6617: Finite Encyclopedia of Integer Sequences
时间限制: 1 Sec 内存限制: 128 MB
提交: 239 解决: 42
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题目描述
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.
输入
Input is given from Standard Input in the following format:
K N
输出
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.
样例输入
3 2
样例输出
2 1
提示
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).
来源/分类
题意：给你个k和一个n，现在要组成序列，k表示能用1-k的数字，n表示序列的长度可以是1-n。如：k=3，n=2，则可以组成的序列是(1),(1,1),(1,2),(1,3),(2),(2,1),(2,2),(2,3),(3),(3,1),(3,2),(3,3)共12个，设总数有X个，问你按照字典序排序第X/2个是那个序列。
小结：规律题要耐心去找，慢慢试总会有收获的。
思路：这题有点幸运。。写上几组示例后发现:
k为偶数的时候答案就是k/2，k,k........
k为奇数时候写了多组然后各种瞎猜后竟然发现就是(k+1)/2,(k+1)/2,(k+1)/2......往前移动n/2个...当然也不能全靠猜，总感觉可能跟中间那个值有关于是竟然真的有关。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define FIN freopen("D://code//in.txt", "r", stdin)
#define ppr(i,x,n) for(int i = x;i <= n;i++)
#define rpp(i,n,x) for(int i = n;i >= x;i--)
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
inline int read() {//读入挂
int ret = 0, c, f = 1;
for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar());
if(c == '-') f = -1, c = getchar();
for(; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';
if(f < 0) ret = -ret;
return ret;
}
ll a[maxn];
int main()
{
IO;
ll k,n;
cin>>k>>n;
if(k%2 == 0)
{
ppr(i,1,n)
{
if(i == 1)
printf("%lld ",k/2);
else
printf("%lld%c",k,i==n?'\n':' ');
}
}
else
{
ll number = (k+1)/2;
ll num = n;
ppr(i,1,n) a[i]=number;
ppr(i,1,n/2)
{
if(a[num] == 1)
{
a[num] = k;
num--;
}
else
{
a[num]--;
num = n;
}
}
ppr(i,1,num)
{
printf("%lld%c",a[i],i==num?'\n':' ');
}
}
return 0;
}