【Codeforces Round #369 (Div. 2)】Codeforces 711B Chris and Magic Square

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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 (), each column of the grid (), and the two long diagonals
of the grid (the main diagonal — and the secondary diagonal — ) are
equal.

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? Input

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

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.

模拟。
细节比较容易写错。
特判n=1。

#include<cstdio>#include<cstring>#define LL long longLL a[510][510],sumi[510],sumj[510],sum1,sum2;int main(){    int i,j,k,m,n,p,q,xx,yy;    LL x,y,z,sum,temp,ans=-1;    scanf("%d",&n);    if (n==1)    {        printf("1\n");        return 0;    }    for (i=1;i<=n;i++)      for (j=1;j<=n;j++)      {        scanf("%I64d",&a[i][j]);        if (a[i][j]==0)        {            xx=i;            yy=j;        }      }    for (i=1;i<=n;i++)      for (j=1;j<=n;j++)      {        sumi[i]+=a[i][j];        sumj[j]+=a[i][j];      }    for (i=1;i<=n;i++)    {        sum1+=a[i][i];        sum2+=a[i][n-i+1];    }    if (xx==1) sum=sumi[2];    else sum=sumi[1];    for (i=1;i<=n;i++)      if (xx==i)      {        if (sumi[i]>=sum)        {            printf("-1\n");            return 0;        }        if (ans==-1) ans=sum-sumi[i];        else        {            if (sumi[i]+ans!=sum)            {                printf("-1\n");                return 0;            }        }      }      else      {        if (sumi[i]!=sum)        {            printf("-1\n");            return 0;        }      }    for (i=1;i<=n;i++)      if (yy==i)      {        if (sumj[i]>=sum)        {            printf("-1\n");            return 0;        }        if (ans==-1) ans=sum-sumj[i];        else        {            if (sumj[i]+ans!=sum)            {                printf("-1\n");                return 0;            }        }      }      else      {        if (sumj[i]!=sum)        {            printf("-1\n");            return 0;        }      }    if ((xx==yy&&sum1+ans!=sum)||(xx!=yy&&sum1!=sum))    {        printf("-1\n");        return 0;    }    if ((xx+yy==n+1&&sum2+ans!=sum)||(xx+yy!=n+1&&sum2!=sum))    {        printf("-1\n");        return 0;    }    printf("%I64d\n",ans);}
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