CodeForces - 711B Chris and Magic Square

题意：给你一个N*N矩阵，其中0代表未知的那个数，让你在这个位置填上一个数使整个矩阵的每一行、每一列、主对角线、次对角线的和都相等。

思路：一开始一直没理解题意，到底是行和=列和=对角线和，还是所以行的和相等，所有列的和相等，两条对角线的和相等。看到了are单词，明白了是第一种情况，之后就是简单的模拟了，特别判断n为1的情况，一定要注意细节，仔细仔细再仔细

#include<iostream>
using namespace std;
typedef long long ll;
const ll INF = 1e18;
ll r[505],c[505],angle[3];
ll a[505][505];
int main(void)
{int n;cin>>n;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)cin>>a[i][j];if(n==1){cout<<1<<endl;return 0;}int u,v;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++){r[i]+=a[i][j];c[i]+=a[j][i];if(a[i][j]==0){u=i;v=j;}if(i==j) angle[1]+=a[i][j];if((i+j)==(n+1)) angle[2]+=a[i][j];}int flag=1;ll R;   for(int i=1;i<=n;i++){if(i==u) continue;if(flag){R=r[i];flag=0;continue;}if(r[i]!=R){cout<<-1<<endl;return 0;}}ll C;flag=1;for(int i=1;i<=n;i++){if(i==v) continue;if(flag){C=c[i];flag=0;continue;}if(c[i]!=C){cout<<-1<<endl;return 0;       }}if(R!=C){cout<<-1<<endl;return 0;}if(u!=v&&(u+v)!=(n+1)){if(angle[1]==angle[2]&&angle[2]==R&&r[u]==c[v]&&R-r[u]>=1&&R-r[u]<=INF){cout<<R-r[u]<<endl;return 0;}else{cout<<-1<<endl;return 0;}}if(u==v&&(u+v)!=(n+1)){if(r[u]==c[v]&&angle[1]+R-r[u]==angle[2]&&angle[2]==R&&R-r[u]>=1&&R-r[u]<=INF){cout<<R-r[u]<<endl;return 0;}else{cout<<-1<<endl;return 0;}}if(u!=v&&(u+v)==(n+1)){if(r[u]==c[v]&&angle[2]+R-r[u]==angle[1]&&angle[1]==R&&R-r[u]>=1&&R-r[u]<=INF){cout<<R-r[u]<<endl;return 0;}else {cout<<-1<<endl;return 0;}}if(u==v&&(u+v)==(n+1)){if(r[u]==c[v]&&angle[2]==angle[1]&&angle[1]==c[v]&&R-r[u]>=1&&R-r[u]<=INF){cout<<R-r[u]<<endl;return 0;}else{cout<<-1<<endl;return 0;}}return 0;
}

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