zoj 2835 Magic Square(set)

Magic Square

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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In recreational mathematics, a magic square of n-degree is an arrangement of n² numbers, distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant. For example, the picture below shows a 3-degree magic square using the integers of 1 to 9.

Given a finished number square, we need you to judge whether it is a magic square.

Input

The input contains multiple test cases.

The first line of each case stands an only integer N (0 < N < 10), indicating the degree of the number square and then N lines follows, with N positive integers in each line to describe the number square. All the numbers in the input do not exceed 1000.

A case with N = 0 denotes the end of input, which should not be processed.

Output

For each test case, print "Yes" if it's a magic square in a single line, otherwise print "No".

Sample Input

2
1 2
3 4
2
4 4
4 4
3
8 1 6
3 5 7
4 9 2
4
16 9 6 3
5 4 15 10
11 14 1 8
2 7 12 13
0

Sample Output

No
No
Yes
Yes
分析：根据幻方矩阵，可以计算出行和（列和，对角线和）为总和/行数；

 #include <iostream>
 #include <cstdio>
 #include <set>
 using namespace std;
 int m[][];
 int main(){
     int n, i, j;
     int row_sum, col_sum;//行和，列和
     int main_diagonal_sum, counter_diagonal_sum;//主对角线元素和，副对角线元素和
     int sum;
     set<int> st;
     while(cin >> n){
         if(n == )
             break;
         st.clear();
         main_diagonal_sum = , counter_diagonal_sum = , sum = ;
         for(i = ; i < n; i++){
             for(j = ; j < n; j++){
                 cin >> m[i][j];
                 sum += m[i][j];
                 st.insert(m[i][j]);
             }
         }
         if(st.size() != n * n){//很重要，矩阵中的数有可能重复，有重数的矩阵直接输出"No"
             cout << "No" << endl;
             continue;
         }
         int aver = sum / n;
         //cout << aver << "a" << endl;
         for(i = ; i < n; i++){
             row_sum = ;
             col_sum = ;
             for(j = ; j < n; j++){
                 row_sum += m[i][j];
                 col_sum += m[j][i];
             }
             if(row_sum != aver || col_sum != aver){
                 cout << "No" << endl;
                 goto RL;
             }
         }
         for(i = ; i < n; i++){
             main_diagonal_sum += m[i][i];
             counter_diagonal_sum += m[i][n -  - i];
         }
         if(main_diagonal_sum != aver || counter_diagonal_sum != aver){
             cout << "No" << endl;
             continue;
         }
         cout << "Yes" << endl;
         RL:
             continue;
     }
     return ;
 }

还有一种方法是将所有的和放到一个set集合，最后判断集合大小是不是1，若为1，则yes，否则no

 #include <iostream>
 #include <cstdio>
 #include <set>
 using namespace std;
 int m[][];
 int main(){
     int n, i, j;
     int row_sum, col_sum;//行和，列和
     int main_diagonal_sum, counter_diagonal_sum;//主对角线元素和，副对角线元素和
     set<int> st;
     while(cin >> n){
         if(n == )
             break;
         st.clear();
         main_diagonal_sum = , counter_diagonal_sum = ;
         for(i = ; i < n; i++){
             for(j = ; j < n; j++){
                 cin >> m[i][j];
                 st.insert(m[i][j]);
             }
         }
         if(st.size() != n * n){//很重要，矩阵中的数有可能重复，有重数的矩阵直接输出"No"
             cout << "No" << endl;
             continue;
         }
         st.clear();
         for(i = ; i < n; i++){
             row_sum = ;
             col_sum = ;
             for(j = ; j < n; j++){
                 row_sum += m[i][j];
                 col_sum += m[j][i];
             }
             st.insert(row_sum);
             st.insert(col_sum);
         }
         for(i = ; i < n; i++){
             main_diagonal_sum += m[i][i];
             counter_diagonal_sum += m[i][n -  - i];
         }
         st.insert(main_diagonal_sum);
         st.insert(counter_diagonal_sum);
         if(st.size() != )
             cout << "No" << endl;
         else
             cout << "Yes" << endl;
     }
     return ;
 }