UVa 11536 Smallest Sub-Array

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本文介绍了一个寻找特定整数序列中包含指定整数集合的最小子序列的问题，并提供了一段C++代码实现。该问题要求从给定序列中找到长度最短的子序列，该子序列需包含所有1到K之间的整数。

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11536 - Smallest Sub-Array

Time limit: 8.000 seconds

Consider an integer sequence consisting of N elements where
X1 = 1
X2 = 2
X3 = 3
Xi = (Xi−1 + Xi−2 + Xi−3)%M + 1 for i = 4 to N
Find 2 values a and b so that the sequence (Xa Xa+1 Xa+2 . . . Xb−1Xb) contains all the integers
from [1, K]. If there are multiple solutions then make sure (b − a) is as low as possible.
In other words, find the smallest subsequence from the given sequence that contains all the integers
from 1 to K.
Consider an example where N = 20, M = 12 and K = 4.
The sequence is {1 2 3 7 1 12 9 11 9 6 3 7 5 4 5 3 1 10 3 3}.
The smallest subsequence that contains all the integers {1 2 3 4} has length 13 and is highlighted
in the following sequence:
{1 2 3 7 1 12 9 11 9 6 3 7 5 4 5 3 1 10 3 3}.

Input
First line of input is an integer T (T < 100) that represents the number of test cases. Each case consists
of a line containing 3 integers N (2 < N < 1000001), M (0 < M < 1001) and K (1 < K < 101). The
meaning of these variables is mentioned above.

Output
For each case, output the case number followed by the minimum length of the subsequence. If there is
no valid subsequence, output ‘sequence nai’ instead. Look at the sample for exact format.

Sample Input
2
20 12 4
20 12 8

Sample Output
Case 1: 13
Case 2: sequence nai

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=1000005;
const int N=105;

int a[maxn]= {0,1,2,3},b[N],pre[maxn],n,m,k;
queue <int> q;

void initial()
{
    memset(b,0,sizeof(b));
    memset(pre,0,sizeof(pre));
    while(!q.empty())  q.pop();
}

void input()
{
    scanf("%d %d %d",&n,&m,&k);
    for(int i=4; i<=n; i++)  a[i]=(a[i-1]+a[i-2]+a[i-3])%m+1;
}

void solve(int co)
{
    int cnt=0,ans=maxn,Max=-1;
    for(int i=1; i<=n; i++)
    {
        while(cnt==k)
        {
            int t=q.front();
            ans=min(ans,Max+1-t);
            b[pre[t]]--;
            if(b[pre[t]]==0)  cnt--;
            q.pop();
        }
        if(a[i]<=k && a[i]>=1)
        {
            if(!b[a[i]])  cnt++;
            b[a[i]]++;
            pre[i]=a[i];
            q.push(i);
            Max=max(i,Max);
        }
    }
    if(ans!=maxn)  printf("Case %d: %d\n",co,ans);
    else  printf("Case %d: sequence nai\n",co);
}

int main()
{
    int T;
    scanf("%d",&T);
    for(int co=1; co<=T; co++)
    {
        initial();
        input();
        solve(co);
    }
    return 0;
}