[51NOD1105]第k大的数(二分答案)

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1105

先排序，二分上下界分别是最小的两个数和最大的两个数的乘积。注意到一个性质，就是a[i]*b[j]，i从左到右，j从右到左。假如遇到一组a[i]*b[j]<=x，j就不必再减小了。这时候比x小的数一定是j个。下一次继续找i，这个b[j]一定不可能大于b[j+1]。问题就由找第k大的数，转换成找第n*n-k+1小的数。每次二分这个乘积x，看看是否符合条件即可。

 /*
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 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
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 ┛┗┛┗┛┃
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 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%lld", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f, sizeof(a))
 #define lp p << 1
 #define rp p << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 LL n, k, ans;
 LL a[maxn], b[maxn];

 bool ok(LL x) {
     LL ret = ;
     LL j = n;
     For(i, , n+) {
         while(j) {
             if(a[i] * b[j] > x) j--;
             else break;
         }
         ret += j;
     }
     return ret >= k;
 }

 int main() {
     // FRead();
     while(cin >> n >> k) {
         k = n * n - k + ;
         For(i, , n+) {
             cin >> a[i] >> b[i];
         }
         sort(a+, a+n+); sort(b+, b+n+);
         LL l = a[] * b[], r = a[n] * b[n];
         while(l <= r) {
             LL m = (l + r) >> ;
             if(ok(m)) {
                 r = m - ;
                 ans = m;
             }
             else l = m + ;
         }
         cout << ans << endl;
     }
     RT ;
 }