zoukankan      html  css  js  c++  java
  • CodeForces

    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 nn flowers in a row in the exhibition. Sonya can put either a rose or a lily in the ii-th position. Thus each of nn positions should contain exactly one flower: a rose or a lily.

    She knows that exactly mm people will visit this exhibition. The ii-th visitor will visit all flowers from lili to riri 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 nn and mm (1n,m1031≤n,m≤103) — the number of flowers and visitors respectively.

    Each of the next mm lines contains two integers lili and riri (1lirin1≤li≤ri≤n), meaning that ii-th visitor will visit all flowers from lili to riri inclusive.

    Output

    Print the string of nn characters. The ii-th symbol should be «0» if you want to put a rose in the ii-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 [13][1…3], there are one rose and two lilies, so the beauty is equal to 12=21⋅2=2;
    • in the segment [24][2…4], there are one rose and two lilies, so the beauty is equal to 12=21⋅2=2;
    • in the segment [25][2…5], there are two roses and two lilies, so the beauty is equal to 22=42⋅2=4.

    The total beauty is equal to 2+2+4=82+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 [56][5…6], there are one rose and one lily, so the beauty is equal to 11=11⋅1=1;
    • in the segment [14][1…4], there are two roses and two lilies, so the beauty is equal to 22=42⋅2=4;
    • in the segment [46][4…6], there are two roses and one lily, so the beauty is equal to 21=22⋅1=2.

    The total beauty is equal to 1+4+2=71+4+2=7.

      一种新的思维方式,不是 根据查询要求构造满足查询区间要求的值 ,而是 构造一个完美的序列,使得它对任意的查询,都满足要求。 (不要为了活着而生活,因为生活所以活着)

      题目要求的是求一个最大的数,区间中两种不同花的数量的乘积,不难发现,应当尽可能的使区间中两种花的数量都变大,不能让其中一种花的数目很大,而另一种很小,因为这样对一些查询区间是不合适的。为避免这种情况,所以应该是让两种花交替的种植。

     1 #include<cstdio>
     2 #include<cstdlib>
     3 #include<cstring>
     4 #include<string>
     5 #include<cmath>
     6 #include<algorithm>
     7 #include<queue>
     8 #include<stack>
     9 #include<deque>
    10 #include<map>
    11 #include<iostream>
    12 using namespace std;
    13 typedef long long  LL;
    14 const double pi=acos(-1.0);
    15 const double e=exp(1);
    16 const int N = 10;
    17 
    18 int main()
    19 {
    20     int n,m,t;
    21     int x,y,i,p,j;
    22     scanf("%d%d",&n,&m);
    23     for(i=1;i<=m;i++)
    24     {
    25         scanf("%d%d",&x,&y);
    26     }
    27     for(i=1;i<=n;i++)
    28     {
    29         if(i%2)
    30             printf("1");
    31         else
    32             printf("0");
    33     }
    34     putchar('
    ');
    35     return 0;
    36 }
    View Code
  • 相关阅读:
    极速南瓜(北京)品牌管理顾问有限公司
    许涛芳_百度百科
    个性化品牌开始繁荣?为设计师和代工厂牵线的平台Maker's Row获得100万美元融资 | 36氪
    薛蟠_百度百科
    莫龙丹_百度百科
    尊履·尚品|手工鞋|固特异|定制鞋
    联系我们_你我想法_【有男度】UNANDU 100%进口 全球设计师品牌精汇 男装_男装搭配_时尚男装_品牌男装_男装搭配技巧_男装网站
    全球高级定制鞋完全指南
    百年定制老字号落户扬城 服装“私人定制”悄然兴起
    享受私人定制服装(下)
  • 原文地址:https://www.cnblogs.com/daybreaking/p/9694801.html
Copyright © 2011-2022 走看看