ZS the Coder and Chris the Baboon arrived at the entrance of Udayland. There is a n × n magic grid on the entrance which is filled with integers. Chris noticed that exactly one of the cells in the grid is empty, and to enter Udayland, they need to fill a positive integer into the empty cell.
Chris tried filling in random numbers but it didn't work. ZS the Coder realizes that they need to fill in a positive integer such that the numbers in the grid form a magic square. This means that he has to fill in a positive integer so that the sum of the numbers in each row of the grid (
Chris doesn't know what number to fill in. Can you help Chris find the correct positive integer to fill in or determine that it is impossible?
The first line of the input contains a single integer n (1 ≤ n ≤ 500) — the number of rows and columns of the magic grid.
n lines follow, each of them contains n integers. The j-th number in the i-th of them denotes ai, j (1 ≤ ai, j ≤ 109 or ai, j = 0), the number in the i-th row and j-th column of the magic grid. If the corresponding cell is empty, ai, j will be equal to 0. Otherwise, ai, j is positive.
It is guaranteed that there is exactly one pair of integers i, j (1 ≤ i, j ≤ n) such that ai, j = 0.
Output a single integer, the positive integer x (1 ≤ x ≤ 1018) that should be filled in the empty cell so that the whole grid becomes a magic square. If such positive integer x does not exist, output - 1 instead.
If there are multiple solutions, you may print any of them.
3 4 0 2 3 5 7 8 1 6
9
4 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1
1
4 1 1 1 1 1 1 0 1 1 1 2 1 1 1 1 1
-1
In the first sample case, we can fill in 9 into the empty cell to make the resulting grid a magic square. Indeed,
The sum of numbers in each row is:
4 + 9 + 2 = 3 + 5 + 7 = 8 + 1 + 6 = 15.
The sum of numbers in each column is:
4 + 3 + 8 = 9 + 5 + 1 = 2 + 7 + 6 = 15.
The sum of numbers in the two diagonals is:
4 + 5 + 6 = 2 + 5 + 8 = 15.
In the third sample case, it is impossible to fill a number in the empty square such that the resulting grid is a magic square.
思路：
题意是，给你一个N*N矩阵，其中0代表未知的那个数，让你在这个位置填上一个数使整个矩阵的 每一行、每一列、主对角线、次对角线的和都相等。只需要分情况讨论，n=1的时候这个数可以是任意正整数，n=2的时候就暴力求就好，n>=3的时候，输入时候先保存0这个点的横纵坐标，然后再选择三列求和，其中相等的两列的值就是所有和都要满足的那个值。然后找到0所在列，把列里的所有元素减去就得到了0这个点应该放进去的数。（不用担心会减去0这个点，因为0对计算结果无影响）。然后就是要检验了，先吧0所在点换为要放进去的数，然后对每一行每一列还有两个对角线都和之前算出来的那个和进行比较，全部相等的时候才说明能放进去。而且要注意这里放下去的数不能是非整数。我的bug出现在最后的一个flag标志上，判断条件if（flag==true）我少打了一个等号结果WA了...粗心害死人啊。
代码：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
long long a[500][500];
int main()
{
//freopen("in.txt","r",stdin);
int n;
scanf("%d",&n);
if(n==1)
{
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
scanf("ll%d",&a[i][j]);
}
printf("1\n");
}
if(n==2)
{
long long flag1=0,flag2=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
scanf("%lld",&a[i][j]);
if(a[i][j]==0)
{
flag1=i;
flag2=j;
}
}
long long result=0;
long long sum[2]={0,0};
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
{
sum[i]+=a[i][j];
}
if(sum[0]>sum[1])
{
result=sum[0]-sum[1];
a[flag1][flag2]=result;
}
else
{
result=sum[1]-sum[0];
a[flag1][flag2]=result;
}
long long a1=a[0][0]+a[1][1];
long long a2=a[0][0]+a[0][1];
long long a3=a[1][0]+a[1][1];
long long a4=a[1][0]+a[0][0];
long long a5=a[1][0]+a[0][1];
long long a6=a[0][1]+a[1][1];
if(a1==a2&&a2==a3&&a3==a4&&a4==a5&&a5==a6&&a[flag1][flag2]>0)printf("%lld\n",result);
else printf("-1\n");
}
else if(n>=3)
{
long long flag1=0,flag2=0;
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
scanf("%lld",&a[i][j]);
if(a[i][j]==0)
{
flag1=i;
flag2=j;
}
}
long long sum[3]={0,0,0};
for(int i=0;i<3;i++)
{
for(int j=0;j<n;j++)
{
sum[i]+=a[i][j];
}
}
long long result=0;
if(sum[0]==sum[1])result=sum[0];
else if(sum[0]==sum[2])result=sum[0];
else if(sum[1]==sum[2])result=sum[1];
for(int j=0;j<n;j++)
{
result=result-a[flag1][j];
}
a[flag1][flag2]=result;
long long jianyan=0;
for(int j=0;j<n;j++)
jianyan+=a[flag1][j];
bool rrr=true;
for(int i=0;i<n;i++)
{
long long e=0;
for(int j=0;j<n;j++)
{
e+=a[i][j];
}
if(e!=jianyan)
{
rrr=false;
break;
}
}
if(rrr==true)
{
for(int j=0;j<n;j++)
{
long long e=0;
for(int i=0;i<n;i++)
{
e+=a[i][j];
}
if(e!=jianyan)
{
rrr=false;
break;
}
}
}
if(rrr==true)
{
long long e=0;
for(int i=0;i<n;i++)
{
e+=a[i][i];
}
if(e!=jianyan)
{
rrr=false;
}
}
if(rrr==true)
{
long long e=0;
for(int i=0;i<n;i++)
{
e+=a[i][n-i-1];
}
if(e!=jianyan)
{
rrr=false;
}
}
if(rrr==true&&a[flag1][flag2]>0)printf("%lld\n",result);
else printf("-1\n");
}
}
京公网安备 11010502036488号