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; }