AtCoder Beginner Contest 086

A

模拟题，也可以当成讨论题做。

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#include<bits/stdc++.h>
using namespace std;

int main()
{
    int a,b;
    while(cin>>a>>b){
        if((a*b)%2==0) cout<<"Even"<<endl;
        else cout<<"Odd"<<endl;
    }
    return 0;
}

B

模拟题。

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#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int a, b;

int get(int num) {
    int res = 0;
    while (num) {
        res++;
        num /= 10;
    }
    return res;
}
int main() {
    cin >> a >> b;
    int c = a * pow(10,get(b) ) + b;
    for (int i = 1; i <= sqrt(c); i++) {
        if (i * i == c) {
            cout << "Yes" << endl;
            return 0;
        }
    }
    cout << "No" << endl;
    return 0;
}

C

需要判断步数够不够，步数够的前提下需要保整奇偶数性质满足。自己手玩一下即可。

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#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n;
int main()
{
    cin >> n;
    int nowx = 0, nowy = 0, nowt = 0;
    int flag = 0;
    while (n--)
    {
        int x, y, t;
        cin >> t >> x >> y;
        if ((abs(x - nowx) + abs(y - nowy) > (t - nowt)) ||
            (((t - nowt) - abs(x - nowx) - abs(y - nowy)) % 2 == 1))
        {
            flag = 1;
        }
        nowx = x, nowy = y, nowt = t;
    }
    if (flag)
        cout << "No" << endl;
    else
        cout << "Yes" << endl;
    return 0;
}

D

由于是以2k为周期不断重复的，因此可以直接取模，然后由于黑白又是相间的，因此可以只计算黑色的，之后枚举每个矩形，考虑优化的话就是直接使用前缀和去枚举分界点。

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#include <iostream>
#include <cstdio>
using namespace std;
const int K = 2005;
int sum[K][K];
inline int calc(int x1, int y1, int x2, int y2) {return sum[x2][y2] - sum[x2][y1 - 1] - sum[x1 - 1][y2] + sum[x1 - 1][y1 - 1];}
int main()
{
	ios::sync_with_stdio(false), cin.tie(nullptr);
	int n, k, ans = 0;
	cin >> n >> k;
	for (int i = 1; i <= n; i++)
	{
		int x, y; char op;
		cin >> x >> y >> op;
		if (op == 'W') x += k;
		sum[x % (2 * k) + 1][y % (2 * k) + 1]++;
	}
	for (int i = 1; i <= 2 * k; i++)
		for (int j = 1; j <= 2 * k; j++)
			sum[i][j] += sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
	for (int i = 1; i <= k; i++)
		for (int j = 1; j <= k; j++)
		{
			int tmp = calc(1, 1, i, j) + calc(i + 1, j + 1, i + k, j + k) + calc(i + k + 1, j + k + 1, 2 * k, 2 * k) + calc(i + k + 1, 1, 2 * k, j) + calc(1, j + k + 1, i, 2 * k);
			ans = max(ans, max(tmp, n - tmp));
		}
	cout << ans;
	return 0;
}