In this section i will share several problem sets programming that I’ve ever done, you will see the problem and the solution in C language programming on the bottom. In another occasion i will share another problem sets but NOT include the solution. I hope my post today can bring some references. Thanks have nice day.
Problem bellow I was found in my test selection for joining some occasion, I think this is good experience can done this test and i share with the problem. Begin with Diagonal Disproportion Problem and Solution
Diagonal Disproportion
Time limit: 2000 ms
Memory limit: 65536 KB
Description
You are to calculate the diagonal disproportion of a square matrix. The diagonal disproportion of a square matrix is the sum of the elements of its main diagonal minus the sum of the elements of its collateral diagonal. The main and collateral diagonals of a square matrix are shown in figures 1 and 2 respectively.
The elements of the main diagonal are shown in green in figure 1, and the elements of the collateral diagonal are shown in cyan in figure 2.
Given a matrix, return its diagonal disproportion. The j’th character of the i’th element of matrix should be treated as the element in the i’th row and j’th column of the matrix.
Input Format
First line consists of one integer N (1 <= N <= 50), the size of the matrix. The next N lines consist of N 1-digit numbers
Output Format
One integer in one line, that is the diagonal disproportion of that matrix.
Sample Input 1
3 190 828 373
Sample Output 1
1
Note
The sum of the elements of the main diagonal is 1+2+3 = 6. The sum of the elements of the collateral diagonal is 0+2+3 = 5. So, the answer is 6-5 = 1.
Sample Input 2
4 9000 0120 0000 9000
Sample Output 2
-1
Sample Input 3
10 7748297018 8395414567 7006199788 5446757413 2972498628 0508396790 9986085827 2386063041 5687189519 7729785238
Sample Output 3
-24
Solution
#include <stdio.h>
#include <stdlib.h>
int main(){
int N;
scanf("%d",&N);
if(N>=1 && N <=50){
char temp[N];
char x;
int i,a, result1 = 0, result2 = 0;
for(i = 0 ; i < N;i++){
fflush(stdin);
scanf("%s",&temp);
x = temp[i];
a = x - '0';
result1 += a;
x = temp[N-(i+1)];
a = x - '0';
result2 += a;
}
printf("%d\n",result1-result2);
}
return 0;
}