简单构造，设数列为1,1,...,1,2,2,...,2,3,3,....,3

设有 x 个1，y 个 2， z 个 3，枚举x，y即可。

不同的连续子序列有x + y + z + x*y + y*z + x*z。。。。

因为事实上K<=10^9时，最小的合法的 x 也不超过100.。。

所以复杂度远远没有想象中那么高。。。。。

Virtual Participation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Special Judge

Problem Description

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?

Input

There are at most 20 testcases，each testcase only contains a single integer K (1≤K≤109)

Output

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.

Sample Input

10

Sample Output

4
1 2 3 4

Author

XJZX

Source

2015 Multi-University Training Contest 4

#include <bits/stdc++.h>
using namespace std;
#define prt(k) cerr<<#k" = "<<k<<endl
const int N = 100000;
int main()
{
    int K;
    while (scanf("%d", &K)==1) {
        if (K<=2)
        {
            if (K==1) puts("1\n1");
            if (K==2) puts("2\n1 1");
            continue;
        }
        bool f = 1;
        for (int x=0;f && x<=1e5;x++) {
            for (int y=0;x+y<=1e5&&x+y+x*y<=K&&y<=sqrt(K+0.5);y++)
            {
                int t = K - x - y - x * y;
                if (t % (x+y+1)==0) {
                    int z = t / (x + y + 1);
                    if (z < 0 || x+y+z>min(N, K)) continue;
                    assert(x+y+z+x*y+y*z+x*z==K);

                    int n = x + y + z;
                    printf("%d\n", n);
                    for (int i=1;i<=n;i++) {
                        int c;
                        if (i<=x) c=1;
                        else {
                            if (i<=x+y) c = 2;
                            else c = 3;
                        }
                        printf("%d%c", c, i==n ? 10 : ' ' );
                    }
                    f =false;
                    break;
                }
            }
        }
        assert(!f);
    }
}

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