CF600 B. Queries about less or equal elements

题目传送门:https://codeforces.com/problemset/problem/600/B

题目大意:
给定长度为\(n\)的序列\(A\)和长度为\(m\)的序列\(B\),对于\(B_i\)找到\(A_j\leqslant B_i\)的个数


排序,二分

乱搞就完事

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=2e5;
int A[N+10];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int n=read(0),m=read(0);
	for (int i=1;i<=n;i++)	A[i]=read(0);
	sort(A+1,A+1+n);
	for (int i=1;i<=m;i++){
		int pos=upper_bound(A+1,A+1+n,read(0))-A;
		printf("%d%c",pos-1,i==m?'\n':' ');
	}
	return 0;
}
posted @ 2021-07-02 15:51  Wolfycz  阅读(14)  评论(0编辑  收藏  举报