Codeforces Round #440 (Div. 2, based on Technocup 2018 Elimination Round 2)

最新推荐文章于 2022-12-16 16:19:21 发布

原创最新推荐文章于 2022-12-16 16:19:21 发布 · 596 阅读

0 ·

CC 4.0 BY-SA版权

数学知识专栏收录该内容

4 篇文章

订阅专栏

给定两个非零数字列表，任务是找到最小的正整数，该数的十进制表示至少包含一个来自第一个列表的数字和一个来自第二个列表的数字。输出这个最小的漂亮整数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

You are given two lists of non-zero digits.

Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 9) — the lengths of the first and the second lists, respectively.

The second line contains n distinct digits a₁, a₂, ..., a_n (1 ≤ a_i ≤ 9) — the elements of the first list.

The third line contains m distinct digits b₁, b₂, ..., b_m (1 ≤ b_i ≤ 9) — the elements of the second list.

Output

Print the smallest pretty integer.

Examples

input

2 3
4 2
5 7 6

output

input

8 8
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1

output

Note

In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.

In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer.

题意就是输出一个最小数，这个数最少要包含一个第一序列的数，和一个第二序列的数。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 0;
inline int get(){
	char c;
	while((c = getchar()) < '0' || c > '9');
	int cnt = c - '0';
	while((c = getchar()) >= '0' && c <= '9') cnt = cnt * 10 + c - '0';
	return cnt;
}
bool a[20],b[20];
int mina,minb;
int n,m,ans;

int main(){
/*	#ifdef lwy
	#else
		freopen(".in","r",stdin);
		freopen(".out","w",stdout);
	#endif*/
	memset(a,false,sizeof(a));
	memset(b,false,sizeof(b));
	n = get(); m = get();
	mina = 10; minb = 10;
	for(int i = 1; i <= n; i++){
		int x; x = get();
		mina = min(mina,x);
		a[x] = true;
	}
	for(int i = 1; i <= m; i++){
		int x; x = get();
		minb = min(minb,x);
		b[x] = true;
	}
	for(int i = 1; i <= 9; i++){
		if(a[i] && b[i]){
			ans = i;
			printf("%d",ans);
			return 0;
		}
	}
	ans = min(mina,minb) * 10 + max(mina,minb);
	printf("%d",ans);
	return 0;
}

You are given an array a₁, a₂, ..., a_n consisting of n integers, and an integer k. You have to split the array into exactly k non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the k obtained minimums. What is the maximum possible integer you can get?

Definitions of subsegment and array splitting are given in notes.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 10⁵) — the size of the array a and the number of subsegments you have to split the array to.

The second line contains n integers a₁, a₂, ..., a_n ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

Print single integer — the maximum possible integer you can get if you split the array into k non-empty subsegments and take maximum of minimums on the subsegments.

Examples

input
5 2
1 2 3 4 5

output
5

input
5 1
-4 -5 -3 -2 -1

output
-5

Note

A subsegment [l, r] (l ≤ r) of array a is the sequence a_l, a_l + 1, ..., a_r.

Splitting of array a of n elements into k subsegments [l₁, r₁], [l₂, r₂], ..., [l_k, r_k] (l₁ = 1, r_k = n, l_i = r_i - 1 + 1 for all i > 1) is k sequences (a_l₁, ..., a_r₁), ..., (a_{l_k}, ..., a_{r_k}).

In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are min(1, 2, 3, 4) = 1 and min(5) = 5. The resulting maximum is max(1, 5) = 5. It is obvious that you can't reach greater result.

In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4, - 5, - 3, - 2, - 1). The only minimum is min( - 4, - 5, - 3, - 2, - 1) = - 5. The resulting maximum is - 5.

题意是将一个序列分成k段，要求每段的最小值最大，并输出这个最大值。

一开始以为是个简单的二分。

但再仔细一看其实只要把k分成 == 1 ，== 2 ，>= 3，3类分类讨论就行了。

然而一开始没有考虑到 == 2 与 >=3 的区别，被hacked了。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1E5;
inline int get(){
	char c;
	while((c = getchar()) < '0' || c > '9');
	int cnt = c - '0';
	while((c = getchar()) >= '0' && c <= '9') cnt = cnt * 10 + c - '0';
	return cnt;
}
int N,K;
int a[MAXN+10];
int b[MAXN+10],c[MAXN+10];
int maxa,mina,ans;
int main(){
/*	#ifdef lwy
	#else
		freopen(".in","r",stdin);
		freopen(".out","w",stdout);
	#endif*/
	N = get(); K = get();
	for(int i = 1; i <= N; i++){
		scanf("%d",&a[i]);
	}
	mina = maxa = a[1];
	for(int i = 1; i <= N; i++){
		mina = min(mina,a[i]);
		maxa = max(maxa,a[i]);
	}
	if(K == 1){
		ans = mina;
	}
	if(K >= 3){
		ans = maxa;
	}
	if(K == 2){
		ans = mina;
		for(int i = 1; i <= N; i++){
			b[i] = c[i] = a[i];
		}
		for(int i = 2; i <= N; i++){
			b[i] = min(b[i],b[i-1]);
		}
		for(int i = N-1; i >= 1; i--){
			c[i] = min(c[i],c[i+1]);
		}
		for(int i = 1; i <= N-1; i++){
			ans = max(max(b[i],c[i+1]),ans);
		}
	}
	printf("%d",ans);
	return 0;
}

You are given several queries. In the i-th query you are given a single positive integer n_i. You are to represent n_i as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.

An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.

Input

The first line contains single integer q (1 ≤ q ≤ 10⁵) — the number of queries.

q lines follow. The (i + 1)-th line contains single integer n_i (1 ≤ n_i ≤ 10⁹) — the i-th query.

Output

For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.

Examples

input

1
12

output

input

2
6
8

output

1
2

input

output

-1
-1
-1

Note

12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.

8 = 4 + 4, 6 can't be split into several composite summands.

1, 2, 3 are less than any composite number, so they do not have valid splittings.

题意就是要将一个数分成几个合数之和，求最大的合数个数。

这题显然一个数分成4k+n时会最多

把这个数取4的模

分4类分别讨论就行了。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 0;
inline int get(){
	char c;
	while((c = getchar()) < '0' || c > '9');
	int cnt = c - '0';
	while((c = getchar()) >= '0' && c <= '9') cnt = cnt * 10 + c - '0';
	return cnt;
}
int q,n;
int k,t,ans;
int main(){
/*	#ifdef lwy
	#else
		freopen(".in","r",stdin);
		freopen(".out","w",stdout);
	#endif*/
	q = get();
	while(q--){
		n = get();
		k = n % 4;
		t = n / 4;
		if(k == 0) ans = t;
		if(k == 1) {
			t -= 2;
			if(t < 0) ans = -1;
			if(t == 0) ans = 1;
			if(t > 0) ans = t + 1;
		}
		if(k == 2){
			t -= 1;
			if(t < 0) ans = -1;
			if(t == 0) ans = 1;
			if(t > 0) ans = t + 1;
		}
		if(k == 3){
			t -= 3;
			if(t < 0) ans = -1;
			if(t == 0) ans = 2;
			if(t > 0) ans = t + 2;
		}
		printf("%d\n",ans);
	}
	return 0;
}