$CF\ 638\ (Div2)$

$A.$

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

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

#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], " \n"[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\n", (1LL << (k + 1)) - 2);
    }
    return 0;
}

$B.$

给定长度为 $n$ 的序列 $a$，其中 $1\leq a_{i}\leq n$
可以在任意位置插入 $1\sim n$ 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，$1\leq k\leq n\leq 100$

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 $1\sim n$ 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 $n\times k$

#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], " \n"[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", n * k);
            for(int i = 1; i <= n; ++i) {
                for(int j = 0; j < k; ++j)
                    printf("%d%c", sub[j], " \n"[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 $minimize\ \left\{s_{i}\mid i = 1,\ 2,\ ...,\ k\right\}$，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 $s[k - 1] \neq s[0]$，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 $s[k - 1]\neq s[k]$，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], " \n"[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
*/

# $CF\ 638\ (Div2)$

\(CF\ 638\ (Div2)\)

\(A.\)

\(B.\)

\(C.\)

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