UCF Local Programming Contest 2017

A

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

int a, b , n, c;

int main()
{
    cin >> a >> b >> n;
    for (int i = 1; i <= n; ++i)
    {
        cin >> c; cout << c << ' ';
        int res = 0;
        if (c <= 1000) res = c * a;
        else res = 1000 * a + (c - 1000) * b;
        cout << res << '\n';
    }
    return 0;
}

B

#include <bits/stdc++.h>
#define P pair<int, int>
using namespace std;

P p[26];
int n;
char a[25], b[25];

int is(int a, int b)
{
    int x = abs(p[a].first - p[b].first), 
        y = abs(p[a].second - p[b].second);
    if (x + y == 0) return 1;
    if (x <= 1 && y <= 1) return 2;
    return 3;
}

int main ()
{
    for (int i = 0; i <= 8; ++i) p[i] = {0, i};
    for (int i = 9; i <= 17; ++i) p[i] = {1, i - 9};
    for (int i = 18; i <= 25; ++i) p[i] = {2, i - 18};
    cin >> n;
    while (n--)
    {
        int ans = 0; cin >> a + 1 >> b + 1;
        for (int i = 1; a[i] || b[i]; ++i)
        {
            if ((!a[i] && b[i]) || (!b[i] && a[i])) ans = 3;
            if (ans == 3) break;
            ans = max(ans, is(a[i] - 'a', b[i] - 'a'));
        }
        cout << ans << '\n';
    }
    return 0;
}

C

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

int t, n, k;

int main()
{
    cin >> t;
    while (t--)
    {
        ll ans = 0, c, d;
        cin >> n >> k >> c;
        while (--k)
        {
            cin >> d;
            ans += min((c + n - d + 1) % n, (d - c - 1 + n) % n);
            c = d;
        }
        cout << ans << '\n';
    }
    return 0;
}

D(广搜)

#include <bits/stdc++.h>
#define P pair<int, int>
#define PP pair<int, P>
#define inf 1e9
using namespace std;

int t, n, a[121], mp[121][81];
priority_queue<PP, vector<PP>, greater<PP> > q;

void updown(int x, int y, int aa, int cur, int c)
{
    if (x < 1 || x > n) return;
    y = min(y, a[x]);
    if (mp[x][y] > cur + 1)
    {
        mp[x][y] = cur + 1;
        q.push({ mp[x][y], {x, y} });
    }
    y = a[x] * aa;
    if (mp[x][y] > c + cur + 1)
    {
        mp[x][y] = c + cur + 1;
        q.push({ mp[x][y], {x, y} });
    }
}

int main()
{
    cin >> t;
    while (t--)
    {
        cin >> n; memset(mp, 0x7f7f, sizeof mp);
        priority_queue<PP, vector<PP>, greater<PP> >().swap(q);
        for (int i = 1; i <= n; ++i) cin >> a[i];
        int x, y, lx, ly, ans = inf;
        cin >> x >> y >> lx >> ly;
        q.push({ 0, {x, y} }); mp[x][y] = 0;
        while (!q.empty())
        {
            int cur = q.top().first;
            if (cur >= ans) break;
            P p = q.top().second; q.pop();
            if (p.first == lx) ans = min(ans, abs(ly - p.second) + cur);
            updown(p.first - 1, p.second, 1, cur, p.second);
            updown(p.first + 1, p.second, 0, cur, a[p.first] - p.second);
        }
        cout << ans << '\n';
    }
    return 0;
}

E(就是对三角函数的应用)

#include <bits/stdc++.h>
#define PI 3.141592653589793
using namespace std;

int t, w, a, b, c, n;

int calc(double x, double y)
{
    double r = x * x + y * y;
    if (r < a * a) return 1;
    if (r < b * b) return 2;
    if (r < c * c) return 3;
    return 4;
}

int main()
{
    cin >> t;
    while (t--)
    {
        cin >> w >> a >> b >> c >> n;
        double g = 2 * PI / w;
        int ans = 0;
        while (n--)
        {
            double x, y; cin >> x >> y;
            double angle = atan(abs(y/x));    
            if (x < 0 && y > 0) angle = PI - angle;
            else if (x < 0 && y < 0) angle += PI;
            else if (x > 0 && y < 0) angle = PI * 2 - angle;
            int ww = ceil(angle / g), r = calc(x, y);
            if (r == 1) {ans += 50; continue;}
            else if (r == 2) ans += 2 * ww;
            else if (r == 3) ans += ww;
        }
        cout << ans << '\n';
    }
    return 0;
}

I

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

const int maxn = 1e5 + 5;

int t, n, a[maxn], p[maxn];
ll c[maxn];

void add(int x, int y)
{
    for (; x <= n; x += -x & x) c[x] += y;
}

ll ask(int x)
{
    ll ans = 0;
    for (; x; x -= x & -x) ans += c[x];
    return ans;
}

int main()
{
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) cin >> a[i], p[a[i]] = i, add(i, a[i]);
        int cur = a[1];
        ll ans = 0;
        for (int i = 1; i <= n; ++i)
        {
            ll x = ask(p[i] - 1) - ask(p[cur] - 1);
            ll y =  ask(p[cur] - 1) - ask(p[i] - 1);
            if (p[i] < p[cur])  x += ask(n);
            if (p[i] > p[cur])  y += ask(n);
            add(p[i], -i); cur = i;
            ans += min(x, y);
        }
        printf("%lld\n", ans);
    }
    return 0;
}