CodeForces 711B Chris and Magic Square (暴力,水题)
and Codeforces 暴力 水题 Magic Square
2023-09-11 14:17:18 时间
题意:给定n*n个矩阵,其中只有一个格子是0,让你填上一个数,使得所有的行列的对角线的和都相等。
析:首先n为1,就随便填,然后就是除了0这一行或者这一列,那么一定有其他的行列是完整的,所以,先把其他的算出来,然后再作差就算这个数了,
然后再去验证其他的对不对就好了。除了n为1,其他的都是唯一解应该。或者没有。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <list> #include <sstream> #define frer freopen("in.txt", "r", stdin) #define frew freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 5e2 + 5; const int mod = 1e9 + 7; const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1}; const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } LL a[maxn][maxn]; int main(){ while(cin >> n){ memset(a, 0, sizeof a); LL sum1 = 0, sum2 = 0; int x, y; for(int i = 0; i < n; ++i){ for(int j = 0; j < n; ++j){ scanf("%I64d", &a[i][j]); a[i][n] += a[i][j]; a[n][j] += a[i][j]; if(a[i][j] == 0) x = i, y = j; if(i == j) sum1 += a[i][j]; if(i + j == n - 1) sum2 += a[i][j]; } } if(1 == n) printf("1\n"); else{ LL ans = 0; for(int i = 0; i < n; ++i){ if(i != x){ ans = a[i][n] - a[x][n]; a[x][n] += ans; a[n][y] += ans; if(x == y) sum1 += ans; if(x + y == n - 1) sum2 += ans; break; } } bool ok = true; if(sum1 != sum2 || ans < 1 || ans > 1e18) ok = false; if(!ok){ printf("-1\n"); continue; } for(int i = 0; i < n; ++i){ if(a[i][n] != sum1 || a[n][i] != sum1){ ok = false; break; } } if(!ok){ printf("-1\n"); continue; } else cout << ans << endl; } } return 0; }
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