HDU5270Wool (區間掃描)

Wool

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 918    Accepted Submission(s): 270

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.

 

題意：

給定一個大區間[L,R]，以及n根木棒的長度，問大區間內有多少個數可以不與現有的木棒構成三角形。

題解：

我們考察任意的兩根木棒x,y(x>y)，我們分析反面，可以構成三角形的情況，那麼能讓x,y,z構成三角形的條件是x+y>z且x-y<z，也就是說當z落在(x-y,x+y)區間內的時候，會構成三角形，那麼對於某個x,在比他小的y裏面我們找一個最大的y即可快速計算出一個區間爲[x-y+1,x+y-1]，之後我們再在大區間裏抹去這些小區間即可。

代碼：

//************************************************************************//
//*Author : Handsome How                                                 *//
//************************************************************************//
#include <cstdio>
#include <algorithm>
#define fur(i,a,b) for(int i=(a);i<=(b);i++)
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
using namespace std;
typedef long long ll;
inline void gn(long long&x){
    int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
const int maxn = 1e5+5;
ll len[maxn], ans;
void del(ll L, ll R, ll l, ll r){	//從[L,R]裏面抹去[l,r] 
	if(l>R||r<L) return;
	l = max(l,L);
	r = min(r,R);
	ans = ans - (r-l+1);
}

int main(){
	int T;
	gn(T);
	while(T--){
		int n;
		ll L,R;
		gn(n);gn(L);gn(R);
		fur(i,1,n)gn(len[i]);
		sort(len+1,len+1+n);
		ans = R - L + 1ll;		
		ll l = len[2]-len[1]+1ll, r = len[2]+len[1]-1ll;
		int now = 3;
		for(;now<=n;now++){
			if(len[now]-len[now-1]+1ll>r){
				del(L,R,l,r);
				l = len[now]-len[now-1]+1ll;
			}
			else{
				l = min(l,len[now]-len[now-1]+1ll);
			}
			r = len[now]+len[now-1]-1ll;
		}
		del(L,R,l,r);
		printf("%I64d\n",ans);
	}
	return 0;
}