Translation of l1-7 matrix columns in 2022 group programming ladder Simulation Competition (20 points)

Given a  n×n  The integer matrix of . For any given positive integer  k<n, We translate the elements of the even column of the matrix down in turn 1、……、k、1、……、k、…… A place , Translate the empty position with an integer  x  repair . You need to calculate the sum of each row element of the result matrix .

Input format ：

The first line of input is given 3 A positive integer ：n（<100）、k（<n）、x（<100）, As described in the question surface .

Next  n  That's ok , Each line gives  n  No more than one. 100 The positive integer , Is the value of the matrix element . Numbers are separated by spaces .

Output format ：

Output the... After translation in one line 1 To  n  Sum of row elements . Between the numbers 1 Space separation , There must be no extra space at the beginning and end of the line .

sample input ：

7 2 99
11 87 23 67 20 75 89
37 94 27 91 63 50 11
44 38 50 26 40 26 24
73 85 63 28 62 18 68
15 83 27 97 88 25 43
23 78 98 20 30 81 99
77 36 48 59 25 34 22

sample output ：

440 399 369 421 302 386 428

Example interpretation

What needs to be translated is 2、4、6 Column . Given  k=2, These three columns should be translated down in sequence 1、2、1 position （ If there are more columns , You should follow 1、2、1、2 …… This Law translates downward in sequence ）, For the empty space at the top 99 Fill in . The translated matrix becomes ：

11 99 23 99 20 99 89
37 87 27 99 63 75 11
44 94 50 67 40 50 24
73 38 63 91 62 26 68
15 85 27 26 88 18 43
23 83 98 28 30 25 99
77 78 48 97 25 81 22

One time in place , It's not easy to time out

#include <bits/stdc++.h>
using namespace std;
int main(void){
	int n,k,x,**p,*q;
	cin>>n>>k>>x;
	p=(int**)malloc(sizeof(int*)*n);
	q=(int*)malloc(sizeof(int)*n);
	for(int i=0;i<n;i++){
		q[i]=0;
		p[i]=(int*)malloc(sizeof(int)*n);
	}
	for(int i=0;i<n;i++){
		for(int s,j=0;j<n;j++){
			cin>>s;
			if(j%2){
				int t=(j/2)%k+1;
				if(i<t)p[i][j]=x;
				if(i+t<n)p[i+t][j]=s;
			}
			else p[i][j]=s;
			q[i]+=p[i][j];
		}
	}
	for(int i=0;i<n;i++){
		if(i)cout<<' ';
		cout<<q[i];
	}
	
}