Wool

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 686    Accepted Submission(s): 192

Problem Description

At dawn, Venus sets a second task for Psyche.

She is to cross a river and fetch golden wool from violent sheep who graze on the other side.

The sheep are wild and tameless, so Psyche keeps on throwing sticks to keep them away.

There are n
sticks on the ground, the length of the i-th
stick is ai.

If the new stick she throws forms a triangle with any two sticks on the ground, the sheep will be irritated and attack her.

Psyche wants to throw a new stick whose length is within the interval
[L,R].
Help her calculate the number of valid sticks she can throw next time.

Input

The first line of input contains an integer
T
(1≤T≤10),
which denotes the number of test cases.

For each test case, the first line of input contains single integer
n,L,R
(2≤n≤105,1≤L≤R≤1018).

The second line contains n
integers, the i-th
integer denotes ai
(1≤ai≤1018).

Output

For each test case, print the number of ways to throw a stick.

Sample Input

2
2 1 3
1 1
4 3 10
1 1 2 4

Sample Output

2
5

Hint

In the first example, $ 2, 3 $ are available.

In the second example, $ 6, 7, 8, 9, 10 $ are available.

Source

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题意：我做这道题时WA了很多发，发现是想太多了。

给你n个数和L,R。问你在【L，R】的区间内有多少个数不能和a数组中的任意两个数构成三角形。

题解：首先给n个数sort一下，你会发现，在这n个数中，最不能取的且最大的数就是a[n]（n从1开始）。对于i：1~n-1中，最不能取的区间就是【a[i+1]-a[i]+1，a[i+1]+a[i]-1】，枚举这个区间的i：1~n-1，再和【L,R】相比，取其相对于【L,R】的补集即可。

AC代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
typedef long long LL;
typedef unsigned long long uLL;
const int N=100010;
LL a[N];
int main()
{
	int n,t;
	LL R ,L;
	LL ans;
	scanf("%d",&t);
	while(t--)
	{
		ans=0;
		scanf("%d%I64d%I64d",&n,&L,&R);
		for(LL i=1;i<=n;i++)
		scanf("%I64d",&a[i]);
		std::sort(a,a+n+1);
		/*
		LL res=a[n]+a[n-1];
		LL sd=std::max(L,res);
		ans=R-sd+1;
		*/
		LL l,r;
		for(LL i=n-1;i>=1;--i)
		{    

			 l=a[i+1]-a[i]+1;
			 r=a[i+1]+a[i]-1;
			if(l>R||r<L)continue;
			if(R>r) ans+=R-r;
			R=l-1;
			if(L>R)break;
		}
		if(L<=R) ans+=R-L+1;

		printf("%I64d\n",ans); 

	}
}