Codeforces Round #263 (Div. 2) proA
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题目:
A. Appleman and Easy Task
time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputToastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?
Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.
Input
The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces.
Output
Print "YES" or "NO" (without the quotes) depending on the answer to the problem.
Sample test(s)
input
3xxoxoxoxx
output
YES
input
4xxxoxoxooxoxxxxx
output
NO
题意分析:
判断每个位置相邻位置为o的数字,如果全为偶数就yes,否者no。简单模拟题。Orz
代码:
#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <string>#include <iostream>using namespace std;char a[105][105];int main(){ int n; char c; int flag; cin >> n; memset(a, 0, sizeof(a)); for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) { cin >> c; if (c == 'o') a[i][j] = 1; } flag = 1; for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) { int cnt = 0; if (a[i-1][j]) ++cnt; if (a[i+1][j]) ++cnt; if (a[i][j-1]) ++cnt; if (a[i][j+1]) ++cnt; if (cnt & 1) { flag = false; break; } } if (flag) cout << "YES" << endl; else cout << "NO" << endl; return 0;}
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