文章目录
*4.6 PAT B1030/A1085
B1030
给定一个正整数数列，和正整数 p，设这个数列中的最大值是 M，最小值是 m，如果 M≤m**p，则称这个数列是完美数列。
现在给定参数 p 和一些正整数，请你从中选择尽可能多的数构成一个完美数列。
输入格式：
输入第一行给出两个正整数 N 和 p，其中 N（≤105）是输入的正整数的个数，p（≤109）是给定的参数。第二行给出 N 个正整数，每个数不超过 109。
输出格式：
在一行中输出最多可以选择多少个数可以用它们组成一个完美数列。
输入样例：
10 8
2 3 20 4 5 1 6 7 8 9
输出样例：
8
A1085
Given a sequence of positive integers and another positive integer p. The sequence is said to be a perfect sequence if M≤m×p where M and m are the maximum and minimum numbers in the sequence, respectively.
Now given a sequence and a parameter p, you are supposed to find from the sequence as many numbers as possible to form a perfect subsequence.
Input Specification:
Each input file contains one test case. For each case, the first line contains two positive integers N and p, where N (≤105) is the number of integers in the sequence, and p (≤109) is the parameter. In the second line there are N positive integers, each is no greater than 109.
Output Specification:
For each test case, print in one line the maximum number of integers that can be chosen to form a perfect subsequence.
Sample Input:
10 8
2 3 20 4 5 1 6 7 8 9
Sample Output:
8
//双指针法
#include<stdio.h>
#include<algorithm>
using namespace std;
int main() {
int n, p, arr[100010];
scanf("%d%d", &n, &p);
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}
sort(arr, arr + n);
int i = 0, j = 0, count = 1;
while (i < n && j < n) {
while (j < n && arr[j] <= (long long)arr[i] * p) {
count = max(count, j - i + 1);
j++;
}
i++;
}
printf("%d", count);
return 0;
}
? *4.6 B1035/A1089
B1035
根据维基百科的定义：
插入排序是迭代算法，逐一获得输入数据，逐步产生有序的输出序列。每步迭代中，算法从输入序列中取出一元素，将之插入有序序列中正确的位置。如此迭代直到全部元素有序。
归并排序进行如下迭代操作：首先将原始序列看成 N 个只包含 1 个元素的有序子序列，然后每次迭代归并两个相邻的有序子序列，直到最后只剩下 1 个有序的序列。
现给定原始序列和由某排序算法产生的中间序列，请你判断该算法究竟是哪种排序算法？
输入格式：
输入在第一行给出正整数 N (≤100)；随后一行给出原始序列的 N 个整数；最后一行给出由某排序算法产生的中间序列。这里假设排序的目标序列是升序。数字间以空格分隔。
输出格式：
首先在第 1 行中输出Insertion Sort表示插入排序、或Merge Sort表示归并排序；然后在第 2 行中输出用该排序算法再迭代一轮的结果序列。题目保证每组测试的结果是唯一的。数字间以空格分隔，且行首尾不得有多余空格。
输入样例 1：
10
3 1 2 8 7 5 9 4 6 0
1 2 3 7 8 5 9 4 6 0
输出样例 1：
Insertion Sort
1 2 3 5 7 8 9 4 6 0
输入样例 2：
10
3 1 2 8 7 5 9 4 0 6
1 3 2 8 5 7 4 9 0 6
输出样例 2：
Merge Sort
1 2 3 8 4 5 7 9 0 6
A1089
According to Wikipedia:
Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there. It repeats until no input elements remain.
Merge sort works as follows: Divide the unsorted list into N sublists, each containing 1 element (a list of 1 element is considered sorted). Then repeatedly merge two adjacent sublists to produce new sorted sublists until there is only 1 sublist remaining.
Now given the initial sequence of integers, together with a sequence which is a result of several iterations of some sorting method, can you tell which sorting method we are using?
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤100). Then in the next line, N integers are given as the initial sequence. The last line contains the partially sorted sequence of the N numbers. It is assumed that the target sequence is always ascending. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in the first line either “Insertion Sort” or “Merge Sort” to indicate the method used to obtain the partial result. Then run this method for one more iteration and output in the second line the resuling sequence. It is guaranteed that the answer is unique for each test case. 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:
10
3 1 2 8 7 5 9 4 6 0
1 2 3 7 8 5 9 4 6 0
Sample Output 1:
Insertion Sort
1 2 3 5 7 8 9 4 6 0
Sample Input 2:
10
3 1 2 8 7 5 9 4 0 6
1 3 2 8 5 7 4 9 0 6
Sample Output 2:
Merge Sort
1 2 3 8 4 5 7 9 0 6
#include<stdio.h>
#include<algorithm>
using namespace std;
int origin[111], temparr[111], changed[111];
int n;
bool isSame(int a[], int b[]) {
for (int i = 0; i < n; i++) {
if (a[i] != b[i])
return false;
}
return true;
}
bool showArray(int a[]) {
for (int i = 0; i < n; i++) {
printf("%d", a[i]);
if (i < n - 1)
printf(" ");
}
printf("\n");
}
bool insertSort() {
bool flag = false;
for (int i = 1; i < n; i++) {
if (i != 1 && isSame(temparr, changed)) {
flag = true;
}
int temp = temparr[i], j = i;
while (j > 0 && temparr[j - 1] > temp) {
temparr[j] = temparr[j - 1];
j--;
}
temparr[j] = temp;
if (flag == true) {
return true;
}
}
return false;
}
void mergeSort() {
bool flag = false;
for (int step = 2; step / 2 <= n; step *= 2) {
if (step != 2 && isSame(temparr, changed)) {
flag = true;
}
for (int i = 0; i < n; i += step) {
sort(temparr + i, temparr + min(i + step, n));
}
if (flag == true) {
showArray(temparr);
return;
}
}
}
int main() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &origin[i]);
temparr[i] = origin[i];
}
for (int i = 0; i < n; i++) {
scanf("%d", &changed[i]);
}
if (insertSort()) {
printf("Insertion Sort\n");
showArray(temparr);
}
else {
printf("Merge Sort\n");
for (int i = 0; i < n; i++) {
temparr[i] = origin[i];
}
mergeSort();
}
return 0;
}
*4.6 PAT A1029
Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.
Given two increasing sequences of integers, you are asked to find their median.
Input Specification:
Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (≤2×105) is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.
Output Specification:
For each test case you should output the median of the two given sequences in a line.
Sample Input:
4 11 12 13 14
5 9 10 15 16 17
Sample Output:
13
#include<stdio.h>
int arr1[1000010], arr2[1000010];
int main() {
int n, m ;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &arr1[i]);
}
scanf("%d", &m);
for (int i = 0; i < m; i++) {
scanf("%d", &arr2[i]);
}
//不加这一行,会发生数组越界的问题,0x7fffffff是int型最大值
arr1[n] = arr2[m] = 0x7fffffff;
int mid = (n + m - 1) / 2;
int i = 0, j = 0;
while (i + j < mid) {
if (arr1[i] < arr2[j]) {
i++;
}
else {
j++;
}
}
if (arr1[i] < arr2[j])
{
printf("%d", arr1[i]);
}
else {
printf("%d", arr2[j]);
}
return 0;
}
*4.6 PAT B1040/A1093
B1040
字符串 APPAPT 中包含了两个单词 PAT，其中第一个 PAT 是第 2 位(P)，第 4 位(A)，第 6 位(T)；第二个 PAT 是第 3 位(P)，第 4 位(A)，第 6 位(T)。
现给定字符串，问一共可以形成多少个 PAT？
输入格式：
输入只有一行，包含一个字符串，长度不超过105，只包含 P、A、T 三种字母。
输出格式：
在一行中输出给定字符串中包含多少个 PAT。由于结果可能比较大，只输出对 1000000007 取余数的结果。
输入样例：
APPAPT
输出样例：
2
A1093
The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters.
Now given any string, you are supposed to tell the number of PAT's contained in the string.
Input Specification:
Each input file contains one test case. For each case, there is only one line giving a string of no more than 105 characters containing only P, A, or T.
Output Specification:
For each test case, print in one line the number of PAT's contained in the string. Since the result may be a huge number, you only have to output the result moded by 1000000007.
Sample Input:
APPAPT
Sample Output:
2
#include<stdio.h>
#include<string.h>
char str[100010];
int leftNumP[100010] = { 0 };
int main() {
gets_s(str);
int len = strlen(str);
for (int i = 0; i < len; i++) {
if (i > 0) {
leftNumP[i] = leftNumP[i - 1];
}
if (str[i] == 'P') {
leftNumP[i]++;
}
}
int ans = 0, rightNum = 0;
for (int i = len - 1; i >= 0; i--) {
if (str[i] == 'T') {
rightNum++;
}
else if (str[i] == 'A') {
ans = (ans + leftNumP[i] * rightNum) % 1000000007;
}
}
printf("%d", ans);
return 0;
}
*4.6 PAT B1045/A1101
B1045
著名的快速排序算法里有一个经典的划分过程：我们通常采用某种方法取一个元素作为主元，通过交换，把比主元小的元素放到它的左边，比主元大的元素放到它的右边。 给定划分后的 N 个互不相同的正整数的排列，请问有多少个元素可能是划分前选取的主元？
例如给定 N=5, 排列是1、3、2、4、5。则：
- 1 的左边没有元素，右边的元素都比它大，所以它可能是主元；
- 尽管 3 的左边元素都比它小，但其右边的 2 比它小，所以它不能是主元；
- 尽管 2 的右边元素都比它大，但其左边的 3 比它大，所以它不能是主元；
- 类似原因，4 和 5 都可能是主元。
因此，有 3 个元素可能是主元。
输入格式：
输入在第 1 行中给出一个正整数 N（≤105）； 第 2 行是空格分隔的 N 个不同的正整数，每个数不超过 109。
输出格式：
在第 1 行中输出有可能是主元的元素个数；在第 2 行中按递增顺序输出这些元素，其间以 1 个空格分隔，行首尾不得有多余空格。
输入样例：
5
1 3 2 4 5
输出样例：
3
1 4 5
A1101
There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its left and those larger than the pivot to its right. Given N distinct positive integers after a run of partition, could you tell how many elements could be the selected pivot for this partition?
For example, given N=5 and the numbers 1, 3, 2, 4, and 5. We have:
- 1 could be the pivot since there is no element to its left and all the elements to its right are larger than it;
- 3 must not be the pivot since although all the elements to its left are smaller, the number 2 to its right is less than it as well;
- 2 must not be the pivot since although all the elements to its right are larger, the number 3 to its left is larger than it as well;
- and for the similar reason, 4 and 5 could also be the pivot.
Hence in total there are 3 pivot candidates.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤105). Then the next line contains N distinct positive integers no larger than 109. The numbers in a line are separated by spaces.
Output Specification:
For each test case, output in the first line the number of pivot candidates. Then in the next line print these candidates in increasing order. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.
Sample Input:
5
1 3 2 4 5
Sample Output:
3
1 4 5
算法笔记 4.1.1.2
