An abandoned sentiment from past

A. An abandoned sentiment from past

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 sequenceb, whose lengthk equals the number of lost elements ina (i.e. the number of zeros). Hitagi is to replace each zero ina with an element fromb so that each element in b should be used exactly once. Hitagi knows, however, that,apart from0, no integer occurs ina andb 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 fromb is used exactly once, and the resulting sequence isnot 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 sequencea andb respectively.

The second line contains n space-separated integersa₁, a₂, ..., a_n (0 ≤ a_i ≤ 200) — Hitagi's broken sequence with exactlyk zero elements.

The third line contains k space-separated integersb₁, b₂, ..., b_k (1 ≤ b_i ≤ 200) — the elements to fill into Hitagi's sequence.

Input guarantees that apart from 0, no integer occurs ina andb more than once in total.

Output

Output "Yes" if it's possible to replace zeros ina with elements inb and make the resulting sequence not increasing, and "No" otherwise.

Examples

Input

4 211 0 0 145 4

Output

Yes

Input

6 12 3 0 8 9 105

Output

No

Input

4 18 94 0 489

Output

Yes

Input

7 70 0 0 0 0 0 01 2 3 4 5 6 7

Output

Yes

Note

In the first sample:

• Sequence a is 11, 0, 0, 14.
• Two of the elements are lost, and the candidates in b are5 and4.
• There are two possible resulting sequences: 11, 5, 4, 14 and11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".

In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.

题意：输入数组a ,b。将数组b中的数填入到数组a中元素为0的位置（一个数只能填一次），如果数组a是递增的就输出No,否则就是Yes。

思路：

一开始想的用深搜来解决，结果耽误了一个半小时。。。仔细想想其实就是一个小技巧，只有k>1,一定就是yes，因为那k个数可以任意摆放，一定有不上升的顺序。然后再单独处理k==1时的情况。

代码：

#include<iostream>#include<algorithm>using namespace std;bool flag;int main(){    int n, k;    int mark;    while(cin >> n >> k)    {        flag=0;        int a[205], b[205];        for (int i = 0; i < n; i++)        {            cin >> a[i];            if (a[i] == 0)            {                mark = i;            }        }        for (int i = 0; i < k; i++)            cin >> b[i];        if (k > 1)        {             cout << "Yes" << endl;        }        else        {            a[mark]=b[0];            for (int i =1; i<n;i++)            {                if (a[i]<a[i-1])                {                    flag =1;                    break;                }            }            if (flag)                cout << "Yes" << endl;            else                cout << "No" << endl;        }    }    return 0;}

dfs

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>using namespace std;int n,k,a[105],b[105],c[105];bool flag,ok[205];bool work(){    for(int i=1;i<=n;i++)    {        if(a[i]<a[i-1]) return true;    }    return false;}void dfs(int i){    if(flag) return;    for(int j=1;j<=k;j++)    {        if(!ok[b[j]])        {            ok[b[j]]=true;            a[c[i]]=b[j];            if(i==k)            {                if(work())                {                    flag=true;                    return;                }            }            else            {                dfs(i+1);            }            ok[b[j]]=false;           // a[c[i]]=b[j];        }    }}int main(){    scanf("%d%d",&n,&k);    int i,num=0;    a[0]=0;    for(i=1;i<=n;i++)    {        scanf("%d",&a[i]);        if(a[i]==0)        {            num++;            c[num]=i;        }    }    for(i=1;i<=k;i++)    {        scanf("%d",&b[i]);    }    flag=false;    memset(ok,false,sizeof(ok));    dfs(1);    if(flag) printf("Yes\n");    else printf("No\n");    return 0;}

心得：

读准题目很重要。。。方法也很重要。。。。唉。。。自己能力是真的不行。。。这么一个水题，捯饬了一个半小时、。。。。。。。