CodeForces 740C - Alyona and mex(思维)

最新推荐文章于 2019-08-28 12:56:00 发布

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本文介绍了一道算法题目，要求构造一个数组使得特定条件下的最小mex尽可能大。mex定义为区间内未出现的最小非负整数。文章提供了解题思路及AC代码。

C. Alyona and mex
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Examples
input
5 3
1 3
2 5
4 5
output
2
1 0 2 1 0
input
4 2
1 4
2 4
output
3
5 2 0 1
Note
The first example: the mex of the subarray (1, 3) is equal to 3, the mex of the subarray (2, 5) is equal to 3, the mex of the subarray (4, 5) is equal to 2 as well, thus the minumal mex among the subarrays chosen by Alyona is equal to 2.

题意：
给出长度为n的数组，m个区间。要求构造一个数组（元素大于等于零），求出m个mex里面的最小值。

mex是区间里没有出现的最小元素（大于等于零）。

解题思路：
对于每一个区间，构造的方式为0123…，那么这个区间的mex就是区间的长度。
找出mex的最小值，然后对数组进行构造，令a[i] = i%res.
因为res是找的最小长度，所以其他的区间都可以包含0-res元素。

AC代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m;
    cin>>n>>m;
    int minn = INT_MAX;
    while(m--)
    {
        int l,r;
        cin>>l>>r;  minn = min(minn,r-l+1);
    }
    cout<<minn<<endl;
    for(int i = 0;i < n;i++)    printf("%s%d",i?" ":"",i%minn);
    return 0;
}