atcoder ABC 355-C题详解

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Problem Statement

There is an N×N grid, where the cell at the i-th row from the top and the j-th column from the left contains the integer N×(i−1)+j.

Over T turns, integers will be announced. On Turn i, the integer Ai​ is announced, and the cell containing Ai​ is marked. Determine the turn on which Bingo is achieved for the first time. If Bingo is not achieved within T turns, print -1.

Here, achieving Bingo means satisfying at least one of the following conditions:

There exists a row in which all N cells are marked.
There exists a column in which all N cells are marked.
There exists a diagonal line (from top-left to bottom-right or from top-right to bottom-left) in which all N cells are marked.

Constraints

2≤N≤2×103
1≤T≤min(N2,2×105)
1≤Ai​≤N2
Ai​=Aj​ if i=j.
All input values are integers.

Input

The input is given from Standard Input in the following format:

Output

If Bingo is achieved within T turns, print the turn number on which Bingo is achieved for the first time; otherwise, print -1.

Sample Input 1

3 5
5 1 8 9 7

Sample Output 1

4

The state of the grid changes as follows. Bingo is achieved for the first time on Turn 4.
​

Sample Input 2

​3 5
4 2 9 7 5

Sample Output 2

-1
Bingo is not achieved within five turns, so print -1.

Sample Input 3

4 12
13 9 6 5 2 7 16 14 8 3 10 11

Sample Output 3

9

思路分析：

先判断输入的数字在哪一行那一列，然后判断是否符合宾果游戏规则，判断主对角线和次对角线以及行、列。

code：

#include <iostream>
using namespace std;
const int N = 2010;
int n, t;
int row[N];
int col[N];
int ans1, ans2;
int main() {cin >> n >> t;for (int i = 0; i < t; i++) {int a;cin >> a;a--;//要减一比如5，5/3=1,5%3=2int row1 = a / n;int col1 = a % n;
//        if (row1 >= n || col1 >= n) {不需要比如第一个样例9/3=3 直接输出-1结束程序了
//            cout << "-1" << endl;
//            return 0;
//        }row[row1]++;col[col1]++;// 检查主对角线if (row1 == col1) {ans1++;}// 检查次对角线if (row1 + col1 == n - 1) {ans2++;}if(ans1 == n || ans2 == n) {cout << i + 1 << endl;return 0;}// 检查行和列if (row[row1] == n || col[col1] == n) {cout << i + 1 << endl;return 0;}}cout << "-1" << endl;return 0;
}