uva 11536 - Smallest Sub-Array（two pointers）

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

2 20 12 4 20 12 8

Case 1: 13 Case 2: sequence nai

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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;
}