题意
题目链接
Sol
由于阶乘的数量增长非常迅速，而(k)又非常小，那么显然最后的序列只有最后几位会发生改变。
前面的位置都是(i = a[i])。那么前面的可以直接数位dp/爆搜，后面的部分是经典问题，可以用逆康托展开计算。
代码语言：javascript复制#include<bits/stdc .h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 1, mod = 1e9 7, INF = 1e9 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x y < 0) return x y mod; return x y >= mod ? x y - mod : x y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x y < 0) x = x y mod; else x = (x y >= mod ? x y - mod : x y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod mod) % mod;}
template <typename A> inline void debug(A a){cout << a << 'n';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 c - '0', c = getchar();
return x * f;
}
int N, K, fac[MAXN];
vector<int> res;
int find(int x) {
sort(res.begin(), res.end());
int t = res[x];
res.erase(res.begin() x);
return t;
}
bool check(int x) {
while(x) {
if((x % 10) != 4 && (x % 10) != 7) return 0;
x /= 10;
}
return 1;
}
int ans;
void dfs(int x, int Lim) {//计算1 - lim中只包含4 7的数量
if(x > Lim) return ;
if(x != 0) ans ;
dfs(x * 10 4, Lim);
dfs(x * 10 7, Lim);
}
signed main() {
N = read(); K = read() - 1;
int T = -1; fac[0] = 1;
for(int i = 1; i <= N;i ) {
fac[i] = i * fac[i - 1];
res.push_back(N - i 1);
if(fac[i] > K) {T = i; break;}
}
if(T == -1) {puts("-1"); return 0;}
dfs(0, N - T);
for(int i = T; i >= 1; i--) {
int t = find(K / fac[i - 1]), pos = N - i 1;
if(check(pos) && check(t)) ans ;
K = K % fac[i - 1];
}
cout << ans;
return 0;
}
/*
*/
