Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.
The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has nn hotels, where the ii-th hotel is located in the city with coordinate xixi. Sonya is a smart girl, so she does not open two or more hotels in the same city.
Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to dd. The girl understands that there are many possible locations to construct such a hotel. Thus she wants to know the number of possible coordinates of the cities where she can build a new hotel.
Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original nn hotels to the new one is equal to dd.
The first line contains two integers nn and dd (1≤n≤1001≤n≤100, 1≤d≤1091≤d≤109) — the number of Sonya's hotels and the needed minimum distance from a new hotel to all others.
The second line contains nn different integers in strictly increasing order x1,x2,…,xnx1,x2,…,xn (?109≤xi≤109?109≤xi≤109) — coordinates of Sonya's hotels.
Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to dd.
4 3 -3 2 9 16
6
5 2 4 8 11 18 19
5
In the first example, there are 66 possible cities where Sonya can build a hotel. These cities have coordinates ?6?6, 55, 66, 1212, 1313, and 1919.
In the second example, there are 55 possible cities where Sonya can build a hotel. These cities have coordinates 22, 66, 1313, 1616, and 2121
有一个长为n的地方可以放rose和lily两种花,有d位游客,每位游览不同的区间,现要求每位游客游览的区间内rose数乘Lily数最大,想了好久,以为是爆搜或者是动态规划一类的难题,果断放弃。
晚上搜题解,才发现是一道考验智商的大水题,只需保证0101010这样的顺序就可以保证每位游客经历的区间不同花的个数最多了。深感遗憾和震惊!!!!
#include<bits/stdc++.h>
using namespace std;int main()
{int a,b,t;scanf("%d%d",&a,&b);for(int i=0;i<b;i++){cin>>t;cin>>t;}for(int i=0;i<a;i++){if(i&1)cout<<'1';elsecout<<'0';}cout<<endl;return 0;
}