Missing number in matrix

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Missing number in matrix

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matrix
Given a matrix A[][] of size N such that it has only one 0, Find the number to be placed in place of the 0 such that sum of the numbers in every row, column and two diagonals become equal.

Note: Diagonals should be only of the form A[i][i] and A[i][N-i-1]. If there is a matrix of size 1, with an element 0, then print 1.

Input:
First line contains the test cases, T . Then T test cases follow. Each test case contains a single integer N which represents the number of rows and columns of the matrix. Then in the next line are N*N space separated of the matrix A.

Output:
Display the number in place of 0 if possible, otherwise display -1. If there is a matrix of size 1, with an element 0, then print 1.

Constraints:

1<=T<=25
1<=n<=100
1<=ai<=1000000000

Example:
Input

225 5 5 031 2 0 3 1 2 2 3 1

Output:

5-1

Explanation:

(i)In the first test case the matrix is

5 55 0

Therefore If we place 5 instead of 0, all the element of matrix will become 5. Therefore row 5+5=10, column 5+5=10 and diagonal 5+5=10, all are equal.

(ii)Second sample, it is not possible to insert an element in place of 0 so that the condition is satisfied.thus result is -1.

简单题，不解释，好久没练了，熟悉下手感。代码如下：

#include<iostream>#include<string.h>#include<vector>//#include <cassert>#include <algorithm>#include <string>#include <iterator>using namespace std;class Cite_Sum{public:    unsigned long long sum;    bool Inf_Used;    Cite_Sum(){sum=0;Inf_Used=0;}    //bool operator == (const Cite_Sum & s);};bool operator == ( Cite_Sum & a,Cite_Sum & b){return a.sum==b.sum;}  bool operator <( Cite_Sum& a,Cite_Sum& b){    return a.sum<b.sum;}bool compare(Cite_Sum& a,Cite_Sum& b){    return a.Inf_Used<b.Inf_Used;  }int main(){    int Count,row,column;       cin>>Count;     while(Count--)    {   int N;        int Infu_num=2;        cin>>N;        if (N==1)        {            unsigned long c;            cin>>c;            if (c==0)            cout<<"1"<<endl;            else cout<<"-1"<<endl;        }        else        {            vector<Cite_Sum> Su(2*N+2);            int ps=1;           unsigned long **p=new unsigned long*[N];           for (int i=0;i<N;i++)             p[i]=new unsigned long[N];           for(int i=0;i<N;i++)          {    for(int j=0;j<N;j++)             { cin>>p[i][j];                if (p[i][j]==0)                {                   row=i;                   column=j;                   //行列对于哪些SUM有影响                   Su[i].Inf_Used=1;                   Su[j+N].Inf_Used=1;                   if(i==N-j-1)                   {  Su[2*N+1].Inf_Used=1;                      Infu_num++;                   }                   if(i==j)                   { Su[2*N].Inf_Used=1;                      Infu_num++;                   }                 }                    if(i==j)               Su[2*N].sum+=p[i][j];               if(i==N-j-1)                           Su[2*N+1].sum+=p[i][j];               Su[i].sum+= p[i][j];               Su[j+N].sum+= p[i][j];        }          }        unsigned long long s=Su[row].sum;        if(Su[row].sum==Su[column+N].sum)        {                 ps=1;            if(Su[2*N].Inf_Used==1&&Su[row].sum!=Su[2*N].sum)                  ps=0;            if(Su[2*N+1].Inf_Used==1&&Su[row].sum!=Su[2*N+1].sum)                  ps=0;          }//被影响行列的SUM是否相等          if (ps!=0)          {             //int s=Su[row].sum;             vector<Cite_Sum>::iterator new_end;             sort(Su.begin(),Su.end(),compare);             while (Infu_num--)             Su.pop_back();             new_end=unique(Su.begin(),Su.end());             //assert(Su.size()==N);             Su.erase(new_end,Su.end());               int h=Su.size();             if (h!=1) ps=0;           }           if (ps!=0&&Su[0].sum>s)          {            unsigned long long re=Su[0].sum-s;              cout<<re<<endl;           }          else cout<<"-1"<<endl;          for (int i=0;i<N;i++)              delete p[i];        }           }       return 0; }