# A.Two Subsequences

### Topic Essentials

Give you a string s. You need to find two non-empty strings a and b to satisfy the following conditions:

(1) Strings a and b are subsequences of s.

(2) For each index i, the character s [i] in the string s must exactly belong to one of the strings a or b.

(3) String a is the smallest possible dictionary order; String b can be any possible string.

For a given string s, print any valid strings a and b.

### Solving problems

Because the title requires the minimum dictionary order of string a, we can make string a the smallest character in string s, and string b the string s to delete string a.

### AC Code

#include <bits/stdc++.h> using namespace std; typedef long long ll; int main() { int t; cin >> t; while (t--) { string s, c; cin >> s; c = s; sort(c.begin(), c.end()); //Sort string c int k = s.find(c[0]); //Find the subscript for the smallest character in string s s.erase(s.begin() + k); //Delete the character at that position in string s cout << c[0] << ' ' << s << endl; //Output a and b } return 0; }

# B. Divine Array

### Topic Essentials

Give you a n array a whose length is n, and array a can be manipulated countless times.

For each operation, at each position J in the array, a j becomes equal to the number of times that a j appears in array a before this operation.

There are m queries asking, what is the value of the Xth number after k operations?

### Solving problems

Based on the simulation of several examples, it is found that after an array passes a certain operand, the value of the array does not change and n<=2000, so we might as well calculate the values of the array from 1 to n operations without timeout. (Proved, at most log 2 ( n ) \log_2(n) The value of the array will not change after log2 (n) operations)

### AC Code

#include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 2010; int cnt[N], d[N][N]; //The cnt array records the number of times each array appears in the array int main() //d[i][j] denotes the value of the jth position after the ith operation { int T; scanf("%d", &T); while (T--) { int n; scanf("%d", &n); for (int i = 1; i <= n; i++) { scanf("%d", &d[0][i]); } for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) cnt[j] = 0; //Initialize cnt array before each operation for (int j = 1; j <= n; j++) cnt[d[i - 1][j]]++; //Update cnt array for (int j = 1; j <= n; j++) d[i][j] = cnt[d[i - 1][j]]; //Modify the value after the first operation } int m; scanf("%d", &m); while (m--) { int k, x; scanf("%d %d", &x, &k); printf("%d\n", d[min(k, n)][x]); } } return 0; }

## C. Array Elimination

### Topic Essentials

Give you n numbers, you can choose any number of K (1<=k<=n) to operate on any number of times, assign the result of this number to x, subtract the number of K from x, and eventually make the number of N zero to find out how many such K are.

### Solving problems

First, we need to know that when both numbers are 1, &together results in 1.

Secondly, the title requires that the result of N numbers be zero at the end. It is possible to deduce that each of the binary representations of the last n numbers is zero. Then, how to change the number of n to the number of zeros i n the first k, the number of first k must all be 1, and the number of second I must be an integer multiple of k i n order to make all 1 i n the first I become 0. Finally, how to make every one zero, find out the number of each one, and find out the maximum common factor res of these numbers. So K is all the factors of res and is less than or equal to n.

### AC Code

#include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 2e5 + 10; int p[N], cnt[32]; //cnt[i] record number of position 1 int gcd(int a, int b) { return b ? gcd(b, a % b) : a; } int main() { int t; scanf("%d", &t); while (t--) { int n; scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%d", &p[i]); memset(cnt, 0, sizeof(cnt)); //Initialize cnt array for (int i = 1; i <= n; i++) //Traversal n Numbers { for (int j = 0; j < 30; j++) //Traverse every bit of every number { int k = (p[i] >> j) & 1; cnt[j] += k; } } int res = 0; for (int i = 0; i < 30; i++) res = gcd(res, cnt[i]); for (int i = 1; i <= n; i++) { if (res % i == 0) printf("%d ", i); } puts(""); } return 0; }