AtCoder Beginner Contest 326

B - 326-like Numbers

题意：找到一个不小于n的数是326数，定义是

思路：简单的模拟循环即可

#include<bits/stdc++.h>
using namespace std;
bool check(int x){
	vector<int>a;
	while(x){
		a.push_back(x%10);
		x/=10;
	}
	if(a[1]*a[2]==a[0])return true;
	else return false;
}
void solve(){
	int n;
	cin>>n;
	for(int i=n;i<=919;i++){
		if(check(i)){
			cout<<i;
			return;
		}
	}
}
int main(){
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int t=1;
	//cin>>t;
	for(int i=1;i<=t;i++)solve();
	return 0;
} 

C - Peak

题意：有n个礼物在A[i]上，找到一个区间[A[i],A[i]+m]使得包含的礼物最多，求包含最多的个数

思路：用尺取法模拟，满足条件j指针就++，不满足i指针++

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=3e5+10;
int a[N];
void solve(){
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++)cin>>a[i];
	sort(a+1,a+1+n);
	int ans=0;
	for(int i=1,j=1;j<=n;){
		if(a[j]-a[i]>=m){
			ans=max(ans,j-1-i+1);
			i++;
			continue;
		}else{
			j++;
		}
		ans=max(ans,j-i);
	}
	cout<<ans;
}
signed main(){
	ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int t=1;
	//cin>>t;
	for(int i=1;i<=t;i++)solve();
	return 0;
}