SCU2016-05 G题双重二分

Analyse:
先二分中位数，然后用二分去判断这个中位数是不是可以成立，双重二分。

/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 三  7/13 17:17:20 2016**Problem**:**Get**:**Code**:***********************[e-mail]**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;typedef vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a))#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3f3f3f3f;const int maxn = 1e5 + 13;int A[maxn];int n;ll k;bool judge(int mid) {    ll low = 0;    for (int i = 0; i < n; i++) {        int t = (lower_bound(A + i, A + n, mid + A[i]) - (A + i));        t--;        //pr(i), pl(t);        low += t;    }    return low < k;}int main(void){#ifdef LOCAL    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    while (~sa(n)) {        for (int i = 0; i < n; i++) sa(A[i]);        sort(A, A + n);        ll l = 0, r = int(2e9);        k = ((ll(n) * (n - 1)) / 2 + 1) / 2;        while (l < r) {            ll mid = (l + r + 1) / 2;        //    pr(l), pr(r), pl(mid);            if (judge(mid)) {                l = mid;            } else r  = mid - 1;        }        printf("%lld\n", l);    }    return 0;}