本文主要是介绍cf 259 b幻方，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
B. Little Elephant and Magic Square
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Little Elephant loves magic squares very much.
A magic square is a 3 × 3 table, each cell contains some positive integer. At that the sums of integers in all rows, columns and diagonals of the table are equal. The figure below shows the magic square, the sum of integers in all its rows, columns and diagonals equals 15.
The Little Elephant remembered one magic square. He started writing this square on a piece of paper, but as he wrote, he forgot all three elements of the main diagonal of the magic square. Fortunately, the Little Elephant clearly remembered that all elements of the magic square did not exceed 105.
Help the Little Elephant, restore the original magic square, given the Elephant's notes.
Input
The first three lines of the input contain the Little Elephant's notes. The first line contains elements of the first row of the magic square. The second line contains the elements of the second row, the third line is for the third row. The main diagonal elements that have been forgotten by the Elephant are represented by zeroes.
It is guaranteed that the notes contain exactly three zeroes and they are all located on the main diagonal. It is guaranteed that all positive numbers in the table do not exceed 105.
Output
Print three lines, in each line print three integers — the Little Elephant's magic square. If there are multiple magic squares, you are allowed to print any of them. Note that all numbers you print must be positive and not exceed 105.
It is guaranteed that there exists at least one magic square that meets the conditions.
Examples
input
Copy
0 1 1 1 0 1 1 1 0
output
Copy
1 1 1 1 1 1 1 1 1
input
Copy
0 3 6 5 0 5 4 7 0
output
Copy
6 3 6 5 5 5 4 7 4
题意是给出一个空着对角线的幻方让你把对角线填上首先幻方有着一个规律每一行的和都相等每一列的和都相等等于对角线的和
假设表示幻方是
a b c
d e f
g h i
所以 假设行的和是x，那么 x=a+b+c,x=d+e+f,x=g+h+i;3*x=矩阵各位置和，然而a+e+i=x,所以可以约掉一个x这样就能求出x了
#include<bits/stdc++.h>
using namespace std;
map<int,int>mm;
const int N=1e5;
int a[5][5];
int b[5];
int main()
{int n;int ans=0;for(int i=1;i<=3;i++)for(int j=1;j<=3;j++){cin>>a[i][j];ans+=a[i][j];b[i]+=a[i][j];}ans/=2;for(int i=1;i<=3;i++){for(int j=1;j<=3;j++){if(a[i][j]==0)cout<<ans-b[i]<<' ';elsecout<<a[i][j]<<' ';}cout<<endl;}
}
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