CodeForces 1166C A Tale of Two Lands

题目链接：http://codeforces.com/problemset/problem/1166/C

题目大意

　　给定 n 个数，任选其中两个数 x，y，使得区间 [min(|x - y|, |x + y|), max(|x - y|, |x + y|)] 能完全盖过区间 [min(|x|, |y|), max(|x|, |y|)]，请问一共有多少种选法？

分析

　　先随便举个例子，比如 |x| = 3, |y| = 5，可以发现，不管是 [3, 5]，还是 [-3, 5]， [3, -5]， [-3, -5]， [min(|x - y|, |x + y|), max(|x - y|, |x + y|)] 都是 [2, 8]，因此，负数对答案没有任何影响。

　　不妨设 x <= y，若要 [y - x, x + y] 覆盖 [x, y]，则必有 2*x >= y。这就很明显了，排完序之后对于每个数二分查找即可。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< string, int > PSI;
typedef set< int > SI;
typedef vector< int > VI;
typedef map< int, int > MII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
const double EPS = 1e-10;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 2e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

LL n, a[maxN], ans; 

int main(){
    INIT(); 
    cin >> n;
    Rep(i, n) {
        cin >> a[i];
        if(a[i] < 0) a[i] = -a[i];
    }
    sort(a, a + n, greater< int >());
    
    rFor(i, n - 1, 0) {
        LL tmp = lower_bound(a, a + i, 2 * a[i], greater< int >()) - a;
        ans += i - tmp;
    }
    cout << ans << endl;
    return 0;
}