Codeforces Round #381 (Div. 1) A. Alyona and mex 构造

A. Alyona and mex

题目连接：

http://codeforces.com/contest/739/problem/A

Description

Alyona's mother wants to present an array of n non-negative integers to Alyona. The array should be special.

Alyona is a capricious girl so after she gets the array, she inspects m of its subarrays. Subarray is a set of some subsequent elements of the array. The i-th subarray is described with two integers li and ri, and its elements are a[li], a[li + 1], ..., a[ri].

Alyona is going to find mex for each of the chosen subarrays. Among these m mexes the girl is going to find the smallest. She wants this minimum mex to be as large as possible.

You are to find an array a of n elements so that the minimum mex among those chosen by Alyona subarrays is as large as possible.

The mex of a set S is a minimum possible non-negative integer that is not in S.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105).

The next m lines contain information about the subarrays chosen by Alyona. The i-th of these lines contains two integers li and ri (1 ≤ li ≤ ri ≤ n), that describe the subarray a[li], a[li + 1], ..., a[ri].

Output

In the first line print single integer — the maximum possible minimum mex.

In the second line print n integers — the array a. All the elements in a should be between 0 and 109.

It is guaranteed that there is an optimal answer in which all the elements in a are between 0 and 109.

If there are multiple solutions, print any of them.

Sample Input

5 3
1 3
2 5
4 5

Sample Output

2
1 0 2 1 0

Hint

题意

给你n个数，然后m个区间，你需要构造n个数，使得这m个区间的mex最小值最大。

题解：

其实只和区间长度有关，你按照0123......0123这样去构造，那个区间总能够包含0到那么多的数的。

代码

#include<bits/stdc++.h>
using namespace std;

int n,m;
int main()
{
    scanf("%d%d",&n,&m);
    int ans = n;
    for(int i=1;i<=m;i++){
        int l,r;
        scanf("%d%d",&l,&r);
        ans = min(ans,r-l+1);
    }
    cout<<ans<<endl;
    for(int i=0;i<n;i++){
        cout<<i%ans<<" ";
    }
    cout<<endl;
}