AtCoder Beginner Contest 088

目录

• A - Infinite Coins
• B - Card Game for Two
• C - Takahashi's Information
• D - Grid Repainting

A - Infinite Coins

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n, a;
int main() {
    cin >> n >> a;
    if (n <= a)
        cout << "Yes" << endl;
    else {
        for (int i = 0; i <= a; i++) {
            if ((n - i) % 500 == 0) {
                cout << "Yes" << endl;
                return 0;
            }
        }
        cout << "No" << endl;
        return 0;
    }
    return 0;
}

B - Card Game for Two

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n, a[N];
int main() {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    int sum1 = 0, sum2 = 0;
    int k = 1;
    for (int i = n - 1; i >= 0; i--) {
        if (k) {
            sum1 += a[i];
            k = 0;
        } else {
            sum2 += a[i];
            k = 1;
        }
    }
    cout << sum1 - sum2 << endl;
    return 0;
}

C - Takahashi's Information

给出一个3x3的数组C，问能否找到6个数a1 b1 a2 b2 a3 b3，使得(C_{i,j}==a_i+b_j)

其中所有的数都在0-100之内

因为数据范围很小，足以支持O(n^3)的算法，所以直接枚举a，去计算b，最后看能不能符合条件即可

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int a[3][3];
int main() {
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            cin >> a[i][j];
        }
    }
    for (int a1 = 0; a1 <= 100; a1++) {
        int b1 = a[0][0] - a1;
        if (b1 < 0) break;
        for (int a2 = 0; a2 <= 100; a2++) {
            int b2 = a[1][1] - a2;
            if (b2 < 0) break;
            if (a1 + b2 != a[0][1]) continue;
            if (a2 + b1 != a[1][0]) continue;
            for (int a3 = 0; a3 <= 100; a3++) {
                int b3 = a[2][2] - a3;
                if (b3 < 0) break;
                if (a1 + b3 != a[0][2]) continue;
                if (a2 + b3 != a[1][2]) continue;
                if (a3 + b1 != a[2][0]) continue;
                if (a3 + b2 != a[2][1]) continue;
                cout << "Yes" << endl;
                return 0;
            }
        }
    }
    cout << "No" << endl;
    return 0;
}

D - Grid Repainting

一个NxM的矩阵，要求从（1，1）点走到（n,m)点，只能走白块

现在可以在走之前将某些白块变成黑块，问最多能将多少白块变成黑块

直接求最短路，然后保留最短路，剩下的白块全部变成黑色即可

#include <bits/stdc++.h>

using namespace std;

const int N = 50 + 5;
typedef long long LL;
int n, m;
char mp[N][N];
int vis[N][N];
int f[2][4] = {0, 0, 1, -1, 1, -1, 0, 0};
int main() {
    cin >> n >> m;
    int num = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> mp[i][j];
            if (mp[i][j] == '.') num++;
        }
    }
    queue<int> q;
    q.push(1);
    q.push(1);
    q.push(1);
    int res = -1;
    while (!q.empty()) {
        int x = q.front();
        q.pop();
        int y = q.front();
        q.pop();
        int step = q.front();
        q.pop();
        if (x == n && y == m) {
            res = step;
            break;
        }
        if (vis[x][y]) continue;
        vis[x][y] = 1;
        for (int i = 0; i < 4; i++) {
            int xx = x + f[0][i];
            int yy = y + f[1][i];
            if (xx >= 1 && xx <= n && yy >= 1 && yy <= m &&
                (vis[xx][yy] == 0) && (mp[xx][yy] == '.')) {
                q.push(xx);
                q.push(yy);
                q.push(step + 1);
            }
        }
    }
    if(res!=-1){
        cout << num - res << endl;
    }
    else
        cout << -1 << endl;
    return 0;
}