标签:click color com with attr mes nbsp e30 pre
https://www.acwing.com/problem/content/1326/
思路:
枚举起点
正反方向
如果满足.输出
#include <bits/stdc++.h> using namespace std; int n; int g[16][16]; int dx[4] = {-1,-1,0,1}; int dy[4] = {0,1,1,1}; bool sucessful; int state,step; int main(){ ios::sync_with_stdio(0); cin >> n; for(int i = 1; i <= n; i++){ int x,y; cin >> x >> y; if(i % 2) g[x][y] = 1; else g[x][y] = 2; for(int j = 0; j < 4; j++) { int l = 0, r = 0; int a, b; while (true) { a = x + dx[j] * (l + 1);b = y + dy[j] * (l + 1); if(a < 1 || a > 15 || b < 1 || b > 15) break; if(g[x][y] != g[a][b]) break; l++; } while (true) { a = x - dx[j] * (r + 1);b = y - dy[j] * (r + 1); if(a < 1 || a > 15 || b < 1 || b > 15) break; if(g[x][y] != g[a][b]) break; r++; } if(l + r + 1>= 5){ sucessful = true; break; } } if(sucessful){ state = g[x][y]; step = i; break; } } if(state == 2) cout << "B" << " " << step; else if(state == 1) cout << "A" << " " << step; else cout << "Tie"; return 0; }
标签:click color com with attr mes nbsp e30 pre
原文地址:https://www.cnblogs.com/xcfxcf/p/12358290.html