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Mike and Matrices CodeChef Solution
Problem -Mike and Matrices CodeChef Solution
“This website is dedicated for CodeChef solution where we will publish right solution of all your favourite CodeChef problems along with detailed explanatory of different competitive programming concepts and languages.
Mike and Matrices CodeChef Solution in C++14
“#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m;
cin >> n >> m;
int a[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> a[i][j];
}
}
long long s1 = 0, s2 = 0;
int k, i, j;
cin >> k;
while (k-- > 0) {
cin >> i >> j;
i -= 1;
j -= 1;
if (i >= n || j >= m) {
s1 = -1;
}
if (s1 != -1) {
s1 += a[i][j];
}
if (i >= m || j >= n) {
s2 = -1;
}
if (s2 != -1) {
s2 += a[j][i];
}
}
cout << max(s1, s2) << endl;
return 0;
}Mike and Matrices CodeChef Solution in PYTH 3
“# cook your dish here
# cook your dish here
n, m = map(int, input().split())
matrix = []
for _ in range (n):
matrix.append([int(x) for x in input().split()])
k = int(input())
pairs = []
for _ in range (k):
pairs.append([int(x) for x in input().split()])
E1, E2 = 0, 0
for i in range(k):
x, y = pairs[i][0], pairs[i][1]
if x > n or y > m:
E1 = -1
break
else :
E1 += matrix[x-1][y-1]
for i in range(k):
x, y = pairs[i][0], pairs[i][1]
if x > m or y > n:
E2 = -1
break
else :
E2 += matrix[y-1][x-1]
print(max(E1, E2))Mike and Matrices CodeChef Solution in C
“#include <stdio.h>
int main ()
{
int N,M;
scanf( "%d%d", &N, &M );
long int A[N][M];
for ( int i = 0 ; i <= N - 1 ; i++ )
{
for ( int j = 0 ; j <= M - 1 ; j++ )
{
scanf( "%ld", &A[i][j] );
}
}
long int L;
scanf( "%ld", &L );
long long int e1 = 0 , e2 = 0;
int f1 = 0 , f2 = 0;
while(L--)
{
int i ,j;
scanf( "%d%d", &i, &j );
i--;
j--;
if ( ( i > N - 1) || (j > M - 1) )
{
e1 = -1;
f1 = 1;
}
if ( (j > N -1) || (i > M - 1) )
{
e2 = -1;
f2 = 1;
}
if ( f1 == 0 )
{
e1 += A[i][j];
}
if ( f2 == 0 )
{
e2 += A[j][i];
}
}
printf( "%lld\n", ( e1 >= e2 ? e1 : e2 ) );
} Mike and Matrices CodeChef Solution in JAVA
“/* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));
BufferedWriter writer=new BufferedWriter(new OutputStreamWriter(System.out));
int[] input=parseInt(reader.readLine());
int n=input[0], m=input[1];
int[][] grid=new int[n][m];
for(int i=0;i<n;i++){
grid[i]=parseInt(reader.readLine());
}
int l=Integer.parseInt(reader.readLine());
int[][] pairs=new int[l][2];
for(int i=0;i<l;i++){
pairs[i]=parseInt(reader.readLine());
}
long s1=0L, s2=0L;
for(int[] p: pairs){
int x=p[0]-1, y=p[1]-1;
if(x>=0 && x<grid.length && y>=0 && y<grid[0].length && s1>=0){
s1+=grid[x][y];
}else{
s1=-1;
}
if(s2>=0 && x>=0 && y>=0 && x<grid[0].length && y<grid.length){
s2+=grid[y][x];
}else{
s2=-1;
}
}
writer.write(Math.max(s1, s2)+"\n");
writer.flush();
}
private static int[] parseInt(String str) {
String[] parts=str.split("\\s+");
int[] ans=new int[parts.length];
for(int i=0;i<ans.length;i++){
ans[i]=Integer.parseInt(parts[i]);
}
return ans;
}
private static long[] parseLong(String str) {
String[] parts=str.split("\\s+");
long[] ans=new long[parts.length];
for(int i=0;i<ans.length;i++){
ans[i]=Long.parseLong(parts[i]);
}
return ans;
}
}Mike and Matrices CodeChef Solution in PYPY 3
“A = []
n,m = map(int,input().split())
for i in range(n):
A_row = list(map(int,input().split()))
A.append(A_row)
e1,e2 = 0,0
l = int(input())
flag1,flag2 = True,True
for i in range(l):
a,b = map(int,input().split())
if a<=n and b<=m and flag1:
e1+=A[a-1][b-1]
else:
e1 = -1
flag1 = False
if b<=n and a<=m and flag2:
e2+=A[b-1][a-1]
else:
e2 = -1
flag2 = False
print(max(e1,e2))Mike and Matrices CodeChef Solution in PYTH
“st = raw_input().split()
N = int(st[0])
M = int(st[1])
Z = min(N,M)
A = [[0 for x in range(M+1)] for y in range(N+1)]
for y in range(N):
st = raw_input().split()
for x in range(M):
n = int(st[x])
A[y+1][x+1] = n
# endfor x
# endfor y
L = int(raw_input())
E1 = 0
E2 = 0
for k in range(L):
st = raw_input().split()
a = int(st[0])
b = int(st[1])
if a > Z:
if a > N:
E1 = -1
else:
E2 = -1
# endif
# endif
if b > Z:
if b > N:
E2 = -1
else:
E1 = -1
# endif
# endif
if E1 > -1:
E1 += A[a][b]
# endif
if E2 > -1:
E2 += A[b][a]
# endif
# endfor k
r = max(E1,E2)
print r
Mike and Matrices CodeChef Solution in C#
“using System;
public class Test
{
public static void Main()
{
var nm = Array.ConvertAll(Console.ReadLine().Split(), int.Parse);
var n = nm[0];
var m = nm[1];
var data = new long[n, m];
for (int i = 0; i < n; i++)
{
var row = Array.ConvertAll(Console.ReadLine().Split(), long.Parse);
for (int j = 0; j < m; j++)
{
data[i, j] = row[j];
}
}
var l = int.Parse(Console.ReadLine());
long e1 = 0;
long e2 = 0;
while (l-- > 0)
{
var pair = Array.ConvertAll(Console.ReadLine().Split(), long.Parse);
long i = pair[0];
long j = pair[1];
if (e1 != -1 && i>0 && j>0 && i < n + 1 && j < m + 1)
{
e1 += data[i-1, j-1];
}
else
{
e1 = -1;
}
if (e2 != -1 && i>0 && j> 0 && j < n +1 && i < m + 1)
{
e2 += data[j-1, i-1];
}
else
{
e2 = -1;
}
}
Console.WriteLine(e1 > e2 ? e1 : e2);
}
}Mike and Matrices CodeChef Solution Review:
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“Conclusion:
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