Sonya and Exhibition（简单思维题）

Sonya and Exhibition

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 li to ri

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≤n,m≤103

) — the number of flowers and visitors respectively.

Each of the next m
lines contains two integers li and ri (1≤li≤ri≤n), meaning that i-th visitor will visit all flowers from li to ri

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
Input

5 3
1 3
2 4
2 5

Output

01100

Input

6 3
5 6
1 4
4 6

Output

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…3]

, there are one rose and two lilies, so the beauty is equal to 1⋅2=2
;
in the segment [2…4]
, there are one rose and two lilies, so the beauty is equal to 1⋅2=2
;
in the segment [2…5]
, there are two roses and two lilies, so the beauty is equal to 2⋅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…6]

, there are one rose and one lily, so the beauty is equal to 1⋅1=1
;
in the segment [1…4]
, there are two roses and two lilies, so the beauty is equal to 2⋅2=4
;
in the segment [4…6]
, there are two roses and one lily, so the beauty is equal to 2⋅1=2

.

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

题意：

要在长度为n的花坛种n颗花（只有玫瑰花和百合花），一共有m个顾客赏花，每个顾客只看(l,r)区间内的花

每个顾客的满意度是所看到的玫瑰花 X 所看的百合花，求如何安排玫瑰花和百合花的位置使得顾客的满意度和最大

分析：

对于每个顾客满意度最大是当所看区间玫瑰花数量与百合花数量相等，所以玫瑰花和百合花可以间隔摆放，和所给区间无关

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i = 0; i < m; i++){
        int l,r;
        scanf("%d%d",&l,&r);
    }
    for(int i = 0; i < n; i++){
        printf("%d", i & 1 ? 1 : 0);
    }
    return 0;
}