An abandoned sentiment from past

Codeforces Round #418 (Div. 2) A.An abandoned sentiment from past

time limit per test : 1 second

题目描述

  A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.

  To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi’s sequence has been broken for a long time, but now Kaiki provides an opportunity.

  Hitagi’s sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.

  If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki’s sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.

Input

  The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.

  The second line contains n space-separated integers a1, a2, …, an (0 ≤ ai ≤ 200) — Hitagi’s broken sequence with exactly k zero elements.

  The third line contains k space-separated integers b1, b2, …, bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi’s sequence.

  Input guarantees that apart from 0, no integer occurs in a and b more than once in total.

Output

  Output “Yes” if it’s possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and “No” otherwise.

Examples

input

4 2

11 0 0 14

5 4

output

Yes

input

6 1

2 3 0 8 9 10

5

output

No

input

4 1

8 94 0 4

89

output

Yes

input

7 7

0 0 0 0 0 0 0

1 2 3 4 5 6 7

output

Yes

本题题意：

  本题题意是给你l两个序列a,b。用b中的数取代a序列中为0的数。如果能组成非递增序列，就输出Yes，否则输出No。

解题思路：

  如果a序列中0的个数大于1，那么我们可以让两个数递减取代，就一定可以组成非递增序列，所以我们分为两种情况来看。

  如果a中0的个数大于1，直接输出Yes。

  如果a中0的个数为1，就把b中的数填进去，判断是否为非递增序列即可。

注意：

  如果两个数相等，这不属于递增序列。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
using namespace std;
int main()
{
int n,k;
int a[150],b[150];
while(scanf("%d%d",&n,&k)!=EOF)
{
int temp;
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
if(a[i]==0)
temp=i;
}
for(int j=0; j<k; j++)
{
scanf("%d",&b[j]);
}
if(k>1)
{
printf("Yes\n");
continue;
}
else
{
a[temp]=b[0];
int ans=-1;
bool flag=true;
for(int i=0; i<n; i++)
{
if(a[i]<=ans)
flag=false;
ans=a[i];
}
if(flag)
printf("No\n");
else
printf("Yes\n");
}
}
return 0;
}