本文共 2461 字，大约阅读时间需要 8 分钟。

1462A. Favorite Sequence

简单看懂题即可，左边输出一个然后右边输出一个。

void solve() {    int n;    cin >> n;    vector
   
     a(n + 1);    // ll a[n + 1]; //两种写法均可    for (int i = 1; i <= n; ++i)        cin >> a[i];    int l = 1, r = n;    bool f = true;    for (int i = 1; i <= n; ++i) {        if (f)            cout << a[l++] << " ", f = false;        else            cout << a[r--] << " ", f = true;    }    cout << endl;}
   

1462B. Last Year's Substring

一开始想错了，正确的思路是拆分字符串看是否能组成 2020

void solve() {    int n;    string s;    cin >> n >> s;    bool f = false;    for (int fir = 0; fir <= 4 && !f; fir++) {        int sec = 4 - fir;  //定位        if (s.substr(0, fir) + s.substr(n - sec) == "2020")            f = true;    }    cout << (f ? "YES\n" : "NO\n");}

1462C. Unique Number

😂打表。注意一下n > 45的话直接输出-1（因为0~9都被使用了最多到45）

ll a[51] = {    0,       1,       2,       3,        4,        5,         6,      7,    8,       9,       19,      29,       39,       49,        59,     69,    79,      89,      189,     289,      389,      489,       589,    689,    789,     1789,    2789,    3789,     4789,     5789,      6789,   16789,    26789,   36789,   46789,   56789,    156789,   256789,    356789, 456789,    1456789, 2456789, 3456789, 13456789, 23456789, 123456789, -1,      -1,    -1,       -1,       -1,};void solve() {    int n, cnt = 0;    string s;    cin >> n;    cout << a[n] << endl;}

当然也可以正常分析：

void solve() {    int n;    cin >> n;    if (n > 45) {        cout << -1 << endl;        return;    }    string s;    int nxt = 9;    while (n > 0) {        if (n >= nxt) {            s += '0' + nxt;            n -= nxt--;        } else {            s += '0' + n;            break;        }    }    reverse(s.begin(), s.end());    cout << s << endl;}

D. Add to Neighbour and Remove

void solve() {    int n;    cin >> n;    vector
   
     a(n);    for (int i = 0; i < n; ++i)        cin >> a[i];    ll ans = INT_MAX, sum = 0;    for (int i = 0; i < n; ++i) {        sum += a[i];        ll cur = 0, toadd = 0;        bool f = true;        for (int j = 0; j < n; ++j) {            if (cur)                toadd++;            cur += a[j];            if (cur > sum) {                f = false;                break;            } else if (cur == sum)                cur = 0;        }        if (f && cur == 0) {            // cout << toadd << " " << i << endl;            ans = min(ans, toadd);        }    }    cout << ans << endl;}
   

E1. Close Tuples (easy version)

//unsolved