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 << "]" << "\n";
     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, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , 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-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e9 + ;
 const int maxN = 2e5 + ;
 const LL ONE = ;
 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] < ) a[i] = -a[i];
     }
     sort(a, a + n, greater< int >());

     rFor(i, n - , ) {
         LL tmp = lower_bound(a, a + i,  * a[i], greater< int >()) - a;
         ans += i - tmp;
     }
     cout << ans << endl;
     return ;
 }