目录

• A - Traveling Budget
• B - Chocolate
• C - Traveling Plan
• D - Grid Components

A - Traveling Budget

#include<bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int a, b, c, d;
int main(){
    cin >> a >> b >> c >> d;
    cout << min(a, b) + min(c, d) << endl;
    return 0;
}

B - Chocolate

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n, x, d;
int main() {
    cin >> n >> d >> x;
    int res = x;
    for (int i = 1; i <= n; i++) {
        int a;
        cin >> a;
        res++;
        for (int j = 2; j <= d; j++) {
            if ((j - 1) % a == 0) res++;
        }
    }
    cout << res << endl;
    return 0;
}

C - Traveling Plan

#include <bits/stdc++.h>

using namespace std;

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

D - Grid Components

给出两个数a和b，要求构造一个地图，使得#联通块的数量为a，.联通块的数量为b

a,b<=500,地图大小小于100x100

构造出一个很大的固定长100 宽100的矩形来完成它，可以将\(100*100\)分为2个\(50*100\) 一个为黑色背景 一个为白色背景，然后往上放

不同的颜色即可

#include <bits/stdc++.h>

using namespace std;

int main() {
    int a, b;
    cin >> a >> b;
    a--, b--;
    printf("100 100\n");
    for (int i = 0; i < 50; i++) {
        for (int j = 0; j < 100; j++) {
            if (i % 2 && j % 2 && a) {
                a--;
                printf(".");
            } else
                printf("#");
        }
        printf("\n");
    }
    for (int i = 50; i < 100; i++) {
        for (int j = 0; j < 100; j++) {
            if (i % 2 && j % 2 && b) {
                b--;
                printf("#");
            } else
                printf(".");
        }
        printf("\n");
    }
    return 0;
}