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 thateach 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.
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
In the first sample:
- Sequence a is 11, 0, 0, 14.
- Two of the elements are lost, and the candidates in b are 5 and 4.
- There are two possible resulting sequences: 11, 5, 4, 14 and 11, 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：就是给你两个数组，一个数组a，另一个数组b。现在的问题就是如果用数组b中的数字替换数组a中的为0的数字，如果存在一种方法使替换后的a数组不单调递增则输出“Yes”，否则的话输出“No”(值得注意的是在a,b数组中除了0不会出现其他有两个一模一样的数字)
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int m,n;
while(~scanf("%d %d",&m,&n))
{
int a[105],b[105];
int i,j;
for(i=0; i<m; i++)
scanf("%d",&a[i]);
for(j=0; j<n; j++)
scanf("%d",&b[j]);
sort(b,b+n);//这样都不单调递增的话那肯定就不会有条件符合了，接下来就依次替换到数组中再判断是否符合就好了
j=n-1;
for(i=0; i<m; i++)
{
if(a[i]==0)
{
a[i]=b[j];
j--;
}
}
int flag=0;
for(i=1; i<m; i++)
{
if(a[i-1]>a[i])
{
flag=1;
break;
}
}
if(flag==1)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}Sengoku still remembers the mysterious "colourful meteoroids" she discovered with Lala-chan when they were little. In particular, one of the nights impressed her deeply, giving her the illusion that all her fancies would be realized.
On that night, Sengoku constructed a permutation p1, p2, ..., pn of integers from 1 to n inclusive, with each integer representing a colour, wishing for the colours to see in the coming meteor outburst. Two incredible outbursts then arrived, each with n meteorids, colours of which being integer sequences a1, a2, ..., an and b1, b2, ..., bn respectively. Meteoroids' colours were also between 1 and ninclusive, and the two sequences were not identical, that is, at least one i (1 ≤ i ≤ n) exists, such that ai ≠ bi holds.
Well, she almost had it all — each of the sequences a and b matched exactly n - 1 elements in Sengoku's permutation. In other words, there is exactly one i (1 ≤ i ≤ n) such that ai ≠ pi, and exactly one j (1 ≤ j ≤ n) such that bj ≠ pj.
For now, Sengoku is able to recover the actual colour sequences a and b through astronomical records, but her wishes have been long forgotten. You are to reconstruct any possible permutation Sengoku could have had on that night.
The first line of input contains a positive integer n (2 ≤ n ≤ 1 000) — the length of Sengoku's permutation, being the length of both meteor outbursts at the same time.
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ n) — the sequence of colours in the first meteor outburst.
The third line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ n) — the sequence of colours in the second meteor outburst. At least one i (1 ≤ i ≤ n) exists, such that ai ≠ bi holds.
Output n space-separated integers p1, p2, ..., pn, denoting a possible permutation Sengoku could have had. If there are more than one possible answer, output any one of them.
Input guarantees that such permutation exists.
5 1 2 3 4 3 1 2 5 4 5
1 2 5 4 3
5 4 4 2 3 1 5 4 5 3 1
5 4 2 3 1
4 1 1 3 4 1 4 3 4
1 2 3 4
In the first sample, both 1, 2, 5, 4, 3 and 1, 2, 3, 4, 5 are acceptable outputs.
In the second sample, 5, 4, 2, 3, 1 is the only permutation to satisfy the constraints.
题意：给你a,b两个数组，两个数组分别有n个数字且每个数字的大小都在1至n之间，现在要求输出一个数字序列p,这个序列p和序列a,b分别都只有一位对应位置的数字不一样，p序列中的数字不会有重复。
//如果有满足条件的p序列，那么a,b数组中最多只有两位数字不一样，
//这时候就要分情况讨论了，有一位数字不一样和有两位数字不一样，
//第一种情况，找到对应的一位，将其置为p数组中从1到n没有出现的数字
//第二种情况，找到对应的两位，这样可能的情况只有两种，相对于第一种就没有那么多种可能
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
int n;
while(~scanf("%d",&n))
{
int i,j,t;
int a[1005],b[1005],s[1005];
memset(s,0,sizeof(s));
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
scanf("%d",&b[i]);
int x=-1,y=-1,flag=0;
for(i=0;i<n;i++)
{
if(a[i]!=b[i])
{
if(flag==1)
{
y=i;
flag++;
}
if(flag==0)
{
x=i;
flag++;
}
}
if(flag==2)
break;
}
for(i=0;i<n;i++)
{
if(i!=x&&i!=y)
s[a[i]]=1;
}
if(flag==2)
{
if(s[a[x]]==0&&s[b[y]]==0)
a[y]=b[y];
else
a[x]=b[x];
printf("%d",a[0]);
for(i=1;i<n;i++)
printf(" %d",a[i]);
printf("\n");
}
else
{
for(i=1;i<=n;i++)
if(s[i]==0)
t=i;
a[x]=t;
printf("%d",a[0]);
for(i=1;i<n;i++)
printf(" %d",a[i]);
printf("\n");
}
}
return 0;
}
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