CodeForces Gym 100735E

给你一个矩阵，他们本来是，每行，每列，两条对角线，的和都是一样的，现在把主对角线的元素都挖去，让你求主对角线上各是什么

以5*5举例吧，假设为a,b,c,d,e，每行的和是sum，然后将所有行加起来就有，a+b+c+d+e+4sum=5sum，然后sum就能求出来，然后根据 每行和为sum，所以每行的那个未知数就能求出来，然后就可以了

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <set>
using namespace std;
#define ll long long
#define maxn 200005
int N;
ll matrix[105][105];
ll row[105];
int main()
{
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
scanf("%d", &N);
ll sum = 0;
for (int i = 0; i < N; ++i)
{
for (int j = 0; j < N; ++j)
{
scanf("%I64d", &matrix[i][j]);
sum += matrix[i][j];
row[i] += matrix[i][j];
}
}
sum /= (N - 1);
for (int i = 0; i < N; ++i)
matrix[i][i] = sum - row[i];
for (int i = 0; i < N; ++i)
{
for (int j = 0; j < N; ++j)
{
printf("%I64d ", matrix[i][j]);
}
printf("\n");
}
//system("pause");
//while (1);
return 0;
}