uva 11536 - Smallest Sub-Array（two pointers）

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探讨了寻找包含特定整数集合的最短子序列算法，输入序列遵循特定递推公式，目标是最小化子序列长度。

Smallest Sub-Array

Input: Standard Input

Output: Standard Output

Consider an integer sequence consisting of N elements where –

X₁ = 1

X₂ = 2

X₃ = 3

X_i = (X_i-1 + X_i-2 + X_i-3) % M + 1 for i = 4 to N

Find 2 values a and b so that the sequence (X_a X_a+1 X_a+2 ... X_b-1 X_b₎ 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 Output for Sample Input

20 12 4

20 12 8

Case 1: 13

Case 2: sequence nai

Problemsetter: Sohel Hafiz

Special Thanks: Md. Arifuzzaman Arif

枚举每个以x[i]为最后一个元素的，最小区间。

#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <iostream>
#include <stack>
#include <set>
#include <cstring>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 1000000 + 5;
const int INF = 1000000000;

int n, m, k;
int x[maxn];
int vis[maxn];

void pre(){
    x[1] = 1;x[2] = 2;x[3] = 3;
    for(int i = 4;i <= n;i++)
        x[i] = (x[i-1]+x[i-2]+x[i-3])%m+1;
}

int main(){
    int t, kase = 0;
    scanf("%d", &t);
    while(t--){
        kase++;
        scanf("%d%d%d", &n, &m, &k);
        pre();

        memset(vis, 0, sizeof(vis));
        int pos = 1;
        int ans = INF;
        int cnt = 0;
        for(int i = 1;i <= n;i++){
            vis[x[i]]++;
            if(x[i] <= k && vis[x[i]] == 1) cnt++;
            while(vis[x[pos]] > 1 || x[pos] > k){// x[pos]>k !!!
                vis[x[pos]]--;
                pos++;
            }
            if(cnt == k) ans = min(ans, i-pos+1);
        }
        if(ans != INF)
            printf("Case %d: %d\n", kase, ans);
        else
            printf("Case %d: sequence nai\n", kase);
    }
    return 0;
}