HDU 5334 Virtual Participation（2015多校第四场）

[align=left] [/align]
[align=left]Problem Description[/align]
[align=left] [/align]
As we know, Rikka is poor at math. Yuta is worrying about this situation, so he asks rikka to
have some practice on codeforces. Then she opens the problem B:

Given an integer K, she needs to come up with an sequence of integers A satisfying that the
number of different continuous subsequence of A is equal to k.

Two continuous subsequences a, b are different if and only if one of the following conditions
is satisfied:

1. The length of a is not equal to the length of b.

2. There is at least one t that at≠bt, where at means the t-th element of a and bt means the
t-th element of b.

Unfortunately, it is too difficult for Rikka. Can you help her?

[align=left]Input[/align]
[align=left] [/align]
There are at most 20 testcases，each testcase only contains a single integer K (1≤K≤109)

[align=left]Output[/align]
[align=left] [/align]
For each testcase print two lines.

The first line contains one integers n (n≤min(K,105)).

The second line contains n space-separated integer Ai (1≤Ai≤n) - the sequence you find.

[align=left]Sample Input[/align]
[align=left] [/align]

10

[align=left]Sample Output[/align]
[align=left] [/align]

4

1 2 3 4

题意：给一个整数k，输出一个长度为n的数列。要求这个数列的所有不为空且不相同的连续子序列的个数为k。

题解：当数列元素都为1时，数列子序列个数为k。所以当k小于等于105时，可以直接输出k个1。

当k大于105时，对于一个1到n的数列，该数列的子序列个数m为n(n+1)/2。如果我们将某个不与1相邻的数替换为1， 那么其m值就会减1；如果是从2的位置开始连续将x个数变为1，其m值减少x(x+1)/2。比如 1 2 3 4 5 6 7 8将3个数变为1后为 1 1 1 1 5 6 4 8，m值减少了6。

对于给定的k，我们可以先找到一个子序列个数大于等于k且最接近的数列n，则这个数列的长度就是我们所要求的数列的长度。当m>k时，求出连续1的个数和不连续1的个数， 将相应位置的数替换成1。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int pos[45000];  //存储数列的子序列的个数
void F()
{
for (int i=1; i<44730; i++)
pos[i] = (i * i + i) / 2;
}
int main()
{
F();
int k;
while (~scanf("%d", &k)){
if (k <= 100000){
printf("%d\n", k);
for (int i=1; i<k; i++)
printf("1 ");
printf("1\n");
}
else{
int x;
//找到子序列个数大于k的数列n
for (int i=1; i<44730; i++)
if (pos[i] >= k && pos[i-1] < k){
x = i;
break;
}
//数列n中不与1相邻元素所要替换成1的个数
int y = pos[x] - k;
printf("%d\n", x);
int temp;
//找到连续1的个数
for (int i=1; i<44730; i++)
if (pos[i] <= y && pos[i+1] > y){
temp = i;
break;
}
printf("1");
for (int i=1; i<=temp; i++)
printf(" 1");
//不与1相邻元素所要替换成1的个数
y = y - pos[temp];
int ans = 1;
for (int i=temp+2; i<=x; i++){
ans++;
if (ans % 2 && y > 0){
printf(" 1");
y--;
}
else
printf(" %d", i);
}
printf("\n");
}
}
return 0;
}

View Code