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[10][10];
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 == 0)
            break;
        st.clear();
        main_diagonal_sum = 0, counter_diagonal_sum = 0, sum = 0;
        for(i = 0; i < n; i++){
            for(j = 0; 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 = 0; i < n; i++){
            row_sum = 0;
            col_sum = 0;
            for(j = 0; 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 = 0; i < n; i++){
            main_diagonal_sum += m[i][i];
            counter_diagonal_sum += m[i][n - 1 - i];
        }
        if(main_diagonal_sum != aver || counter_diagonal_sum != aver){
            cout << "No" << endl;
            continue;
        }
        cout << "Yes" << endl;
        RL:
            continue;
    }
    return 0;
}

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[10][10];
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 == 0)
            break;
        st.clear();
        main_diagonal_sum = 0, counter_diagonal_sum = 0;
        for(i = 0; i < n; i++){
            for(j = 0; 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 = 0; i < n; i++){
            row_sum = 0;
            col_sum = 0;
            for(j = 0; j < n; j++){
                row_sum += m[i][j];
                col_sum += m[j][i];
            }
            st.insert(row_sum);
            st.insert(col_sum);
        }
        for(i = 0; i < n; i++){
            main_diagonal_sum += m[i][i];
            counter_diagonal_sum += m[i][n - 1 - i];
        }
        st.insert(main_diagonal_sum);
        st.insert(counter_diagonal_sum);
        if(st.size() != 1)
            cout << "No" << endl;
        else
            cout << "Yes" << endl;
    }
    return 0;
}