E. Lucky Permutation
time limit per test
memory limit per test
input
output
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467
k-th permutation of integers from 1 to n. Determine how many lucky numbers in the permutation are located on the positions whose indexes are also lucky numbers.
Input
n and k (1 ≤ n, k ≤ 109)
Output
k-th permutation of numbers from 1 to n does not exist, print the single number "-1" (without the quotes). Otherwise, print the answer to the problem: the number of such indexes i, that i and ai
Examples
input
7 4
output
1
input
4 7
output
1
Note
n elements, where each integer from 1 to n occurs exactly once. The element of permutation in position with index i is denoted as ai (1 ≤ i ≤ n). Permutation a is lexicographically smaller that permutation b if there is such a i(1 ≤ i ≤ n), that ai < bi, and for any j (1 ≤ j < i) aj = bj. Let's make a list of all possible permutations of n elements and sort it in the order of lexicographical increasing. Then the lexicographically k-th permutation is the k-th element of this list of permutations.
In the first sample the permutation looks like that:
1 2 3 4 6 7 5
The only suitable position is 4.
In the second sample the permutation looks like that:
2 1 3 4
The only suitable position is 4.
当n大于13时，1到n这个数列真正排列的只有后面13个数字因为13! > k;所以用数位dp求出1到n-13上满足条件的数的个数。在用逆康拓展开求出后面13个数字第k个排列的情况，在求满足条件的数的个数.如果n <=13直接用逆康拓展开求出它第K个排列
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#define maxn 100005
#define INF 1e18
using namespace std;
typedef long long ll;
int dp[15][10], kk[20], vis[20], d[15];
void Init(){
d[0] = 1;
for(int i = 1; i <= 12; i++)
d[i] = d[i-1] * i;
dp[1][4] = dp[1][7] = 1;
for(int i = 2; i <= 10; i++){
dp[i][4] += dp[i-1][4] + dp[i-1][7];
dp[i][7] += dp[i-1][4] + dp[i-1][7];
dp[i][0] += dp[i-1][4] + dp[i-1][7] + dp[i-1][0];
}
}
int Query(int m){
int num[15];
int cnt = 0;
while(m){
num[cnt++] = m % 10;
m /= 10;
}
int ans = dp[cnt][0];
for(int i = cnt-1; i >= 0; i--){
if(num[i] > 7)
ans += dp[i+1][7];
if(num[i] > 4)
ans += dp[i+1][4];
if(num[i] != 4 && num[i] != 7)
break;
}
return ans;
}
bool judge(int m){
while(m){
int d = m % 10;
if(d != 4 && d != 7)
return false;
m /= 10;
}
return true;
}
int main(){
// freopen("in.txt", "r", stdin);
Init();
int n, k;
scanf("%d%d", &n, &k);
if(n <= 12){
ll p = 1;
for(int i = 1; i <= n; i++){
p *= i;
}
if(k > p){
puts("-1");
return 0;
}
}
int ans = 0;
if(n > 13)
ans += Query(n - 12);
int p = 0;
int mins = n - 12 >= 1 ? n - 12 : 1;
for(int i = mins; i <= n; i++)
kk[p++] = i;
k--;
for(int i = p-1; i >= 0; i--){
int h = k / d[i];
int v, cnt = 0;
for(int j = 0; j < p; j++){
if(vis[j] == 0){
if(cnt == h){
vis[j] = 1;
if(judge(kk[j]) && judge(mins+(p-1-i))){
ans++;
}
break;
}
cnt++;
}
}
k %= d[i];
}
printf("%d\n", ans);
return 0;
}
