1105 第K大的数
题目描述
数组A和数组B,里面都有n个整数。数组C共有n^2个整数,分别是A[0] * B[0],A[0] * B[1] ……A[1] * B[0],A[1] * B[1]……A[n - 1] * B[n - 1](数组A同数组B的组合)。求数组C中第K大的数。
例如:A:1 2 3,B:2 3 4。A与B组合成的C包括2 3 4 4 6 8 6 9 12共9个数。
输入
第1行:2个数N和K,中间用空格分隔。N为数组的长度,K对应第K大的数。(2 <= N <= 50000,1 <= K <= 10^9)
第2 - N + 1行:每行2个数,分别是A[i]和B[i]。(1 <= A[i],B[i] <= 10^9)
输出
输出第K大的数。
样例
Input示例
3 2
1 2
2 3
3 4
Output示例
9
题意
排序后二分即可
AC代码
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <ctime>
#include <vector>
#include <fstream>
#include <list>
#include <iomanip>
#include <numeric>
using namespace std;
#define ls st<<1
#define rs st<<1|1
#define fst first
#define snd second
#define MP make_pair
#define PB push_back
#define LL long long
#define PII pair<int,int>
#define VI vector<int>
#define CLR(a,b) memset(a, (b), sizeof(a))
#define ALL(x) x.begin(),x.end()
#define ber(i,s,e) for(int i=(s); i<=(e); i++)
#define rep(i,s,e) for(int i=(s); i>=(e); i--)
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+10;
const int mod = 1e9+7;
const double eps = 1e-8;
void fe() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt","w",stdout);
#endif
}
LL read()
{
LL x=0,f=1;char ch=getchar();
while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
while (ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
return x*f;
}
LL arr[MAXN], brr[MAXN];
LL n, k;
LL check(LL x) {
LL j = n, sum = 0;
for(int i = 1; i <= n; i++) {
while(j) {
if(arr[i]*brr[j] > x)
j--;
else
break;
}
sum += j;
}
return sum;
}
int main() {
ios::sync_with_stdio(false);
cin >> n >> k;
for(int i = 1; i <= n; i++) {
cin >> arr[i] >> brr[i];
}
sort(arr+1,arr+n+1);
sort(brr+1,brr+1+n);
LL l = arr[1]*brr[1], r = arr[n]*brr[n];
LL p = n*n-k+1;
LL mid, ans=0;
while(l <= r) {
mid = l + (r-l)/2;
if(check(mid) >= p)
r = mid-1;
else
l = mid+1;
}
if(check(r+1) >= p)
r++;
cout << r << endl;
return 0;
}