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Wool

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.

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).

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

2
5

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

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

题意大致是：给出已有木棍个数及长度 然后给出可用的木棍的长度范围 现在要再放一根木棍但是要使得任意三根木棍都不能组成三角形 问放的这根木棍的长度共有多少种选择

秉持着两边之和小于等于第三边不能构成三角形

/*

测试数据：
1000
4 1 20
5 6 7 8

4 1 150
2 3 7 100

2 1 1000000000000000000
2 3

4 1 1000000000000000000
2 3 7 100

4 1000000 1000000000000000
2 3 7 100

4 2 3
1 1 1 1

4 1 100
2 3 9 100

2 2 2
1 5

2 2 2
1 1

2 2 2
2 3
*/

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100010
using namespace std;
long long a[maxn];
int main()
{
    int t,n;
    long long l,r,xl,xr,ans;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%I64d%I64d",&n,&l,&r);
        for(int i=1;i<=n;++i)
            scanf("%I64d",&a[i]);
        ans=0;
        sort(a+1,a+n+1);
        xl=l>1?l:1;
        xr=(a[2]-a[1])<=r?(a[2]-a[1]):r;
        if(xl<=xr)
            ans+=xr-xl+1;
        for(int i=3;i<=n;++i)
        {
            if((a[i]-a[i-1])>=l&&(a[i-1]+1)<=r&&(a[i]-a[i-1])>=a[i-1]+a[i-2])
            {
                xl=l>a[i-1]+a[i-2]?l:(a[i-1]+a[i-2]);
                xr=r<a[i]-a[i-1]?r:(a[i]-a[i-1]);
                if(xl<=xr)
                    ans+=xr-xl+1;
            }
        }
        if(a[n]+a[n-1]<=r)
        {
            if(a[n]+a[n-1]>=l)
                ans+=r-(a[n]+a[n-1])+1;
            else
                ans+=r-l+1;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

HDOJ 5720 Wool

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原文地址：http://blog.csdn.net/wyjwyl/article/details/51944044