AtCoder Beginner Contest 092

C

因为是按顺序走的，所以假如第 i个点不用走，我们只需要减去第 i−1个点到第 i个点的距离，以及第 i个点到第 i+1个点的距离，最后再加上第 i−1个点到第 i+1 个点的距离即可

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// LUOGU_RID: 174605005
#include <bits/stdc++.h>
using namespace std;
int a[100001], n, ans;
int main()
{
    cin >> n;
    for (int i = 1; i <= n; ++i)
        cin >> a[i];
    int b = 0;
    for (int i = 1; i <= n + 1; ++i)
        b += abs(a[i] - a[i - 1]);
    for (int i = 1; i <= n; ++i)
    {
        ans = b - abs(a[i] - a[i - 1]) - abs(a[i + 1] - a[i]) + abs(a[i + 1] - a[i - 1]);
        cout << ans << endl;
    }
    cout << endl;
}

D

构造一个100*100的矩阵。

上半部分黑色，下半部分白色， 选择上半部分的B-1个不相邻的涂黑，下半部分类似。

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#include <bits/stdc++.h>
#define int long long
const int N = 155;
using namespace std;

char col[N][N];
const int A = 50, B = 100;

void solve()
{
    int a, b;
    cin >> a >> b;
    a--, b--;
    cout << B << ' ' << B << endl;
    for (int i = 1; i <= A; i++)
        for (int j = 1; j <= B; j++)
        {
            col[i][j] = '.';
            col[i + A][j] = '#';
        }
    for (int i = 2; b; i += 2)
        for (int j = 2; b && j <= B; j += 2)
            col[i][j] = '#', b--;
    for (int i = 52; a; i += 2)
        for (int j = 2; a && j <= B; j += 2)
            col[i][j] = '.', a--;
    for (int i = 1; i <= B; i++)
    {
        for (int j = 1; j <= B; j++)
            cout << col[i][j];
        cout << endl;
    }
}

main()
{
    ios::sync_with_stdio(0);
    int _T = 1; // cin>>_T;
    while (_T--)
        solve();
}