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 kequals 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

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

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 andb 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 "Yes" if it‘s possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.

4 2
11 0 0 14
5 4

Yes

6 1
2 3 0 8 9 10
5

No

4 1
8 94 0 4
89

Yes

7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7

Yes

題意就是把b數組插到a數組的0中
只有有一個不是遞增序列，那就輸出Yes
所以我們把b按照遞減序列排序
只要b遞減後 a還是遞增 那這個序列就遞增 反之遞減

 1 #include <cctype>
 2 #include <cstdio>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 const int MAXN=1010;
 8 
 9 int n,m;
10 
11 int a[MAXN],b[MAXN],q[MAXN];
12 
13 inline void read(int&x) {
14     int f=1;register char c=getchar();
15     for(x=0;!isdigit(c);c==‘-‘&&(f=-1),c=getchar());
16     for(;isdigit(c);x=x*10+c-48,c=getchar());
17     x=x*f;
18 }
19 
20 inline bool cmp(int a,int b) {
21     return a>b;
22 }
23 
24 int hh() {
25     int t=0;
26     read(n);read(m);
27     for(int i=1;i<=n;++i) read(a[i]);
28     for(int i=1;i<=m;++i) read(b[i]);
29     sort(b+1,b+1+m,cmp);
30     for(int i=1;i<=n;++i) if(!a[i]) a[i]=b[++t];
31     for(int i=1;i<n;++i) 
32       if(a[i]>a[i+1]) {
33           printf("Yes\n");
34           return 0;
35       }
36     printf("No\n");
37     return 0;
38 }
39 
40 int sb=hh();
41 int main(int argc,char**argv) {}

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Codeforces Round #418 A--An abandoned sentiment from past