Sonya and Exhibition CodeForces - 1004B(水题)

最新推荐文章于 2019-01-23 12:08:26 发布

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最新推荐文章于 2019-01-23 12:08:26 发布

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本文介绍了一个关于花卉展览的问题，Sonya希望在一条线上摆放玫瑰和百合，使得来访的每位游客看到的花段中，玫瑰与百合数量乘积之和最大化。通过间隔种植玫瑰和百合，可以实现这一目标。

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Sonya decided to organize an exhibition of flowers. Since the girl likes only roses and lilies, she decided that only these two kinds of flowers should be in this exhibition.

There are $n$ flowers in a row in the exhibition. Sonya can put either a rose or a lily in the $i$ -th position. Thus each of $n$ positions should contain exactly one flower: a rose or a lily.

She knows that exactly $m$ people will visit this exhibition. The $i$ -th visitor will visit all flowers from $l_{i}$ to $r_{i}$ inclusive. The girl knows that each segment has its own beauty that is equal to the product of the number of roses and the number of lilies.

Sonya wants her exhibition to be liked by a lot of people. That is why she wants to put the flowers in such way that the sum of beauties of all segments would be maximum possible.

Input

The first line contains two integers $n$ and $m$ ( $1 \leq n, m \leq 10^{3}$ ) — the number of flowers and visitors respectively.

Each of the next $m$ lines contains two integers $l_{i}$ and $r_{i}$ ( $1 \leq l_{i} \leq r_{i} \leq n$ ), meaning that $i$ -th visitor will visit all flowers from $l_{i}$ to $r_{i}$ inclusive.

Output

Print the string of $n$ characters. The $i$ -th symbol should be «0» if you want to put a rose in the $i$ -th position, otherwise «1» if you want to put a lily.

If there are multiple answers, print any.

Examples
inputCopy
5 3
1 3
2 4
2 5
outputCopy
01100
inputCopy
6 3
5 6
1 4
4 6
outputCopy
110010

Note

In the first example, Sonya can put roses in the first, fourth, and fifth positions, and lilies in the second and third positions;

in the segment $[1 \dots 3]$ , there are one rose and two lilies, so the beauty is equal to $1 \cdot 2 = 2$ ;
in the segment $[2 \dots 4]$ , there are one rose and two lilies, so the beauty is equal to $1 \cdot 2 = 2$ ;
in the segment $[2 \dots 5]$ , there are two roses and two lilies, so the beauty is equal to $2 \cdot 2 = 4$ .

The total beauty is equal to $2 + 2 + 4 = 8$ .

In the second example, Sonya can put roses in the third, fourth, and sixth positions, and lilies in the first, second, and fifth positions;

in the segment $[5 \dots 6]$ , there are one rose and one lily, so the beauty is equal to $1 \cdot 1 = 1$ ;
in the segment $[1 \dots 4]$ , there are two roses and two lilies, so the beauty is equal to $2 \cdot 2 = 4$ ;
in the segment $[4 \dots 6]$ , there are two roses and one lily, so the beauty is equal to $2 \cdot 1 = 2$ .

The total beauty is equal to $1 + 4 + 2 = 7$

题意：小明家有一个长为n的花园（只有玫瑰花和百合花，且花只能放在整数位置，每个位置放一朵），有m个小朋友来他家参观，每个小朋友将会以li为起点，参观到ri。小朋友获得的愉悦感等同与从l到r的距离里，玫瑰花与百合花数量的乘积。问小明怎样摆放花才能使所以的小朋友获得的愉悦感最大，分别用01表示玫瑰花和百合花。

思路：高中学过，a+b=t，要使a×b最大，那么，a和b应该尽量接近。也就是说，如果a+b=3，那么a=1，b=2，或者b=1，a=2。如果a+b=4，那么a=2，b=2。这样的话，不管小朋友怎么参观，我们只要将玫瑰和百合间隔种植就可以了。

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++) cout<<(i&1);
    cout<<endl;
    return 0;
}