1540: K large number of two array elements multiplied by the K large (binary answer + input and output optimization)

1540: K Large number
Time Limit:10 Sec Memory limit:128 MB
submit:104 Solved:6
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Description

There are two sequences, a, B, and their lengths are N and M respectively, so what is the K element after multiplying the elements in the two sequence by the N*M elements from the large to the small?

Input

The first behavior of the input is a positive integer t (t<=10), which represents a total of T-Group test data.

The first line of each set of test data has three positive integers n,m and K (1<=n, M<=100000,1<=k<=n*m), respectively, representing the length of a sequence, the length of the b sequence, and the subscript of the element being asked. The second behavior n a positive integer represents sequence a. The third behavior m positive integers represent sequence B. The size of all elements in the sequence satisfies [1,100000].

Output

For each set of test data, the output line contains an integer representing the number of elements of the K-large.

Sample Input
3
3 2 3
1 2 3
1 2
2 2 1
1 1
1 1
2 2 4
1 1
1 1
Sample Output
3
1

1

#include <bits/stdc++.h>using namespace Std;template<class t>inline T Read (t&x) {char C;    while ((C=getchar ()) <=32) if (c==eof) return 0;    BOOL Ok=false;    if (c== '-') Ok=true,c=getchar ();    for (x=0; c>32; C=getchar ()) x=x*10+c-' 0 ';    if (OK) x=-x; return 1;} Template<class t> inline T read_ (t&x,t&y) {return read (x) &&read (y);} Template<class t> inline T read__ (t&x,t&y,t&z) {return read (x) &&read (y) &&read (z);}    Template<class t> inline void Write (T x) {if (x<0) Putchar ('-'), x=-x;    if (x<10) putchar (x+ ' 0 '); else write (X/10), Putchar (x%10+ ' 0 ');}    Template<class t>inline void Writeln (T x) {write (x); Putchar (' \ n ');} -------ZCC IO template------const int MAXN=1E5+1;CONST double inf=999999999; #define Lson (rt<<1), L,m#define Rson (rt<<1|1), M+1,r#define M ((l+r) >>1) #define for (i,t,n) for (int i= (t);i< (n); i++) typedef long Long LL; typedef double DB;TYPEDEF pair<int,int> P; #define BUG printf ("---\ n"); #define MOD 1000000007int a[maxn],b[maxn];int n,m,k;bool pan (LL num) {int    cnt=0; for (int i=0;i<n;i++) cnt+=m-(Upper_bound (B,b+m,num/a[i])-B);//Find out if the number of elements larger than num//printf ("Num=%lld cnt=%d\n", n    UM,CNT); return cnt>k-1;}    void Slove () {ll l= (LL) a[0]*b[0];        ll H= (LL) a[n-1]*b[m-1];//between the maximum minimum value of the two-part answer while (L&LT;=H) {ll mid= (LL) (L+H)/2;        if (Pan (mid)) l=mid+1;    else h=mid-1; } writeln (l);}    int main () {int T;    Freopen ("In.txt", "R", stdin);    Read (T);        while (t--) {read__ (n,m,k);        for (i,0,n) read (A[i]);        for (i,0,m) read (B[i]);        Sort (a,a+n);        Sort (b,b+m);    Slove (); } return 0;}

1540: K large number of two array elements multiplied by the K large (binary answer + input and output optimization)