$CF 638 (Div 2)$

(CF 638 (Div2))

(A.)

给定 (n) 个数，分别为 (2^1, 2^2, ..., 2^n)，保证 (n) 是偶数，是否可以将这些数分成两组，使得两组之间数字和的差最小，输出这个差值

注意到 (2^1 + 2^2 +.. + 2^{n - 1} = 2^n - 1)

如果一组拿到 (2^n)，那么另一组必须拿 (2^{n - 1}, 2^{n - 2}, ..., 2^{frac{n}{2}})

等比数列求和一下

[ans = 2^{n} + frac{2^1cdot (1 - 2^{frac{n}{2} - 1})}{1 - 2} - frac{2^{frac{n}{2}}cdot (1 - 2^{frac{n}{2}})}{1 - 2} = 2^{frac{n}{2} + 1} - 2 ]

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define pb push_back
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%d%c", a[i], " 
"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 2e5 + 7;
const double PI = acos(-1);
const double EPS = 1e-6;
using namespace std;
inline int read()
{
    char c = getchar();
    int ans = 0, f = 1;
    while(!isdigit(c)) {if(c == '-') f = -1; c = getchar();}
    while(isdigit(c)) {ans = ans * 10 + c - '0'; c = getchar();}
    return ans * f;
}

int t, n;
int main()
{
    t = read();
    while(t--) {
        n = read();
        int k = n / 2;
        printf("%lld
", (1LL << (k + 1)) - 2);
    }
    return 0;
}

(B.)

给定长度为 (n) 的序列 (a)，其中 (1leq a_{i}leq n)
可以在任意位置插入 (1sim n) 范围内的数字（包括首尾）
要求新数组中任意长为 (k) 的子数组的和都一样，输出这个新数组
其中，(1leq kleq nleq 100)

如果一个数组任意长为 (k) 的子数组和都一样，那么它必然有长度为 (k) 的循环节

考虑构造这个循环节

将序列 (a) 中数字去重，若去重后的数量大于 (k)，则无法构造长度为 (k) 的循环节；反之可以（长度不够的话填充 (1sim n) 内的任意整数）

最暴力的构造方式即为，输出 (n) 个循环节，新数组长度为 (n imes k)

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define pb push_back
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%d%c", a[i], " 
"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 2e5 + 7;
const double PI = acos(-1);
const double EPS = 1e-6;
using namespace std;
inline int read()
{
    char c = getchar();
    int ans = 0, f = 1;
    while(!isdigit(c)) {if(c == '-') f = -1; c = getchar();}
    while(isdigit(c)) {ans = ans * 10 + c - '0'; c = getchar();}
    return ans * f;
}
 
int t, n, k;
int main()
{
    t = read();
    while(t--) {
        n = read(), k = read();
        set<int> st;
        for(int i = 1; i <= n; ++i) {
            int x = read();
            st.insert(x);
        }
        if(st.size() > k) puts("-1");
        else {
            vector<int> sub;
            for(auto i: st) sub.pb(i);
            for(int i = 1; i <= k - int(st.size()); ++i) sub.pb(1);
            printf("%d
", n * k);
            for(int i = 1; i <= n; ++i) {
                for(int j = 0; j < k; ++j)
                    printf("%d%c", sub[j], " 
"[i == n && j== k - 1]);
            }
        }
 
    }
    return 0;
}
/*
*/

(C.)

给定一个字符串 (s)，要求分配每一个字符，将 (s) 划分成 (k) 个新串 (s_{i})，且每个字符都只分配一次，同时最小化新串中字典序最大的串，即 (minimize left{s_{i}mid i = 1, 2, ..., k ight})，输出这个串

将 (s) 升序排序，记共有 (kind) 个不同字符

考虑平均分配前 (k) 个字符，

• 若 (s[k - 1] eq s[0])，那么 (ans = s[k - 1])，例如

  [s = aabbb, k = 3 Rightarrow s_{1} = abb, s_{2} = a, s_{3} = b ]

• 若 (s[k - 1] = s[0])，考虑后缀 (s[k - 1: n])

  • 若 (kind = 1)，平均分配即可，例如

    [s = aaaaaaa, k = 3 Rightarrow s_{1} = aa, s_{2} = aa, s_{3} = aaa ]

  • 若 (kind = 2)

    • 若 (s[k - 1] = s[k])，直接输出后缀，例如
      [s = aaabbb, k = 2 Rightarrow s_{1} = a, s_{2} = aabbb ]

    • 若 (s[k - 1] eq s[k])，平均分配，例如
      [s = aabbb, k = 2 Rightarrow s_{1} = ab, s_{2} = abb ]

  • 若 (kind > 2)，直接输出后缀，例如
    $$
    s = aabbc, k = 2 Rightarrow s_{1} = a, s_{2} = abbc
    $$

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int, int>
#define pb push_back
#define arrayDebug(a, l, r) for(int i = l; i <= r; ++i) printf("%d%c", a[i], " 
"[i == r])
typedef long long LL;
typedef unsigned long long ULL;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int DX[] = {0, -1, 0, 1, 0, -1, -1, 1, 1};
const int DY[] = {0, 0, 1, 0, -1, -1, 1, 1, -1};
const int MOD = 1e9 + 7;
const int N = 2e5 + 7;
const double PI = acos(-1);
const double EPS = 1e-6;
using namespace std;
inline int read()
{
    char c = getchar();
    int ans = 0, f = 1;
    while(!isdigit(c)) {if(c == '-') f = -1; c = getchar();}
    while(isdigit(c)) {ans = ans * 10 + c - '0'; c = getchar();}
    return ans * f;
}
 
int t, n, k;
string s;
int main()
{
    t = read();
    while(t--) {
        n = read(), k = read();
        cin >> s;
        sort(s.begin(), s.end());
        if(s[k - 1] != s[0]) cout<<s[k - 1]<<endl;
        else {
            set<int> st(s.begin(), s.end());
            if(st.size() == 1) {
                int len = s.length() % k ? s.length() / k + 1 : s.length() / k;
                //cout<<len<<endl;
                for(int i = 1; i <= len; ++i) putchar(s[0]);
                puts("");
            }
            else if(st.size() == 2) {
                if(k < n - 1 && s[k - 1] == s[k]) {cout<<s.substr(k - 1, s.length() - k + 1)<<endl;}
                else {
                    int len = s.length() % k ? s.length() / k + 1 : s.length() / k;
                    putchar(s[k - 1]);
                    if(k < n) for(int i = 1; i < len; ++i) putchar(s[k]);
                    puts("");
                }
            }
            else cout<<s.substr(k - 1, s.length() - k + 1)<<endl;
        }
    }
    return 0;
}
/*
6
4 2
baba
5 2
baacb
5 3
baacb
5 3
aaaaa
6 4
aaxxzz
7 1
phoenix
*/