2023-3-21复习刷题练习

1075 台球碰撞

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
#include <cmath>
#include <iomanip>
#include <iostream>
#define pi = 3.1415926535
using namespace std;

int main()
{
    double L, W, x, y, R, a, v, s;
    while (cin >> L >> W >> x >> y >> R >> a >> v >> s) {
        if (L <= 0 || W <= 0) {
            break;
        }
        L -= 2 * R;
        W -= 2 * R;
        x -= R;
        y -= R;
        double arc = a * acos(-1) / 180;
        // cout << arc << endl;
        double LL = fabs(x + s * cos(arc) * v);
        double WW = fabs(y + s * sin(arc) * v);
        while (LL > 2 * L) {
            LL -= 2 * L;
        }
        while (WW > 2 * W) {
            WW -= 2 * W;
        }
        double resx, resy;
        // cout << LL << " " << WW << endl;
        if (LL < L) {
            resx = LL + R;
        } else {
            resx = 2 * L - LL + R;
        }
        if (WW < W) {
            resy = WW + R;
        } else {
            resy = 2 * W - WW + R;
        }
        cout << fixed << setprecision(2) << resx << " " << resy << endl;
    }
    return 0;
}
1076 内部收益率

题解

NPV(x) = CF0 + CF1/x + CF2/x^2 + … + CFT/x^T

求导后，容易发现NPV是单调递减的，

并且x -> 0时，NPV -> +∞ > 0,

x -> +∞时，NPV -> CF0 < 0,

所以NPV(x)在(0,+∞)有且仅有一个零点。

二分法求出零点。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
#include <cmath>
#include <iomanip>
#include <iostream>
#include <vector>
using namespace std;
double Exp = 1e-9;
double a[26];
int n;
int check(double mid)
{
    double res = 0;
    for (int i = 1; i <= n; i++) {
        res += a[i] / pow(mid, i);
    }
    if (res + a[0] > 0) {
        return 1;
    } else {
        return 0;
    }
}
int main()
{
    while (cin >> n && n) {
        for (int i = 0; i <= n; i++) {
            cin >> a[i];
        }
        double l = 0, r = 1e6;
        while (fabs(r - l) >= Exp) {
            double mid = (r + l) / 2;
            if (check(mid) == 1) {
                l = mid;
            } else {
                r = mid;
            }
        }
        cout << fixed << setprecision(2);
        cout << l - 1 << endl;
    }
    return 0;
}
1077 Biggest Number

题解超时

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
#include <iostream>

#include <queue>

using namespace std;
int n, m, cnt;
// x * m + y;
string ans;
string s[31];
int dx[] = { 0, 0, 1, -1 };
int dy[] = { 1, -1, 0, 0 };
int vis[31][31];
int vis2[31][31];

bool cut(int x, int y, string now)
{
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            vis2[i][j] = 0;
        }
    }
    queue<pair<int, int>> q;
    q.push({ x, y });
    vis2[x][y] = 1;
    int count = 0;
    while (q.size()) {
        auto [xx, yy] = q.front();
        q.pop();
        for (int i = 0; i < 4; i++) {
            int fx = xx + dx[i];
            int fy = yy + dy[i];
            if (fx < 0 || fx >= n || fy < 0 || fy >= m) {
                continue;
            }
            if (s[fx][fy] == '#') {
                continue;
            }
            if (vis2[fx][fy] || vis[fx][fy]) {
                continue;
            }
            q.push({ fx, fy });
            count++;
            vis2[fx][fy] = 1;
            if (count + now.size() > ans.size())
                return false;
        }
    }
    if (count + now.size() < ans.size())
        return true;
    if (count + now.size() == ans.size()) {
        if (ans > now) {
            return true;
        }
    }
    return false;
}

void dfs(int x, int y, string now)
{
    if (cut(x, y, now)) {
        return;
    }
    vis[x][y] = 1;
    for (int i = 0; i < 4; i++) {
        int fx = x + dx[i];
        int fy = y + dy[i];
        if (fx < 0 || fx >= n || fy < 0 || fy >= m) {
            continue;
        }
        if (s[fx][fy] == '#' || vis[fx][fy]) {
            continue;
        }
        string tmp = now + s[fx][fy];
        dfs(fx, fy, tmp);
    }
    vis[x][y] = 0;
    if (ans.size() < now.size()) {
        ans = now;
    } else if (ans.size() == now.size()) {
        ans = max(ans, now);
    }
}
int main()
{
    while (cin >> n >> m && n && m) {
        for (int i = 0; i < n; i++) {
            cin >> s[i];
        }

        ans = "";
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (s[i][j] != '#') {
                    string tmp = "";
                    tmp += s[i][j];
                    dfs(i, j, tmp);
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}

A. Monsters (easy version)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

int main()
{
    int T;
    cin >> T;
    while (T--) {
        int n;
        cin >> n;
        vector<long long> a(n);
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }
        sort(a.begin(), a.end());
        long long cnt = 1, flag = 0, ans = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] == cnt) {
                flag = 1;
            } else {
                x if (flag)
                {
                    cnt++;
                    i--;
                }
                else
                {
                    ans += a[i] - cnt;
                    a[i] = cnt;
                    cnt++;
                }
                flag = 0;
            }
        }
        // for (int i = 0; i < n; i++)
        //     cout << a[i] << " ";
        // cout << endl;
        cout << ans << endl;
    }
    return 0;
}