hdu 5720 - Wool (线段交的长度）

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

22 1 31 14 3 101 1 2 4

 

Sample Output

25
Hint
In the first example, $ 2, 3 $ are available.In the second example, $ 6, 7, 8, 9, 10 $ are available.
 

 

思路： 排序，得出可笑的区间长度并，总长减去可行的，就是不可行的长度。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <bits/stdc++.h>#include <stack>#define LL long longusing namespace std;#define For(i,j,n) for(int i=j;i<=n;i++)#define mst(ss,b) memset(ss,b,sizeof(ss));const int N=1200000;const LL mod=998244353;int T;int n,m;int len;LL ll,rr;LL a[1200000];int b[1200000];LL r[120000000];LL l[120000000];bool cmp(LL x,LL b){    return x>b;}bool can(LL li,LL ri){    LL t=max(li,ll);    LL u=min(ri,rr);    if(t>u)        return false;    return true;}bool cmp2(int x,int y){    return l[x]<l[y];}int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d%I64d%I64d",&n,&ll,&rr);        for(int i=0;i<n;i++)            scanf("%I64d",&a[i]);        sort(a,a+n,cmp);        int cnt=0;        for(int i=0;i<n;i++)        {            r[i] = a[i] + a[i+1] -1;            l[i] = a[i] - a[i+1] +1;            if(can(l[i],r[i]))            {                b[cnt++]=i            }        }        sort(b,b+cnt,cmp2);        LL low,high;        LL ans=0;        for(int i=0;i<cnt;i++)        {            int j;            low  = l[b[i]];            high = r[b[i]];            for(j=i+1;j<cnt;j++)            {                if(l[b[j]]<=high)                {                    high = max(high,r[b[j]]);                }                else break            }            if(low <=ll) low=ll;            if(high>= rr) high=rr;            ans = ans + high - low + 1;            i=j-1;        }        printf("%I64d\n",rr-ll+1 - ans);    }    return 0;}