Pass rate: 218 / 338 (64.50%) Accuracy: 218 / 1249 (17.45%)
Title Description
At the beginning of school, the guide will nn Three students form a temporary class, but they think there are too many students in the class, so they give you a transcript of all students, and let you divide them into different regular classes according to everyone's grades as follows:
- If the total number of the temporary shift is less than kk, then it becomes the last regular class (the class number is s+1s+1 , ss For the number of regular classes that have been assigned, obviously at the beginning s=0s=0 )， And end the shift.
- Start with the first student and put the continuous kk Students are divided into a regular class (if there are not enough students in the last class) kk People, then they will be merged to the previous shift), and the shift numbers are s+1,s+2,s+3,…,s+ms+1,s+2,s+3,…,s+m (suppose it's separated mm Class).
- Separate from the previous step mm The highest scores of each class are moved out in turn to form a temporary class (these students will no longer belong to the original class), and the class is divided again (back to the second class) eleven Step).
You should also pay attention to the following points during shift assignment:
- The relative order of students remains unchanged before and after class division, such as students aa yes eleven The highest score in the class, students bb yes twenty-two The highest score of the class, after they move out to the same temporary class aa Should also be ranked bb front.
- Only regular classes have a class number, and from eleven It starts to increase in turn. There is no shift number for the temporary shift.
- Record the minimum score of each regular shift, and finally output the minimum score of each shift according to the shift number sequence.
input
The first line contains two integers nn and kk
Second line nn Integer xx ， Indicates the grade of each student
output
Multiline output, 2nd ii Row output ii Minimum score of class
sample input
10 3 10 23 21 89 11 43 56 99 60 78
sample output
10 11 56 23 99
Example description
First divide into 1 class: (10, 23, 21), 2 classes: (89, 11, 31), 3 classes: (56, 99, 60, 78), and then take out the highest score of each class to form a temporary class: (23, 89, 99). The number of people in this class is not less than k (k=3), and continue to divide into classes. Divide the class into 4 classes: (23, 89, 99), and take out the highest score to form a temporary class: (99). At this time, there is only 1 person in the temporary class (less than k) At the end of the shift, the temporary shift becomes 5 shifts. Finally, the lowest scores of shifts 1 to 5 are output in sequence: 10, 11, 56, 23 and 99
HINT
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Before simulating the above process, first clarify the idea, think about which values need to be recorded, whether to use an array, what the control conditions of the loop are, etc.
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about 100%100% Data, 2 ≤ n ≤ 1052 ≤ n ≤ 105 ， 2≤k≤1002≤k≤100 ， 0≤x≤1000≤x≤100
AUTHOR: toush1
#include <stdio.h> int max[1000000]; int min[1000000]; int tem[1000000]; int main() { int n, k, i, pos, first, max_num, min_num; //-----------Enter parameters and all grades and initialize-------------- scanf("%d%d", &n, &k); for (i = 1; i <= n; i++) { min[i] = 101; max[i] = -1; } for (i = 1; i <= n; i++) scanf("%d", &tem[i]); //------------Shift Division: cycle when the number of temporary shifts is greater than or equal to k----------- max_num = 1; min_num = 1; while (n >= k) { i = 0; for (first = 1; n - first >= k; first += k)//Find k at a time (k in a group) { for (i = 0; i < k; i++) { if (tem[first + i] > max[max_num]) max[max_num] = tem[first + i]; if (tem[first + i] < min[min_num]) min[min_num] = tem[first + i]; } max_num++; min_num++; } //---------------Divide the rest of the temporary class into the last class------------- if (n - first + 1 == k ) { max_num++; min_num++; } for (i = first; i <= n; i++) { if (tem[i] > max[max_num - 1]) max[max_num - 1] = tem[i]; if (tem[i] < min[min_num - 1]) min[min_num - 1] = tem[i]; } //---------------Change the highest shift into a temporary shift and clear the highest shift--------------- for (i = 1; i <= max_num - 1; i++) { tem[i] = max[i]; max[i] = -1; } n = max_num - 1;//n = 3 max_num = 1; i = 0; } //-----------------Divide the remaining people in the last temporary class into the new class and take the lowest score---------------- for (i = 1; i <= n; i++) { if (tem[i] < min[min_num]) min[min_num] = tem[i]; } for (i = 1; i <= min_num; i ++) printf("%d\n", min[i]); return 0; }