2020 CCPC-Wannafly Winter Camp Day3 Div.1 & 2 (A balloon black) (Equation Solutions)

2020 CCPC-Wannafly Winter Camp Day3 Div.1 & 2 (A balloon black) (Equation Solutions)

Time limit: C / C ++ 1 second, 2 seconds languages other
space restrictions: C / C ++ 262144K, other languages 524288K
64bit the IO the Format: LLD%

Title Description

There in front of a small D ${n}$ black balloons.
The first hypothesis ${i}$ highly black balloons is a positive integer $h_i$ , Small D now know that any two balloons of different height and you help small D to restore the specific height of each balloon black thing?

Enter a description:

The first line of an integer ${n}$
Next ${n}$ lines of ${n}$ integers, where the first ${i}$ Line ${j}$ integers represents ${i}$ balloons and ${j}$ height and a balloon. (when ${i=j}$ When this number is ${0}$)。
$2 \ n \ the 1000$ , each of the input does not exceed the number of $10^5$ . The only answer is to ensure that data.

Output Description:

Row ${n}$ an integer that represents the answer.
The only guarantee an answer.

Example 1

Entry

5
0 3 4 5 6
3 0 5 6 7
4 5 0 7 8
5 6 7 0 9
6 7 8 9 0

Export

1 2 3 4 5

answer

To solve the equation, but there is an ingenious solution:

Three equations of the first term before the first use, reuse of items related to contact with the first equation.

Code

#include<stdio.h>
int main() {
	int n, i, j, x, h1;
	scanf("%d", &n);
	int a[1001][1001];
	for (i = 1; i <= n; i++)
		for (j = 1; j <= n; j++)
			scanf("%d", &a[i][j]);
	x = a[2][3] - a[1][2];
	h1 = (a[1][3] - x) / 2;
	printf("%d ", h1);
	for (i = 2; i <= n; i++)
		printf("%d ", a[1][i] - h1);
	return 0;
}