NOTE
比赛链接：Atcoder Beginner Contest 457

A - Array#

关键词：签到

Code#

1
// Problem: AT ABC457 A
2
// Contest: AtCoder - Polaris.AI Programming Contest 2026（AtCoder Beginner Contest 457）
3
// URL: https://atcoder.jp/contests/abc457/tasks/abc457_a
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// Time: 2026-05-09 20:00:09
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#include <bits/stdc++.h>
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using namespace std;
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// clang-format off
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#define endl '\n'
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#define all(x) (x).begin(), (x).end()
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#define fastio() ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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// clang-format on
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using ll = long long;
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using ull = unsigned long long;
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using pii = pair<int, int>;
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using pdd = pair<double, double>;
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using pll = pair<long long, long long>;
19

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const int dx[] = {-1, 0, 1, 0, -1, 1, 1, -1};
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const int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};
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const int inf = 0x3f3f3f3f;
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const int N = 0;
24

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void solve()
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{
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    int n;
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    cin >> n;
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    vector<ll> a(n + 1);
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    for (int i = 1; i <= n; i++)
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        cin >> a[i];
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    int x;
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    cin >> x;
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    cout << a[x] << endl;
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}
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int main()
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{
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    fastio();
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    int T = 1;
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    // cin >> T;
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    while (T--)
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        solve();
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    return 0;
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}

B - Arrays#

关键词：签到

NOTE
注意使用 resize() 分配空间。

Code#

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// Problem: AT ABC457 B
2
// Contest: AtCoder - Polaris.AI Programming Contest 2026（AtCoder Beginner Contest 457）
3
// URL: https://atcoder.jp/contests/abc457/tasks/abc457_b
4
// Time: 2026-05-09 20:00:10
5

6
#include <bits/stdc++.h>
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using namespace std;
8

9
// clang-format off
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#define endl '\n'
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#define all(x) (x).begin(), (x).end()
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#define fastio() ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
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// clang-format on
14

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using ll = long long;
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using ull = unsigned long long;
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using pii = pair<int, int>;
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using pdd = pair<double, double>;
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using pll = pair<long long, long long>;
20

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const int dx[] = {-1, 0, 1, 0, -1, 1, 1, -1};
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const int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};
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const int inf = 0x3f3f3f3f;
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const int N = 2e5 + 10;
25

26
void solve()
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{
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    int n;
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    cin >> n;
30

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    vector<vector<int>> a(n + 1);
32

33
    for (int i = 1; i <= n; i++)
34
    {
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        int m;
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        cin >> m;
37

38
        a[i].resize(m + 1);
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        for (int j = 1; j <= m; j++)
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            cin >> a[i][j];
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    }
43

44
    int x, y;
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    cin >> x >> y;
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47
    cout << a[x][y] << endl;
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}
49

50
int main()
51
{
52
    fastio();
53

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    int T = 1;
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    // cin >> T;
56

57
    while (T--)
58
        solve();
59

60
    return 0;
61
}

C - Long Sequence#

关键词：模拟

思路#

直接模拟即可。注意计算最后一轮的位置时，可以通过取模来实现。

Code#

1
// Problem: AT ABC457 C
2
// Contest: AtCoder - Polaris.AI Programming Contest 2026（AtCoder Beginner Contest 457）
3
// URL: https://atcoder.jp/contests/abc457/tasks/abc457_c
4
// Time: 2026-05-09 20:00:11
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#include <bits/stdc++.h>
6
using namespace std;
7

8
// clang-format off
9
#define endl '\n'
10
#define all(x) (x).begin(), (x).end()
11
#define fastio() ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
12
// clang-format on
13

14
using ll = long long;
15
using ull = unsigned long long;
16
using pii = pair<int, int>;
17
using pdd = pair<double, double>;
18
using pll = pair<long long, long long>;
19

20
const int dx[] = {-1, 0, 1, 0, -1, 1, 1, -1};
21
const int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};
22
const int inf = 0x3f3f3f3f;
23
const int N = 0;
24

25
void solve()
26
{
27
    ll n, k;
28
    cin >> n >> k;
29

30
    vector<vector<ll>> a(n + 1);
31
    vector<ll> L(n + 1);
32

33
    for (int i = 1; i <= n; i++)
34
    {
35
        int m;
36
        cin >> m;
37

38
        L[i] = m;
39
        a[i].resize(m + 1);
40

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        for (int j = 1; j <= m; j++)
42
            cin >> a[i][j];
43
    }
44

45
    vector<int> c(n + 1);
46

47
    for (int i = 1; i <= n; i++)
48
        cin >> c[i];
49

50
    for (int i = 1; i <= n; i++)
51
    {
52
        ll t = (ll) c[i] * L[i];
53

54
        if (k <= t)
55
        {
56
            ll pos = k % L[i];
57
            if (pos == 0)
58
                pos = L[i];
59

60
            cout << a[i][pos] << endl;
61
            return;
62
        }
63
        else
64
            k -= t;
65
    }
66
}
67

68
int main()
69
{
70
    fastio();
71

72
    int T = 1;
73
    // cin >> T;
74

75
    while (T--)
76
        solve();
77

78
    return 0;
79
}

D - Raise Minimum#

关键词：二分答案

思路#

题目求最小值最大，考虑二分答案。

我们遍历所有的 $A_i$ ，计算“为了让所有数都至少达到 mid，我们总共需要花费多少次操作（存入 tmp）。

NOTE
注意：使用小数计算 ceil 可能会丢精度，最好使用整数。也就是ceil(a / b) == (a + b - 1) / b。
另外这道题需要用 __int128。假设序列长度 $N = 2 \times 10^5$ ，操作次数 $K = 10^{18}$ ，如果我们把这 $10^{18}$ 次操作全部加在最后那个元素 $A_N$ 上（此时 $i = 2 \times 10^5$ ）。那么 $A_N$ 最终的值会变成： $A_N + K \times N \approx 10^{18} + 10^{18} \times 2 \cdot 10^5 \approx 2 \times 10^{23}$ 。
__int128 的用法详见杂项文档。

Code#

1
// Problem: AT ABC457 D
2
// Contest: AtCoder - Polaris.AI Programming Contest 2026（AtCoder Beginner Contest 457）
3
// URL: https://atcoder.jp/contests/abc457/tasks/abc457_d
4
// Time: 2026-05-09 20:00:12
5
#include <bits/stdc++.h>
6
using namespace std;
7

8
// clang-format off
9
#define endl '\n'
10
#define all(x) (x).begin(), (x).end()
11
#define fastio() ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
12
// clang-format on
13

14
using ll = long long;
15
using ull = unsigned long long;
16
using pii = pair<int, int>;
17
using pdd = pair<double, double>;
18
using pll = pair<long long, long long>;
19

20
const int dx[] = {-1, 0, 1, 0, -1, 1, 1, -1};
21
const int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};
22
const int inf = 0x3f3f3f3f;
23
const int N = 2e5 + 10;
24

25
void write(__int128 x)
26
{
27
    if (x < 0)
28
    {
29
        putchar('-');
30
        x = -x;
31
    }
32
    if (x > 9)
33
        write(x / 10);
34
    putchar(x % 10 + '0');
35
}
36

37
void solve()
38
{
39
    ll n, k;
40
    cin >> n >> k;
41

42
    vector<ll> a(n + 1);
43

44
    for (ll i = 1; i <= n; i++)
45
        cin >> a[i];
46

47
    __int128 l = 0, r = 1e24, res;
48
    while (l < r)
49
    {
50
        __int128 mid = (l + r + 1) >> 1;
51
        __int128 tmp = 0;
52
        bool flag = true;
53

54
        for (int i = 1; i <= n; i++)
55
        {
56
            if (a[i] < mid)
57
            {
58
                __int128 cnt = (mid - a[i] + i - 1) / i;
59
                tmp += cnt;
60
            }
61

62
            if (tmp > k)
63
            {
64
                flag = false;
65
                break;
66
            }
67
        }
68

69
        if (flag)
70
            res = mid, l = mid;
71
        else
72
            r = mid - 1;
73
    }
74
    write(res);
75
}
76

77
int main()
78
{
79
    fastio();
80

81
    int T = 1;
82
    // cin >> T;
83

84
    while (T--)
85
        solve();
86

87
    return 0;
88
}