ZOJ2835 Magic Square "magic square Verification"

Topic Links:

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1835

Main topic:

Give you an n-order matrix that determines whether n-order magic squares. The numbers used are not necessarily 1~n^2, but

The numbers must not be duplicated.

Problem Solving Ideas:

According to the definition of magic square, traverse and judge all the rows, vertical lines, oblique and equal, and use the flag[] array

To determine if a duplicate number exists.

AC Code:

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring>using namespace    Std;int map[20][20],sum[40],flag[1100];int Judge (int N) {memset (sum,0,sizeof (Sum));    memset (flag,0,sizeof (Flag)); for (int i = 0, i < n; ++i) for (int j = 0; j < N; ++j) {if (!            FLAG[MAP[I][J]]) flag[map[i][j] = 1;        else return 0;    } for (int i = 0, i < n; ++i) {for (int j = 0; j < N; ++j) sum[i] + = map[i][j];    } for (int i = 0, i < n; ++i) {for (int j = 0; j < N; ++j) sum[i+n] + = Map[j][i];    } for (int i = 0; i < N; ++i) sum[n+n] + = Map[i][i];    for (int i = 0; i < N; ++i) sum[n+n+1] + = map[i][n-1-i];    int tmp = sum[0];    for (int i = 0; I <= 2*n+1; ++i) if (sum[i]! = tmp) return 0; return 1;}    int main () {int N; while (~SCANF ("%d", &n) && N) {for (int i = 0; i < N        ++i) for (int j = 0; j < N; ++j) scanf ("%d", &map[i][j]);        if (Judge (N)) printf ("yes\n");    else printf ("no\n"); } return 0;}

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ZOJ2835 Magic Square "magic square Verification"