CF 561 div2 C

题意：给定数组 a[0]~a[n-1], -1e9<=a[i]<=1e9 在其中找数 x，使得

min( |x-y|, |x+y| ) <= |x| < |y| <= max( |x-y|, |x+y| )

化简后就是 | y |<=2*| x |, x<=y

数据都取abs  

一个简单的二分 

wa在最大值上了

习惯上拿MOD当最大值了

二分最右边界的值设为MOD了 结果数据范围比MOD大

呜 下次MOD就专门做最大余数了

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<list>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> p;
typedef long double ld;
#define mem(x) memset(x, 0, sizeof(x))
#define me(x) memset(x, -1, sizeof(x))
#define fo(i,n) for(ll i=0; i<n; i++)
#define sc(x) scanf("%lf", &x)
#define pr(x) printf("%lld
", x)
#define pri(x) printf("%lld ", x)
#define lowbit(x) x&-x
const ll MOD = 1e9+7;
const ll M = 1e18;
const ll N = 3e5 +5;
ll a[N];
int main()
{
    ll i, j, k;
    ll n, m, t;
    cin>>n;
    for(i=0 ;i<n; i++)
    {
        cin>>a[i];
        a[i]=abs(a[i]);
    }
    sort(a,a+n);
    ll l, r, ans=0;
    a[n]=MOD;
    for(i=0; i<n; i++)
    {
        l=i; r=n;
        k=2*a[i];
        ll mid;
        while(l<=r)
        {
            mid=(l+r)/2;
            if(a[mid]>k) r=mid-1;
            else if(a[mid]<k) l=mid+1;
            else
            {
                if(a[mid]==a[mid+1])
                    {mid++;break;}
                else break;
            }
        }
        //cout<<"aaa"<<mid<<endl;
        if(a[mid]<=k)
        {
            ans+=mid-i;
        }
        else
        {
            if(a[mid-1]<=k)
                ans+=mid-1-i;
        }
    }
    cout<<ans<<endl;
    return 0;
}