[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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 */

 #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 ;

 }

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[51NOD1105]第k大的数(二分答案)

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