test case. Each case contains one positive integer with no more than 20 digits.

Output Specification:

For each test case, first print in a line “Yes” if doubling the input number gives a number that consists of only a permutation of the digits in the original number, or “No” if not. Then in the next line, print the doubled number.

Sample Input:
1234567899
Sample Output:
Yes
2469135798

题目大意：给出一个长度不超过20的整数，问这个整数两倍后的数位是否为原数位的一个排列。不管是yes还是no最后都要输出整数乘以2的结果
分析：使用char数组存储这个数，没个数的数位乘以2 + 进位，同时设立book来标记数位出现的情况。只有最后book的每个元素都是0的时候才说明这两个数字是相等的一个排列结果～～

#include <cstdio>
#include <string.h>
using namespace std;
int book[10];
int main() {
    char num[22];
    scanf("%s", num);
    int flag = 0, len = strlen(num);
    for(int i = len - 1; i >= 0; i--) {
        int temp = num[i] - '0';
        book[temp]++;
        temp = temp * 2 + flag;
        flag = 0;
        if(temp >= 10) {
            temp = temp - 10;
            flag = 1;
        }
        num[i] = (temp + '0');
        book[temp]--;
    }
    int flag1 = 0;
    for(int i = 0; i < 10; i++) {
        if(book[i] != 0)
            flag1 = 1;
    }
    printf("%s", (flag == 1 || flag1 == 1) ? "No\n" : "Yes\n");
    if(flag == 1) printf("1");
    printf("%s", num);
    return 0;
}

#include <cstdio>

#include <string.h>

using namespace std;

int book[10];

int main() {

char num[22];

scanf("%s", num);

int flag = 0, len = strlen(num);

for(int i = len - 1; i >= 0; i--) {

int temp = num[i] - '0';

book[temp]++;

temp = temp * 2 + flag;

flag = 0;

if(temp >= 10) {

temp = temp - 10;

flag = 1;

}

num[i] = (temp + '0');

book[temp]--;

}

int flag1 = 0;

for(int i = 0; i < 10; i++) {

if(book[i] != 0)

flag1 = 1;

}

printf("%s", (flag == 1 || flag1 == 1) ? "No\n" : "Yes\n");

if(flag == 1) printf("1");

printf("%s", num);

return 0;

}

1015. Reversible Primes (20)-PAT甲级真题

A reversible prime in any number system is a prime whose “reverse” in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (< 105) and D (1 < D <= 10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line “Yes” if N is a reversible prime with radix D, or “No” if not.

Sample Input:
73 10
23 2
23 10
-2
Sample Output:
Yes
Yes
No
题目大意：如果一个数本身是素数，而且在d进制下反转后的数在十进制下也是素数，就输出Yes，否则就输出No
分析：判断输入是否为负数，判断n是否为素数，把n转换为d进制再反过来转换为10进制，判断是否为素数

#include <cstdio>
#include <cmath>
using namespace std;
bool isprime(int n) {
    if(n <= 1) return false;
    int sqr = int(sqrt(n * 1.0));
    for(int i = 2; i <= sqr; i++) {
        if(n % i == 0)
            return false;
    }
    return true;
}
int main() {
    int n, d;
    while(scanf("%d", &n) != EOF) {
        if(n < 0) break;
        scanf("%d", &d);
        if(isprime(n) == false) {
            printf("No\n");
            continue;
        }
        int len = 0, arr[100];
        do{
            arr[len++] = n % d;
            n = n / d;
        }while(n != 0);
        for(int i = 0; i < len; i++)
            n = n * d + arr[i];
        printf("%s", isprime(n) ? "Yes\n" : "No\n");
    }
    return 0;
}

#include <cstdio>

#include <cmath>

using namespace std;

bool isprime(int n) {

if(n <= 1) return false;

int sqr = int(sqrt(n * 1.0));

for(int i = 2; i <= sqr; i++) {

if(n % i == 0)

return false;

}

return true;

}

int main() {

int n, d;

while(scanf("%d", &n) != EOF) {

if(n < 0) break;

scanf("%d", &d);

if(isprime(n) == false) {

printf("No\n");

continue;

}

int len = 0, arr[100];

do{

arr[len++] = n % d;

n = n / d;

}while(n != 0);

for(int i = 0; i < len; i++)

n = n * d + arr[i];

printf("%s", isprime(n) ? "Yes\n" : "No\n");

}

return 0;

}

1047. Student List for Course (25)-PAT甲级真题

Zhejiang University has 40000 students and provides 2500 courses. Now given the registered course list of each student, you are supposed to output the student name lists of all the courses.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (<=40000), the total number of students, and K (<=2500), the total number of courses. Then N lines follow, each contains a student’s name (3 capital English letters plus a one-digit number), a positive number C (<=20) which is the number of courses that this student has registered, and then followed by C course numbers. For the sake of simplicity, the courses are numbered from 1 to K.

Output Specification:

For each test case, print the student name lists of all the courses in increasing order of the course numbers. For each course, first print in one line the course number and the number of registered students, separated by a space. Then output the students’ names in alphabetical order. Each name occupies a line.

Sample Input:
10 5
ZOE1 2 4 5
ANN0 3 5 2 1
BOB5 5 3 4 2 1 5
JOE4 1 2
JAY9 4 1 2 5 4
FRA8 3 4 2 5
DON2 2 4 5
AMY7 1 5
KAT3 3 5 4 2
LOR6 4 2 4 1 5
Sample Output:
1 4
ANN0
BOB5
JAY9
LOR6
2 7
ANN0
BOB5
FRA8
JAY9
JOE4
KAT3
LOR6
3 1
BOB5
4 7
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1
5 9
AMY7
ANN0
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1
题目大意：给出选课人数和课程数目，然后再给出每个人的选课情况，请针对每门课程输出选课人数以及所有选该课的学生姓名，按照字典序～
分析：建立int的二维数组，course[i][j] = k表示第i号课程上的人的一个列表，k是上这个课的学生的姓名所在的字符数组name[i][j]的i下标～～
注意：strcmp返回的不一定是-1，0，1这几个数字，要返回bool变量还需要在后面添加strcmp是大于0还是小于0～～

#include <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
using namespace std;
char name[40010][5];
vector<int> course[2510];
bool cmp1(int a, int b) {
    return strcmp(name[a], name[b]) < 0;
}
int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    for(int i = 0; i < n; i++) {
        int cnt, temp;
        scanf("%s %d", name[i], &cnt);
        for(int j = 0; j < cnt; j++) {
            scanf("%d", &temp);
            course[temp].push_back(i);
        }
    }
    for(int i = 1; i <= k; i++) {
        printf("%d %d\n", i, course[i].size());
        sort(course[i].begin(), course[i].end(), cmp1);
        for(int j = 0; j < course[i].size(); j++)
            printf("%s\n", name[course[i][j]]);
    }
    return 0;
}

#include <iostream>

#include <vector>

#include <algorithm>

#include <string.h>

using namespace std;

char name[40010][5];

vector<int> course[2510];

bool cmp1(int a, int b) {

return strcmp(name[a], name[b]) < 0;

}

int main() {

int n, k;

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

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

int cnt, temp;

scanf("%s %d", name[i], &cnt);

for(int j = 0; j < cnt; j++) {

scanf("%d", &temp);

course[temp].push_back(i);

}

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

printf("%d %d\n", i, course[i].size());

sort(course[i].begin(), course[i].end(), cmp1);

for(int j = 0; j < course[i].size(); j++)

printf("%s\n", name[course[i][j]]);

}

return 0;

}

1039. Course List for Student (25)-PAT甲级真题

Zhejiang University has 40000 students and provides 2500 courses. Now given the student name lists of all the courses, you are supposed to output the registered course list for each student who comes for a query.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=40000), the number of students who look for their course lists, and K (<=2500), the total number of courses. Then the student name lists are given for the courses (numbered from 1 to K) in the following format: for each course i, first the course index i and the number of registered students Ni (<= 200) are given in a line. Then in the next line, Ni student names are given. A student name consists of 3 capital English letters plus a one-digit number. Finally the last line contains the N names of students who come for a query. All the names and numbers in a line are separated by a space.

Output Specification:

For each test case, print your results in N lines. Each line corresponds to one student, in the following format: first print the student’s name, then the total number of registered courses of that student, and finally the indices of the courses in increasing order. The query results must be printed in the same order as input. All the data in a line must be separated by a space, with no extra space at the end of the line.

Sample Input:
11 5
4 7
BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
1 4
ANN0 BOB5 JAY9 LOR6
2 7
ANN0 BOB5 FRA8 JAY9 JOE4 KAT3 LOR6
3 1
BOB5
5 9
AMY7 ANN0 BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
ZOE1 ANN0 BOB5 JOE4 JAY9 FRA8 DON2 AMY7 KAT3 LOR6 NON9
Sample Output:
ZOE1 2 4 5
ANN0 3 1 2 5
BOB5 5 1 2 3 4 5
JOE4 1 2
JAY9 4 1 2 4 5
FRA8 3 2 4 5
DON2 2 4 5
AMY7 1 5
KAT3 3 2 4 5
LOR6 4 1 2 4 5
NON9 0
题目大意：有N个学生，K门课，给出选择每门课的学生姓名，最后对于给出的N个学生的姓名的选课情况进行询问，要求按顺序给出每个学生的选课情况～
分析：考虑到string、cin、cout会超时，可以使用hash(26*26*26*10+10)将学生姓名变为int型，然后存储在vector里面～～

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int getid(char *name) {
    int id = 0;
    for(int i = 0; i < 3; i++)
        id = 26 * id + (name[i] - 'A');
    id = id * 10 + (name[3] - '0');
    return id;
}
const int maxn = 26 * 26 * 26 * 10 + 10;
vector<int> v[maxn];

int main() {
    int n, k, no, num, id = 0;
    char name[5];
    scanf("%d %d", &n, &k);
    for(int i = 0; i < k; i++) {
        scanf("%d %d", &no, &num);
        for(int j = 0; j < num; j++) {
            scanf("%s", name);
            id = getid(name);
            v[id].push_back(no);
        }
    }
    for(int i = 0; i < n; i++) {
        scanf("%s", name);
        id = getid(name);
        sort(v[id].begin(), v[id].end());
        printf("%s %lu", name, v[id].size());
        for(int j = 0; j < v[id].size(); j++)
            printf(" %d", v[id][j]);
        printf("\n");
    }
    return 0;
}

#include <cstdio>

#include <vector>

#include <algorithm>

using namespace std;

int getid(char *name) {

int id = 0;

for(int i = 0; i < 3; i++)

id = 26 * id + (name[i] - 'A');

id = id * 10 + (name[3] - '0');

return id;

}

const int maxn = 26 * 26 * 26 * 10 + 10;

vector<int> v[maxn];

int main() {

int n, k, no, num, id = 0;

char name[5];

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

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

scanf("%d %d", &no, &num);

for(int j = 0; j < num; j++) {

scanf("%s", name);

id = getid(name);

v[id].push_back(no);

}

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

scanf("%s", name);

id = getid(name);

sort(v[id].begin(), v[id].end());

printf("%s %lu", name, v[id].size());

for(int j = 0; j < v[id].size(); j++)

printf(" %d", v[id][j]);

printf("\n");

}

return 0;

}