input

61 5 5 5 4 2

output

YES

input

510 20 30 20 10

output

YES

input

41 2 1 2

output

NO

input

73 3 3 3 3 3 3

output

YES

There are two popular keyboard layouts in Berland, they differ only in letters positions. All the other keys are the same. In Berland they use alphabet with 26 letters which coincides with English alphabet.

You are given two strings consisting of 26 distinct letters each: all keys of the first and the second layouts in the same order.

You are also given some text consisting of small and capital English letters and digits. It is known that it was typed in the first layout, but the writer intended to type it in the second layout. Print the text if the same keys were pressed in the second layout.

Since all keys but letters are the same in both layouts, the capitalization of the letters should remain the same, as well as all other characters.

The first line contains a string of length 26 consisting of distinct lowercase English letters. This is the first layout.

The second line contains a string of length 26 consisting of distinct lowercase English letters. This is the second layout.

The third line contains a non-empty string s consisting of lowercase and uppercase English letters and digits. This is the text typed in the first layout. The length of s does not exceed 1000.

Print the text if the same keys were pressed in the second layout.

#include <bits/stdc++.h>
using namespace std;
int b[2010];
set<int>s;
int main() {int k,n,p=0;cin>>n>>k;for(int i=1; i<=n; i++) {int q;cin>>q;p+=q;s.insert(p);}for(int i=1; i<=k; i++) {cin>>b[i];}sort(b+1,b+k+1);int ans=0;for(auto &node:s) {int f=1;for(int j=1; j<=k; j++) {if(!s.count(node-b[1]+b[j])) {f=0;break;}}if(f)ans++;}cout<<ans<<endl;}

先将排序，假设第i个人拿到的钥匙是第k[i]个，那么显然k[i]<k[i+1]。dp[i][j]表示第i个人拿第j个钥匙，前i个人走的最大距离的最小值

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e3+10;
int a[N],b[N],d[N][N];
int main()
{int n,k,p;cin>>n>>k>>p;for(int i=0;i<n;i++)scanf("%d",a+i);for(int i=0;i<k;i++)scanf("%d",b+i);sort(a,a+n);sort(b,b+k);d[0][0]=abs(a[0]-b[0])+abs(b[0]-p);for(int i=1;i<=k-n;i++)d[0][i]=min(d[0][i-1],abs(a[0]-b[i])+abs(b[i]-p));for(int i=1;i<n;i++){d[i][i-1]=2e9+10;for(int j=i;j<=k-n+i;j++)d[i][j]=min(d[i][j-1],max(d[i-1][j-1],abs(a[i]-b[j])+abs(b[j]-p)));}cout<<d[n-1][k-1];return 0;
}

Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this integer is between 1 and 100 000, inclusive. It is possible that some cards have the same integers on them.

Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn't know where this card (or these cards) is.

You are to determine the total number of times Vasily takes the top card from the deck.

The first line contains single integer n (1 ≤ n ≤ 100 000) — the number of cards in the deck.

The second line contains a sequence of n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000), where ai is the number written on the i-th from top card in the deck.

Print the total number of times Vasily takes the top card from the deck.

In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2, 6, 3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6, 3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards.

E是个线段树维护区间最小值，可是我一脸懵bi