Codeforces Global Round 12 C2. Errich-Tac-Toe (Hard Version)（思维）

C2. Errich-Tac-Toe (Hard Version)

Question: Give a matrix with k'X' or'O' marks, and now you need to modify no more than k / 3 marks ('X' is changed to'O','O' is changed to'X'), So that there are no three consecutive identical marks in the matrix

Idea: Modify the position of (i + j)% 3 == opx and (i + j)% 3 == opo separately for two different tags, provided that the number of modifications is less than one-third of the total number of tags One, so enumerate opx and opo, and then determine whether it is legal

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int N = 310;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;

char s[N][N];

int main() {
    int t, n;
    scanf("%d", &t);
    while(t--) {
        int xx = 0, oo = 0;
        scanf("%d", &n);
        int x[3] = {0, 0, 0}, o[3] = {0, 0, 0};
        for(int i = 0; i < n; ++i) scanf("%s", &s[i]);
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < n; ++j) {
                if(s[i][j] == 'X') xx++, x[(i + j) % 3]++;
                else if(s[i][j] == 'O') oo++, o[(i + j) % 3]++;
            }
        }
        int opx = -1, opo = -1;
        for(int i = 0; i < 3; ++i) {
            for(int j = 0; j < 3; ++j) {
                if(i == j) continue;
                if(x[i] + o[j] <= (xx + oo) / 3) {
                    opx = i, opo = j;
                    break;
                }
            }
            if(opx != -1) break;
        }
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < n; ++j) {
                if(s[i][j] == 'X' && (i + j) % 3 == opx) cout << 'O';
                else if(s[i][j] == 'O' && (i + j) % 3 == opo) cout << 'X';
                else cout << s[i][j];
            }
            cout<<'\n';
        }
    }
}