AtCoder Beginner Contest 089

目录

• A - Grouping 2
• B - Hina Arare
• C - March
• D - Practical Skill Test

A - Grouping 2

#include<bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n;
int main(){
    cin >> n;
    cout << n / 3 << endl;
    return 0;
}

B - Hina Arare

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
set<char> s;
int main() {
    int n;
    cin >> n;
    while (n--) {
        char x;
        cin >> x;
        s.insert(x);
    }
    if (s.size() == 3) cout << "Three" << endl;
    else
        cout << "Four" << endl;
        return 0;
}

C - March

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
LL res = 0;
int n;
string s;
LL num[5];
int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> s;
        if (s[0] == 'A') num[0]++;
        if (s[0] == 'C') num[1]++;
        if (s[0] == 'H') num[2]++;
        if (s[0] == 'R') num[3]++;
        if (s[0] == 'M') num[4]++;
    }
    for (int i = 0; i <= 2; i++) {
        for (int j = i + 1; j <= 3; j++) {
            for (int k = j + 1; k <= 4; k++) {
                res += num[i] * num[j] * num[k];
            }
        }
    }
    cout << res << endl;
    return 0;
}

D - Practical Skill Test

给出一个NxM的矩阵，元素是1到NxM的所有数字，以及一个数字d

现在给出q个查询(q<=1e5)，每个查询有一个x和y，要求x变成x+d，花费是x和x+d在矩阵上的曼哈顿距离，直到x变为y为止

对于每个查询，输出花费

因为q很大，每次模拟就会爆炸，所以需要考虑离线的方法

用一个数组res存放从最小的值变到i所花的花费，然后直接输出\(res[y]-res[x]\)即可

#include <bits/stdc++.h>

using namespace std;

const int N = 3e2 + 5;
typedef long long LL;
int n, m, d;
int mp[N][N];
struct node {
    LL x, y;
} add[N * N];
int res[N * N];
int main() {
    cin >> n >> m >> d;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> mp[i][j];
            add[mp[i][j]].x = i;
            add[mp[i][j]].y = j;
        }
    }
    for (int i = d + 1; i <= n * m; i++) {
        res[i] = res[i - d] + abs(add[i].x - add[i - d].x) +
                 abs(add[i].y - add[i - d].y);
    }
    int q;
    cin >> q;
    while(q--){
        int x,y;
        cin >> x >> y;
        cout << res[y] - res[x] << endl;
    }
    return 0;
}