E - Finite Encyclopedia of Integer Sequences
题目传送门
Time limit : 2sec / Memory limit : 256MB
Score : 800 points
Problem Statement
In Finite Encyclopedia of Integer Sequences (FEIS), all integer sequences of lengths between 1 and N (inclusive) consisting of integers between 1 and K (inclusive) are listed.
Let the total number of sequences listed in FEIS be X. Among those sequences, find the (X⁄2)-th (rounded up to the nearest integer) lexicographically smallest one.
Constraints
1≤N,K≤3×105
N and K are integers.
Input
Input is given from Standard Input in the following format:
K N
Output
Print the (X⁄2)-th (rounded up to the nearest integer) lexicographically smallest sequence listed in FEIS, with spaces in between, where X is the total number of sequences listed in FEIS.
Sample Input 1
Copy
3 2
Sample Output 1
Copy
2 1
There are 12 sequences listed in FEIS: (1),(1,1),(1,2),(1,3),(2),(2,1),(2,2),(2,3),(3),(3,1),(3,2),(3,3). The (12⁄2=6)-th lexicographically smallest one among them is (2,1).
Sample Input 2
Copy
2 4
Sample Output 2
Copy
1 2 2 2
Sample Input 3
Copy
5 14
Sample Output 3
Copy
3 3 3 3 3 3 3 3 3 3 3 3 2 2
Solution
显然,当K是偶数的时候,中间的那个序列是k/2,k,k,k…一共n位数字;
那么K是奇数的时候呢?看似有些棘手,毕竟此处x/2还因四舍五入入了一位。找规律呀找规律……
假如n位全填(K+1)/2,此时距离我们想要的结果还差n/2位,向前数一数,暴力可过。
Code
#include <bits/stdc++.h>
#define ll long long
using namespace std;
int n,k;
int a[300000];
int main() {
scanf("%d%d",&k,&n);
if (k%2==0) {
printf("%d",k/2);
for (int i=1;i<n;++i) printf(" %d",k);
printf("\n");
} else {
fill(a,a+n,(k+1)/2);
int t=n-1;
for (int i=0;i<n/2;++i) {
if (a[t]!=0) a[t]--;
else {
a[t-1]--;
a[t]=k;
if (a[t-1]!=0) t=n-1;
else t--;//下一步还会再处理的
}
}
for (int i=0;i<n;++i) {
printf("%d",a[i]);
if (a[i+1]!=0) printf(" ");
else break;
}
printf("\n");
}
return 0;
}