Input
Output
For each one output a string of A’s and B’s giving the order in which the machines are used.
Examples
Input
7
Output
AAA
Input
10
Output
ABB
额。。。其实很简单的一道题。。
本来打算 dfs 的。。。。
然后T在19个点。
int dfs(int x) { if (x > n)return 0; if (x == n)return 1; if (dfs(2 * x + 1)) { vis[++tot] = 1; return 1; } if (dfs(2 * x + 2)) { vis[++tot] = 2; return 1; } return 0; } 其实分个奇偶就可以了。。。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<string> #include<bitset> #include<ctime> #include<deque> typedef long long ll; using namespace std; typedef unsigned long long int ull; #define maxn 30005 #define ms(x) memset(x,0,sizeof(x)) #define Inf 0x7fffffff #define inf 0x3f3f3f3f const long long int mod = 1e9 + 7; #define pi acos(-1.0) #define pii pair<int,int> #define eps 1e-7 #define pll pair<ll,ll> ll quickpow(ll a, ll b) { ll ans = 1; a = a % mod; while (b > 0) { if (b % 2)ans = ans * a; b = b / 2; a = a * a; } return ans; } int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); } int n; int vis[maxn]; int tot; void dfs() { while (n) { if (n % 2 == 0) { vis[++tot] = 2; n = (n - 2) / 2; } else { vis[++tot] = 1; n = (n - 1) / 2; } } } int main() { // ios::sync_with_stdio(false); // cin >> n; scanf("%d", &n); dfs(); for (int i = tot; i >= 1; i--) { if (vis[i] == 1)printf("A"); else if (vis[i] == 2)printf("B"); } printf("\n"); }