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stringlengths 52
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#include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >> n;
vector<int> b(n);
for(int i = 0; i < n; i++){
cin >> b[i];
b[i]--;
}
vector<int> ans;
while(b.size() > 0){
int z = -1;
for(int i = 0; i < b.size(); i++){
if(b[i] == i) z = i;
}
if(z == -1){
cout << -1 << endl;
exit(0);
}
ans.push_back(z);
b.erase(b.begin() + z);
}
reverse(ans.begin(), ans.end());
for(int x : ans){
cout << x + 1 << '\n';
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >> n;
vector<int> b(n);
for(int i = 0; i < n; i++){
cin >> b[i];
b[i]--;
}
vector<int> ans;
while(b.size() > 0){
int z = -1;
for(int i = 0; i < b.size(); i++){
if(b[i] == i) z = i;
}
if(z == -1){
cout << -1 << endl;
exit(0);
}
ans.push_back(z);
b.erase(b.begin() + z);
}
reverse(ans.begin(), ans.end());
for(int x : ans){
cout << x + 1 << '\n';
}
}
``` |
#include<bits/stdc++.h>
using namespace std;
const int N=110;
int n;
int a[N],b[N];
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
for (int i=1;i<=n;i++) {
int ans=0,sum=0;
for (int j=1;j<=n;j++)
if (a[j]) {
sum++;
if (a[j]==sum) ans=j;
}
if (!ans) {
printf("%d\n",-1); exit(0);
}
else
b[i]=a[ans],a[ans]=0;
}
for (int i=n;i>=1;i--) printf("%d\n",b[i]);
} | ### Prompt
Your task is to create a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
const int N=110;
int n;
int a[N],b[N];
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
for (int i=1;i<=n;i++) {
int ans=0,sum=0;
for (int j=1;j<=n;j++)
if (a[j]) {
sum++;
if (a[j]==sum) ans=j;
}
if (!ans) {
printf("%d\n",-1); exit(0);
}
else
b[i]=a[ans],a[ans]=0;
}
for (int i=n;i>=1;i--) printf("%d\n",b[i]);
}
``` |
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#include <cstring>
using namespace std;
#define N 105
int b[N],n,a[N];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%d",&b[i]);
int len=n;
for(int i=n;i;i--)
{
bool flag=0;
for(int j=len;j;j--)
if(b[j]==j)
{
a[i]=b[j];flag=1;
for(int k=j;k<len;k++)b[k]=b[k+1];
break;
}
if(!flag)return puts("-1"),0;len--;
}
for(int i=1;i<=n;i++)printf("%d\n",a[i]);
} | ### Prompt
Generate a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#include <cstring>
using namespace std;
#define N 105
int b[N],n,a[N];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)scanf("%d",&b[i]);
int len=n;
for(int i=n;i;i--)
{
bool flag=0;
for(int j=len;j;j--)
if(b[j]==j)
{
a[i]=b[j];flag=1;
for(int k=j;k<len;k++)b[k]=b[k+1];
break;
}
if(!flag)return puts("-1"),0;len--;
}
for(int i=1;i<=n;i++)printf("%d\n",a[i]);
}
``` |
#include<iostream>
#include<vector>
using namespace std;
int main(){
int N;
cin >> N;
int b[N+1];
for(int i=1;i<=N;i++) cin >> b[i];
vector<int> ans(N);
for(int i=1;i<=N;i++){
if(b[i]>i){
cout << -1 << endl;
return 0;
}
}
for(int i=1;i<=N;i++){
ans.insert(ans.begin()+b[i],b[i]);
}
for(int i=1;i<=N;i++){
cout << ans[i] << endl;
}
} | ### Prompt
Construct a Cpp code solution to the problem outlined:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<iostream>
#include<vector>
using namespace std;
int main(){
int N;
cin >> N;
int b[N+1];
for(int i=1;i<=N;i++) cin >> b[i];
vector<int> ans(N);
for(int i=1;i<=N;i++){
if(b[i]>i){
cout << -1 << endl;
return 0;
}
}
for(int i=1;i<=N;i++){
ans.insert(ans.begin()+b[i],b[i]);
}
for(int i=1;i<=N;i++){
cout << ans[i] << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, b_, ans = 0;
vector<int> a, b;
cin >> n;
for(int i = 0; i < n; ++i) {
cin >> b_;
b.push_back(b_);
}
while(b.size()) {
for(int i = b.size()-1; i >= 0; --i) {
if(b[0] != 1) {
cout << -1 << '\n';
return 0;
}
if(b[i] == i+1) {
a.push_back(b[i]);
b.erase(b.begin()+i);
break;
}
}
}
for(int i = n-1; i >= 0; --i) {
cout << a[i] << '\n';
}
}
| ### Prompt
Please create a solution in CPP to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, b_, ans = 0;
vector<int> a, b;
cin >> n;
for(int i = 0; i < n; ++i) {
cin >> b_;
b.push_back(b_);
}
while(b.size()) {
for(int i = b.size()-1; i >= 0; --i) {
if(b[0] != 1) {
cout << -1 << '\n';
return 0;
}
if(b[i] == i+1) {
a.push_back(b[i]);
b.erase(b.begin()+i);
break;
}
}
}
for(int i = n-1; i >= 0; --i) {
cout << a[i] << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MX = 100;
int a[MX], ans[MX];
int main() {
int n;
ignore = scanf("%d", &n);
for (int i = 0; i < n; i++) ignore = scanf("%d", a + i);
int N = n;
while (n > 0) {
int i = n - 1;
while (i >= 0 && a[i] != i + 1) i--;
if (i == -1) break;
ans[n - 1] = a[i];
for (int j = i; j + 1 < n; j++) a[j] = a[j + 1];
n--;
}
if (n > 0) {
printf("%d\n", -1);
return 0;
}
for (int i = 0; i < N; i++) printf("%d\n", ans[i]);
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MX = 100;
int a[MX], ans[MX];
int main() {
int n;
ignore = scanf("%d", &n);
for (int i = 0; i < n; i++) ignore = scanf("%d", a + i);
int N = n;
while (n > 0) {
int i = n - 1;
while (i >= 0 && a[i] != i + 1) i--;
if (i == -1) break;
ans[n - 1] = a[i];
for (int j = i; j + 1 < n; j++) a[j] = a[j + 1];
n--;
}
if (n > 0) {
printf("%d\n", -1);
return 0;
}
for (int i = 0; i < N; i++) printf("%d\n", ans[i]);
return 0;
}
``` |
#include<iostream>
using namespace std;
const int N=110;
int n,b[N],p[N],Ans[N];
int main()
{
cin>>n;
for(int i=1;i<=n;i++) cin>>b[i];
for(int i=1;i<=n;i++)
{
if(b[i]>i) return puts("-1"),0;
for(int j=1;j<i;j++) if(p[j]>=b[i]) p[j]++;
p[i]=b[i];
}
for(int i=1;i<=n;i++) Ans[p[i]]=b[i];
for(int i=1;i<=n;i++) printf("%d\n",Ans[i]);
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<iostream>
using namespace std;
const int N=110;
int n,b[N],p[N],Ans[N];
int main()
{
cin>>n;
for(int i=1;i<=n;i++) cin>>b[i];
for(int i=1;i<=n;i++)
{
if(b[i]>i) return puts("-1"),0;
for(int j=1;j<i;j++) if(p[j]>=b[i]) p[j]++;
p[i]=b[i];
}
for(int i=1;i<=n;i++) Ans[p[i]]=b[i];
for(int i=1;i<=n;i++) printf("%d\n",Ans[i]);
return 0;
}
``` |
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> b(n),a(n);
for(int i=0; i<n; i++) {
cin >> b[i];
}
for(int i=0; i<n; i++) {
for(int j=b.size()-1; j>=0; j--) {
if(b[j]==j+1) {
a[i]=b[j];
b.erase(b.begin()+j);
break;
}
}
}
if(a[n-1]!=1) {
cout << -1 << endl;
}
else {
for(int i=n-1; i>=0; i--) {
cout << a[i] << endl;
}
}
} | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> b(n),a(n);
for(int i=0; i<n; i++) {
cin >> b[i];
}
for(int i=0; i<n; i++) {
for(int j=b.size()-1; j>=0; j--) {
if(b[j]==j+1) {
a[i]=b[j];
b.erase(b.begin()+j);
break;
}
}
}
if(a[n-1]!=1) {
cout << -1 << endl;
}
else {
for(int i=n-1; i>=0; i--) {
cout << a[i] << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main()
{
int N; cin >> N;
vector<int> b(N);
for (int i = 0; i < N; ++i) {
cin >> b[i];
b[i]--;
}
vector<int> ans;
while (b.size() > 0) {
int z = -1;
for (int i = 0; i < b.size(); ++i) {
if (b[i] == i) z = b[i];
}
if (z == -1) {
cout << -1;
return 0;
}
ans.insert(ans.begin(), z);
b.erase(b.begin()+z);
}
for (int i = 0; i < ans.size(); ++i) {
cout << ans[i]+1 << endl;
}
} | ### Prompt
Develop a solution in cpp to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main()
{
int N; cin >> N;
vector<int> b(N);
for (int i = 0; i < N; ++i) {
cin >> b[i];
b[i]--;
}
vector<int> ans;
while (b.size() > 0) {
int z = -1;
for (int i = 0; i < b.size(); ++i) {
if (b[i] == i) z = b[i];
}
if (z == -1) {
cout << -1;
return 0;
}
ans.insert(ans.begin(), z);
b.erase(b.begin()+z);
}
for (int i = 0; i < ans.size(); ++i) {
cout << ans[i]+1 << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=s;i<t;i++)
#define pii pair<int,int>
#define MAXNUM 111
pii st[MAXNUM];int num[MAXNUM];
int main()
{
int n;scanf("%d",&n);
rep(i,1,n+1)scanf("%d",&num[i]);
vector<int> x;int flag=1;
rep(i,1,n+1)
{
if(num[i]>i){flag=0;break;}
x.insert(x.begin()+num[i]-1,num[i]);
}
if(!flag)printf("-1\n");
else {
for(int k:x)printf("%d\n",k);
}
} | ### Prompt
In CPP, your task is to solve the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,s,t) for(int i=s;i<t;i++)
#define pii pair<int,int>
#define MAXNUM 111
pii st[MAXNUM];int num[MAXNUM];
int main()
{
int n;scanf("%d",&n);
rep(i,1,n+1)scanf("%d",&num[i]);
vector<int> x;int flag=1;
rep(i,1,n+1)
{
if(num[i]>i){flag=0;break;}
x.insert(x.begin()+num[i]-1,num[i]);
}
if(!flag)printf("-1\n");
else {
for(int k:x)printf("%d\n",k);
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int num[105],ans[105];
bool in[105];
int main() {
memset(in,1,sizeof(in));
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&num[i]);
for(int i=n;i>0;i--) {
int cur=0,s=0;
for(int j=1;j<=n;j++)
if (in[j]) {
s++;
if (num[j]==s) cur=j;
}
if (cur&&num[cur]<=i) {
ans[i]=num[cur];
in[cur]=0;
}
else {
puts("-1");
return 0;
}
}
for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int num[105],ans[105];
bool in[105];
int main() {
memset(in,1,sizeof(in));
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&num[i]);
for(int i=n;i>0;i--) {
int cur=0,s=0;
for(int j=1;j<=n;j++)
if (in[j]) {
s++;
if (num[j]==s) cur=j;
}
if (cur&&num[cur]<=i) {
ans[i]=num[cur];
in[cur]=0;
}
else {
puts("-1");
return 0;
}
}
for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N=1e6+10;
int main(){
int n;
cin>>n;
vector<int>a(n);
for(int i=0;i<n;i++)cin>>a[i];
vector<int>b(n);
for(int i=0;i<n;i++){
int k=1,rm=-1;
for(int j=0;j<n;j++){
if(a[j]==k)
rm=j;
if(a[j])k++;
}
if(rm==-1){
cout<<rm<<endl;return 0;
}
b[i]=a[rm];
a[rm]=0;
}
reverse(b.begin(),b.end());
for(auto x:b)
cout<<x<<endl;
}
/*
1 2 1 2
1 2 3 3 2
1 2 2 3 3
*/
| ### Prompt
Please formulate a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N=1e6+10;
int main(){
int n;
cin>>n;
vector<int>a(n);
for(int i=0;i<n;i++)cin>>a[i];
vector<int>b(n);
for(int i=0;i<n;i++){
int k=1,rm=-1;
for(int j=0;j<n;j++){
if(a[j]==k)
rm=j;
if(a[j])k++;
}
if(rm==-1){
cout<<rm<<endl;return 0;
}
b[i]=a[rm];
a[rm]=0;
}
reverse(b.begin(),b.end());
for(auto x:b)
cout<<x<<endl;
}
/*
1 2 1 2
1 2 3 3 2
1 2 2 3 3
*/
``` |
#include <iostream>
#include <vector>
#include <algorithm>
#include <stack>
using namespace std;
int main()
{
int N;cin>>N;
vector<int> b(N);
for(int i=0;i<N;++i)cin>>b[i];
stack<int> ans;
for(int i=0;i<N;i++)
{
for(int j=b.size()-1;1;j--)
{
if(j==-1)
{
cout<<-1<<endl;
return 0;
}
if((j+1)==b[j])
{
//cout<<b[j]<<"を挿入"<<endl;
ans.push(j+1);
b.erase(b.begin()+j);;
break;
}
}
}
for(int i=0;i<N;i++)
{
cout<<ans.top()<<endl;
ans.pop();
}
return 0;
} | ### Prompt
Your challenge is to write a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <iostream>
#include <vector>
#include <algorithm>
#include <stack>
using namespace std;
int main()
{
int N;cin>>N;
vector<int> b(N);
for(int i=0;i<N;++i)cin>>b[i];
stack<int> ans;
for(int i=0;i<N;i++)
{
for(int j=b.size()-1;1;j--)
{
if(j==-1)
{
cout<<-1<<endl;
return 0;
}
if((j+1)==b[j])
{
//cout<<b[j]<<"を挿入"<<endl;
ans.push(j+1);
b.erase(b.begin()+j);;
break;
}
}
}
for(int i=0;i<N;i++)
{
cout<<ans.top()<<endl;
ans.pop();
}
return 0;
}
``` |
#include<iostream>
#include<vector>
using namespace std;
int main() {
int N; cin >> N;
vector<int>b(N), ans;
for (auto&& x : b)cin >> x;
for (int i = 0; i < N; i++) {
int target = -1;
for (int j = 0; j < b.size(); j++) {
if (j + 1 == b[j])target = j + 1;
}
if (target == -1) { cout << -1 << endl; return 0; }
ans.push_back(target);
b.erase(b.begin() + target - 1);
}
for (int i = N - 1; i >= 0;i--)cout << ans[i] << endl;
return 0;
} | ### Prompt
Please create a solution in CPP to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<iostream>
#include<vector>
using namespace std;
int main() {
int N; cin >> N;
vector<int>b(N), ans;
for (auto&& x : b)cin >> x;
for (int i = 0; i < N; i++) {
int target = -1;
for (int j = 0; j < b.size(); j++) {
if (j + 1 == b[j])target = j + 1;
}
if (target == -1) { cout << -1 << endl; return 0; }
ans.push_back(target);
b.erase(b.begin() + target - 1);
}
for (int i = N - 1; i >= 0;i--)cout << ans[i] << endl;
return 0;
}
``` |
#include <cstdio>
#include <vector>
std::vector<int> v;
int main()
{
int n; scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
int x; scanf("%d", &x);
if(x > i) return 0 * puts("-1");
v.insert(v.begin() + x - 1, x);
}
for(int i = 0; i < n; i++) printf("%d\n", v[i]);
} | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <cstdio>
#include <vector>
std::vector<int> v;
int main()
{
int n; scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
int x; scanf("%d", &x);
if(x > i) return 0 * puts("-1");
v.insert(v.begin() + x - 1, x);
}
for(int i = 0; i < n; i++) printf("%d\n", v[i]);
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
stack<int> s, tmp, ans;
for(int i = 0; i < n; ++i){
int num;
cin >> num;
s.push(num);
}
while(!s.empty()){
int num = s.top();
s.pop();
if(num == (int)s.size() + 1){
ans.push(num);
while(!tmp.empty()){
s.push(tmp.top());
tmp.pop();
}
}
else{
tmp.push(num);
}
}
if(tmp.empty()){
while(!ans.empty()){
cout << (ans.top()) << "\n";
ans.pop();
}
}
else{
cout << -1;
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
stack<int> s, tmp, ans;
for(int i = 0; i < n; ++i){
int num;
cin >> num;
s.push(num);
}
while(!s.empty()){
int num = s.top();
s.pop();
if(num == (int)s.size() + 1){
ans.push(num);
while(!tmp.empty()){
s.push(tmp.top());
tmp.pop();
}
}
else{
tmp.push(num);
}
}
if(tmp.empty()){
while(!ans.empty()){
cout << (ans.top()) << "\n";
ans.pop();
}
}
else{
cout << -1;
}
return 0;
}
``` |
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)
using namespace std;
typedef long long ll;
const ll N=12e4;
int a[N];
int p[N];
int n;
int main()
{
scanf("%d",&n);
fo(i,1,n)scanf("%d",&a[i]);
int ok=1;
fd(i,n,1){
int po=0;
fd(l,i,1)if(a[l]==l){
po=l;
break;
}
fo(l,po,i-1)a[l]=a[l+1];
if(!po){
ok=0; break;
}
p[i]=po;
}
if(!ok)puts("-1");
else
fo(i,1,n)printf("%d\n",p[i]);
} | ### Prompt
Develop a solution in cpp to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)
using namespace std;
typedef long long ll;
const ll N=12e4;
int a[N];
int p[N];
int n;
int main()
{
scanf("%d",&n);
fo(i,1,n)scanf("%d",&a[i]);
int ok=1;
fd(i,n,1){
int po=0;
fd(l,i,1)if(a[l]==l){
po=l;
break;
}
fo(l,po,i-1)a[l]=a[l+1];
if(!po){
ok=0; break;
}
p[i]=po;
}
if(!ok)puts("-1");
else
fo(i,1,n)printf("%d\n",p[i]);
}
``` |
// In the name of God
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<int> v(n);
for(int i=0;i<n;i++) {
cin >> v[i];
}
vector<int> ans;
for(int i=0;i<n;i++) {
if(ans.size() < v[i]-1) {
cout << -1;
exit(0);
}
ans.insert(ans.begin()+v[i]-1, v[i]);
}
for(int i=0;i<n;i++) cout << ans[i] << endl;
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
// In the name of God
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<int> v(n);
for(int i=0;i<n;i++) {
cin >> v[i];
}
vector<int> ans;
for(int i=0;i<n;i++) {
if(ans.size() < v[i]-1) {
cout << -1;
exit(0);
}
ans.insert(ans.begin()+v[i]-1, v[i]);
}
for(int i=0;i<n;i++) cout << ans[i] << endl;
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
const int maxn=1050;
int a[maxn],n,tmp[maxn],c[maxn];
int main(){
cin >> n;
for (int i=1;i<=n;i++) cin >> a[i];
for (int i=n;i;i--){
int pos=0;
for (int j=1;j<=i;j++) if (a[j]==j) pos=j;
if (!pos) {puts("-1");return 0;}
c[i]=a[pos];
for (int j=pos;j<i;j++) a[j]=a[j+1];
}
for (int i=1;i<=n;i++) printf("%d\n",c[i]);
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
const int maxn=1050;
int a[maxn],n,tmp[maxn],c[maxn];
int main(){
cin >> n;
for (int i=1;i<=n;i++) cin >> a[i];
for (int i=n;i;i--){
int pos=0;
for (int j=1;j<=i;j++) if (a[j]==j) pos=j;
if (!pos) {puts("-1");return 0;}
c[i]=a[pos];
for (int j=pos;j<i;j++) a[j]=a[j+1];
}
for (int i=1;i<=n;i++) printf("%d\n",c[i]);
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
vector<int>v;
int n,x,i;
int main(){
cin>>n;
for(i=0;++i<=n;){
cin>>x;
if(x>i) return 0*puts("-1");
v.insert(v.begin()+x-1,x);
}
for(i=-1;++i<n;)cout<<v[i]<<endl;
}
| ### Prompt
Please create a solution in CPP to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
vector<int>v;
int n,x,i;
int main(){
cin>>n;
for(i=0;++i<=n;){
cin>>x;
if(x>i) return 0*puts("-1");
v.insert(v.begin()+x-1,x);
}
for(i=-1;++i<n;)cout<<v[i]<<endl;
}
``` |
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin>>n;
int a[101];
for (int i=1; i<=n; i++)
cin>>a[i];
vector<int> ans;
for(; n>0; n--) {
int pos = n;
while(pos > 0 && a[pos] != pos)
pos--;
if (pos > 0) {
ans.push_back(pos);
while(pos+1 <= n) {
a[pos] = a[pos+1];
pos++;
}
}else
break;
}
if (n)
cout<<-1<<endl;
else {
for (int i=ans.size()-1; i>=0; i--)
cout<<ans[i]<<" ";
cout<<endl;
}
return 0;
} | ### Prompt
Construct a CPP code solution to the problem outlined:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin>>n;
int a[101];
for (int i=1; i<=n; i++)
cin>>a[i];
vector<int> ans;
for(; n>0; n--) {
int pos = n;
while(pos > 0 && a[pos] != pos)
pos--;
if (pos > 0) {
ans.push_back(pos);
while(pos+1 <= n) {
a[pos] = a[pos+1];
pos++;
}
}else
break;
}
if (n)
cout<<-1<<endl;
else {
for (int i=ans.size()-1; i>=0; i--)
cout<<ans[i]<<" ";
cout<<endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(){
int N; cin >> N;
vector<int> b(N), ans;
for (int i=0; i<N; i++) cin >> b[i];
while(N>0){
int befN = N;
for (int i=N-1; i>=0; i--){
if (b[i]==i+1){
ans.push_back(i+1);
b.erase(b.begin()+i);
N--;
break;
}
}
if (N==befN){
cout << -1 << endl;
return 0;
}
}
for (int i=ans.size()-1; i>=0; i--){
cout << ans[i] << endl;
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(){
int N; cin >> N;
vector<int> b(N), ans;
for (int i=0; i<N; i++) cin >> b[i];
while(N>0){
int befN = N;
for (int i=N-1; i>=0; i--){
if (b[i]==i+1){
ans.push_back(i+1);
b.erase(b.begin()+i);
N--;
break;
}
}
if (N==befN){
cout << -1 << endl;
return 0;
}
}
for (int i=ans.size()-1; i>=0; i--){
cout << ans[i] << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a[110], b[110];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", b + i);
}
for (int i = n; i; --i) {
bool is = false;
for (int j = i; j; --j) {
if (b[j] > j) break;
if (b[j] == j) {
a[i] = j;
for (int k = j; k < i; ++k) {
b[k] = b[k + 1];
}
is = true;
break;
}
}
if (!is) {
puts("-1");
return 0;
}
}
for (int i = 1; i <= n; ++i) {
printf("%d\n", a[i]);
}
} | ### Prompt
Generate a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[110], b[110];
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", b + i);
}
for (int i = n; i; --i) {
bool is = false;
for (int j = i; j; --j) {
if (b[j] > j) break;
if (b[j] == j) {
a[i] = j;
for (int k = j; k < i; ++k) {
b[k] = b[k + 1];
}
is = true;
break;
}
}
if (!is) {
puts("-1");
return 0;
}
}
for (int i = 1; i <= n; ++i) {
printf("%d\n", a[i]);
}
}
``` |
#include <stdio.h>
#include <iostream>
using namespace std;
int a[110],ans[110];
int main(){
int n,i,j,k;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(k=n;k>=1;k--){
for(i=k;i>=1;i--)
if(a[i]==i){
ans[k]=i;
break;
}
if(i==0) break;
for(j=i+1;j<=k;j++) a[j-1]=a[j];
}
if(k) printf("-1\n");
else for(i=1;i<=n;i++) printf("%d\n",ans[i]);
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <stdio.h>
#include <iostream>
using namespace std;
int a[110],ans[110];
int main(){
int n,i,j,k;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(k=n;k>=1;k--){
for(i=k;i>=1;i--)
if(a[i]==i){
ans[k]=i;
break;
}
if(i==0) break;
for(j=i+1;j<=k;j++) a[j-1]=a[j];
}
if(k) printf("-1\n");
else for(i=1;i<=n;i++) printf("%d\n",ans[i]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(){
int N;
cin>>N;
vector<int> a(N);
for(int i=0;i<N;i++){
cin>>a.at(i);
if(a.at(i) > i+1){
cout << -1 << endl;
return 0;
}
}
int ans[N];
for(int i=N-1;i>=0;i--){
for(int j=i;j>=0;j--){
if( a.at(j) == j+1){
ans[i] = j+1;
a.erase( a.begin()+j);
break;
}
}
}
for(int i=0;i<N;i++){
cout<<ans[i]<<endl;
}
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(){
int N;
cin>>N;
vector<int> a(N);
for(int i=0;i<N;i++){
cin>>a.at(i);
if(a.at(i) > i+1){
cout << -1 << endl;
return 0;
}
}
int ans[N];
for(int i=N-1;i>=0;i--){
for(int j=i;j>=0;j--){
if( a.at(j) == j+1){
ans[i] = j+1;
a.erase( a.begin()+j);
break;
}
}
}
for(int i=0;i<N;i++){
cout<<ans[i]<<endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int c[111];
int main() {
int n;
cin >> n;
list<int> l;
for (int i = 0; i < n; i++) {
int b;
cin >> b;
l.push_back(b);
}
for (int i = n; i--;) {
int k = i + 1;
auto j = l.end();
for (; j != l.begin(); k--)
if (*--j == k) goto p;
cout << -1 << endl;
return 0;
p:
l.erase(j);
c[i] = k;
}
for (int i = 0; i < n; i++)
cout << c[i] << '\n';
cout << flush;
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int c[111];
int main() {
int n;
cin >> n;
list<int> l;
for (int i = 0; i < n; i++) {
int b;
cin >> b;
l.push_back(b);
}
for (int i = n; i--;) {
int k = i + 1;
auto j = l.end();
for (; j != l.begin(); k--)
if (*--j == k) goto p;
cout << -1 << endl;
return 0;
p:
l.erase(j);
c[i] = k;
}
for (int i = 0; i < n; i++)
cout << c[i] << '\n';
cout << flush;
return 0;
}
``` |
#include<cstdio>
#include<vector>
int n;
std::vector <int> ve;
std::vector <int> ret;
int main(){
#ifdef Ezio
freopen("input","r",stdin);
#endif
scanf("%d",&n);
for(int i=0,x;i<n;i++){
scanf("%d",&x);
ve.push_back(x);
}
while(n){
int i;
for(i=n-1;~i;i--)
if(ve[i]==i+1){
ve.erase(ve.begin()+i);
ret.push_back(i+1);
break;
}
if(i==-1)return puts("-1"),0;
n--;
}
while(ret.size()){
printf("%d\n",ret.back());
ret.pop_back();
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<cstdio>
#include<vector>
int n;
std::vector <int> ve;
std::vector <int> ret;
int main(){
#ifdef Ezio
freopen("input","r",stdin);
#endif
scanf("%d",&n);
for(int i=0,x;i<n;i++){
scanf("%d",&x);
ve.push_back(x);
}
while(n){
int i;
for(i=n-1;~i;i--)
if(ve[i]==i+1){
ve.erase(ve.begin()+i);
ret.push_back(i+1);
break;
}
if(i==-1)return puts("-1"),0;
n--;
}
while(ret.size()){
printf("%d\n",ret.back());
ret.pop_back();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> a(n + 1);
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
vector<int> ret(n + 1);
for (int k = n; k >= 1; k--) {
for (int i = k; i >= 1; i--) {
if (a[i] > i) {
cout << -1 << '\n';
return 0;
} else if (a[i] == i) {
ret[k] = i;
a.erase(a.begin() + i);
break;
}
}
}
for (int i = 1; i <= n; i++) {
cout << ret[i] << '\n';
}
return 0;
} | ### Prompt
Generate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> a(n + 1);
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
vector<int> ret(n + 1);
for (int k = n; k >= 1; k--) {
for (int i = k; i >= 1; i--) {
if (a[i] > i) {
cout << -1 << '\n';
return 0;
} else if (a[i] == i) {
ret[k] = i;
a.erase(a.begin() + i);
break;
}
}
}
for (int i = 1; i <= n; i++) {
cout << ret[i] << '\n';
}
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main(){
int n;
cin >> n;
vector<int> b(n);
for(int i=0; i<n; i++){
cin >> b[i];
}
vector<int> ans;
for(int j=0; j<n; j++){
for(int i=b.size()-1; i>=0; i--){
if(b[i] == i+1){
ans.push_back(i+1);
b.erase(b.begin()+i);
break;
}
}
}
if(ans.size() == n){
for(int i=n-1; i>=0; i--){
cout << ans[i] << endl;
}
}else{
cout << -1 << endl;
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main(){
int n;
cin >> n;
vector<int> b(n);
for(int i=0; i<n; i++){
cin >> b[i];
}
vector<int> ans;
for(int j=0; j<n; j++){
for(int i=b.size()-1; i>=0; i--){
if(b[i] == i+1){
ans.push_back(i+1);
b.erase(b.begin()+i);
break;
}
}
}
if(ans.size() == n){
for(int i=n-1; i>=0; i--){
cout << ans[i] << endl;
}
}else{
cout << -1 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
typedef long long int ll;
using namespace std;
const ll MOD = 1e9 + 7;
int main()
{
int n; cin>>n;
vector<int> a,v;
for(int i=0;i<n;i++){
int x; cin>>x; a.push_back(x);
}
bool fok=true;
for(int i=0;i<n;i++){
bool ok=false;
for(int i=a.size()-1;i>=0;i--){
if(a[i]==i+1){
ok=true;
v.push_back(i+1);
a.erase(a.begin()+i); break;
}
}
if(!ok) {fok = false; break;}
}
if(fok){
for(int i=n-1;i>=0;i--) cout<<v[i]<<endl;
}
else cout<<"-1";
} | ### Prompt
Create a solution in Cpp for the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
typedef long long int ll;
using namespace std;
const ll MOD = 1e9 + 7;
int main()
{
int n; cin>>n;
vector<int> a,v;
for(int i=0;i<n;i++){
int x; cin>>x; a.push_back(x);
}
bool fok=true;
for(int i=0;i<n;i++){
bool ok=false;
for(int i=a.size()-1;i>=0;i--){
if(a[i]==i+1){
ok=true;
v.push_back(i+1);
a.erase(a.begin()+i); break;
}
}
if(!ok) {fok = false; break;}
}
if(fok){
for(int i=n-1;i>=0;i--) cout<<v[i]<<endl;
}
else cout<<"-1";
}
``` |
#include<bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long LL;
const int maxn=2e5+5;
int n,b[maxn],b0;
int ans[maxn];
int main()
{
scanf("%d",&n);
fo(i,1,n) scanf("%d",&b[i]);
b0=n;
fo(i,1,n)
{
fd(j,b0,1) if (b[j]==j)
{
ans[n-i+1]=j;
fo(k,j,b0-1) b[k]=b[k+1];
b0--;
break;
}
}
if (b0) puts("-1");
else fo(i,1,n) printf("%d\n",ans[i]);
} | ### Prompt
Create a solution in cpp for the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long LL;
const int maxn=2e5+5;
int n,b[maxn],b0;
int ans[maxn];
int main()
{
scanf("%d",&n);
fo(i,1,n) scanf("%d",&b[i]);
b0=n;
fo(i,1,n)
{
fd(j,b0,1) if (b[j]==j)
{
ans[n-i+1]=j;
fo(k,j,b0-1) b[k]=b[k+1];
b0--;
break;
}
}
if (b0) puts("-1");
else fo(i,1,n) printf("%d\n",ans[i]);
}
``` |
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> v;
for(int i=0;i<n;i++) {
int b;
cin >> b;
if (b > (i+1)) {
cout << -1 << endl;
return 0;
}
// (b-1)番目にbを挿入
auto it = v.begin();
for(int j=0;j<(b-1);j++) {
it++;
}
v.insert(it, b);
}
for(auto i: v) {
cout << i << endl;
}
return 0;
} | ### Prompt
Generate a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> v;
for(int i=0;i<n;i++) {
int b;
cin >> b;
if (b > (i+1)) {
cout << -1 << endl;
return 0;
}
// (b-1)番目にbを挿入
auto it = v.begin();
for(int j=0;j<(b-1);j++) {
it++;
}
v.insert(it, b);
}
for(auto i: v) {
cout << i << endl;
}
return 0;
}
``` |
#include<cstdio>
#define maxn 105
int n,a[maxn],cnt[maxn],ans[maxn];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
for(int i=1;i<=n;i++) cnt[a[i]]++;
for(int i=1;i<=n;i++)
if(cnt[i]>n-i+1) { printf("-1\n"); return 0; }
for(int i=n;i;i--)
{
int pos=0;
for(int j=1;j<=i;j++)
if(a[j]==j&&a[j]>a[pos]) pos=j;
ans[i]=pos;
for(int j=pos;j<i;j++) a[j]=a[j+1];
for(int j=1;j<i;j++)
if(a[j]>=i) { printf("-1\n"); return 0; }
}
for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
} | ### Prompt
Please provide a CPP coded solution to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<cstdio>
#define maxn 105
int n,a[maxn],cnt[maxn],ans[maxn];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
for(int i=1;i<=n;i++) cnt[a[i]]++;
for(int i=1;i<=n;i++)
if(cnt[i]>n-i+1) { printf("-1\n"); return 0; }
for(int i=n;i;i--)
{
int pos=0;
for(int j=1;j<=i;j++)
if(a[j]==j&&a[j]>a[pos]) pos=j;
ans[i]=pos;
for(int j=pos;j<i;j++) a[j]=a[j+1];
for(int j=1;j<i;j++)
if(a[j]>=i) { printf("-1\n"); return 0; }
}
for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
}
``` |
#include<bits/stdc++.h>
using namespace std;
long long n,tmp;
bool ok;
vector<int> v,ans;
int main(){
cin>>n;
for(int i=0;i<n;i++){
cin>>tmp;
v.push_back(tmp);
}
for(int i=0;i<n;i++){
ok=false;
for(int j=v.size()-1;j>=0;j--){
if((j+1)==v[j]){
ok=true;
ans.push_back(v[j]);
v.erase(v.begin()+j);
break;
}
}
if(ok==false){
cout<<"-1"<<endl;
return 0;
}
}
for(int i=ans.size()-1;i>=0;i--)cout<<ans[i]<<endl;
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
long long n,tmp;
bool ok;
vector<int> v,ans;
int main(){
cin>>n;
for(int i=0;i<n;i++){
cin>>tmp;
v.push_back(tmp);
}
for(int i=0;i<n;i++){
ok=false;
for(int j=v.size()-1;j>=0;j--){
if((j+1)==v[j]){
ok=true;
ans.push_back(v[j]);
v.erase(v.begin()+j);
break;
}
}
if(ok==false){
cout<<"-1"<<endl;
return 0;
}
}
for(int i=ans.size()-1;i>=0;i--)cout<<ans[i]<<endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(void){
int N;
cin >> N;
vector<int> b(N+1);
vector<int> ans;
for(int i=0;i<N;i++)cin >> b[i+1];
for(int i=N;i>=1;i--){
for(int j=i;j>=1;j--){
if(b[j]==j){
ans.push_back(j);
b.erase(b.begin()+j);
break;
}
}
}
if(b.size()==1){
reverse(ans.begin(),ans.end());
for(auto x:ans)cout << x << endl;
}else{
cout << -1 << endl;
}
} | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(void){
int N;
cin >> N;
vector<int> b(N+1);
vector<int> ans;
for(int i=0;i<N;i++)cin >> b[i+1];
for(int i=N;i>=1;i--){
for(int j=i;j>=1;j--){
if(b[j]==j){
ans.push_back(j);
b.erase(b.begin()+j);
break;
}
}
}
if(b.size()==1){
reverse(ans.begin(),ans.end());
for(auto x:ans)cout << x << endl;
}else{
cout << -1 << endl;
}
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main(){
int N;cin>>N;
list<int>B;for(int i=0;i<N;i++){int b;cin>>b;B.push_back(b);}
vector<int>ans;
while(B.size()){
int i=B.size();bool OK=false;
auto itr=B.end();itr--;
while(1){
if(*itr==i){ans.push_back(i);B.erase(itr);OK=true;break;}
i--;
if(itr==B.begin())break;
itr--;
}
if(OK==false){cout<<-1;return 0;}
}
for(int i=N-1;i>=0;i--)cout<<ans[i]<<endl;
} | ### Prompt
Please provide a cpp coded solution to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int N;cin>>N;
list<int>B;for(int i=0;i<N;i++){int b;cin>>b;B.push_back(b);}
vector<int>ans;
while(B.size()){
int i=B.size();bool OK=false;
auto itr=B.end();itr--;
while(1){
if(*itr==i){ans.push_back(i);B.erase(itr);OK=true;break;}
i--;
if(itr==B.begin())break;
itr--;
}
if(OK==false){cout<<-1;return 0;}
}
for(int i=N-1;i>=0;i--)cout<<ans[i]<<endl;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main() {
int N; cin >> N;
vector<int> v(N);
for (int i = 0; i < N; i++) cin >> v[i];
vector<int> r(N);
for (int i = N-1; i >= 0; i--) {
int p = -1;
for (int j = 0; j < v.size(); j++) {
if (v[j] == j+1) p = j;
}
if (p == -1) {
cout << -1 << endl; return 0;
}
r[i] = p+1;
v.erase(v.begin()+p);
}
for (int i = 0; i < N; i++) cout << r[i] << endl;
} | ### Prompt
Create a solution in CPP for the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main() {
int N; cin >> N;
vector<int> v(N);
for (int i = 0; i < N; i++) cin >> v[i];
vector<int> r(N);
for (int i = N-1; i >= 0; i--) {
int p = -1;
for (int j = 0; j < v.size(); j++) {
if (v[j] == j+1) p = j;
}
if (p == -1) {
cout << -1 << endl; return 0;
}
r[i] = p+1;
v.erase(v.begin()+p);
}
for (int i = 0; i < N; i++) cout << r[i] << endl;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,i,j,k;
cin >> n;
vector<int> b(n),ans(n);
for(i=0; i<n; i++){
cin >> b[i];
}
for(i=n-1; i>=0; i--){
k=0;
for(j=i; j>=0; j--){
if(b[j]==j+1){
ans[i]=j+1;
for(k=j; k<n-1; k++){
b[k]=b[k+1];
}
j=0;
}
}
if(k==0){
cout << -1; return 0;
}
}
for(i=0; i<n; i++){
cout << ans[i] << endl;
}
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int n,i,j,k;
cin >> n;
vector<int> b(n),ans(n);
for(i=0; i<n; i++){
cin >> b[i];
}
for(i=n-1; i>=0; i--){
k=0;
for(j=i; j>=0; j--){
if(b[j]==j+1){
ans[i]=j+1;
for(k=j; k<n-1; k++){
b[k]=b[k+1];
}
j=0;
}
}
if(k==0){
cout << -1; return 0;
}
}
for(i=0; i<n; i++){
cout << ans[i] << endl;
}
}
``` |
#include<bits/stdc++.h>
using namespace std;
int n,a[105];
int ans[105],x;
int Max;
int main(){
scanf("%d",&n);
x=n;
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
for(int i=1;i<=n;i++){
Max=0;
for(int j=1;j<=n-i+1;j++){
if(a[j]>j){
printf("-1\n");
return 0;
}
if(a[j]==j) Max=max(Max,j);
}
ans[x]=Max;
x--;
for(int j=Max;j<=n-i;j++){
a[j]=a[j+1];
}
}
for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
return 0;
} | ### Prompt
Generate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int n,a[105];
int ans[105],x;
int Max;
int main(){
scanf("%d",&n);
x=n;
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
for(int i=1;i<=n;i++){
Max=0;
for(int j=1;j<=n-i+1;j++){
if(a[j]>j){
printf("-1\n");
return 0;
}
if(a[j]==j) Max=max(Max,j);
}
ans[x]=Max;
x--;
for(int j=Max;j<=n-i;j++){
a[j]=a[j+1];
}
}
for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 102;
int N;
vector<int> A;
void solve()
{
vector<int> ans;
while(!A.empty()){
int n = -1;
for(int i=0;i<(int)A.size();++i){
if(A[i]==i+1){
n = i;
}
}
if(n==-1){
cout << -1 << endl;
return;
}
ans.push_back(n+1);
A.erase(A.begin()+n);
}
reverse(ans.begin(), ans.end());
for(int x : ans) cout << x << endl;
}
int main()
{
cin >> N;
A.resize(N);
for(int i=0;i<N;++i) cin >> A[i];
solve();
return 0;
} | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 102;
int N;
vector<int> A;
void solve()
{
vector<int> ans;
while(!A.empty()){
int n = -1;
for(int i=0;i<(int)A.size();++i){
if(A[i]==i+1){
n = i;
}
}
if(n==-1){
cout << -1 << endl;
return;
}
ans.push_back(n+1);
A.erase(A.begin()+n);
}
reverse(ans.begin(), ans.end());
for(int x : ans) cout << x << endl;
}
int main()
{
cin >> N;
A.resize(N);
for(int i=0;i<N;++i) cin >> A[i];
solve();
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main() {
int n;cin>>n;
vector<int> a(n);
for(int i=0;i<n;i++) cin>>a[i],--a[i];
vector<int> ans;
function<bool(vector<int>)> solve=[&](vector<int> a) {
if(a.empty()) return true;
for(int i=a.size()-1;~i;i--)
if(a[i]==i) {
a.erase(a.begin()+i);
bool ret=solve(a);
ans.push_back(i);
return ret;
}
return false;
};
if(solve(a)) {
for(auto x:ans) printf("%d\n",x+1);
}
else puts("-1");
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main() {
int n;cin>>n;
vector<int> a(n);
for(int i=0;i<n;i++) cin>>a[i],--a[i];
vector<int> ans;
function<bool(vector<int>)> solve=[&](vector<int> a) {
if(a.empty()) return true;
for(int i=a.size()-1;~i;i--)
if(a[i]==i) {
a.erase(a.begin()+i);
bool ret=solve(a);
ans.push_back(i);
return ret;
}
return false;
};
if(solve(a)) {
for(auto x:ans) printf("%d\n",x+1);
}
else puts("-1");
return 0;
}
``` |
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int d[2020];
int main()
{
int num;
scanf("%d", &num);
for (int i = 1; i <= num; i++)scanf("%d", &d[i]);
vector<int>v;
for (int i = num; i >= 1; i--)
{
int r = -1;
for (int j = 1; j <= i; j++)if (d[j] == j)r = j;
if (r == -1)
{
printf("-1\n");
return 0;
}
v.push_back(r);
for (int j = r; j < i; j++)d[j] = d[j + 1];
}
for (int i = num - 1; i >= 0; i--)printf("%d\n", v[i]);
} | ### Prompt
Generate a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int d[2020];
int main()
{
int num;
scanf("%d", &num);
for (int i = 1; i <= num; i++)scanf("%d", &d[i]);
vector<int>v;
for (int i = num; i >= 1; i--)
{
int r = -1;
for (int j = 1; j <= i; j++)if (d[j] == j)r = j;
if (r == -1)
{
printf("-1\n");
return 0;
}
v.push_back(r);
for (int j = r; j < i; j++)d[j] = d[j + 1];
}
for (int i = num - 1; i >= 0; i--)printf("%d\n", v[i]);
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(){
int n,x,i;cin>>n;
vector<int>v;
for(i=1;i<=n;i++){
cin>>x;
if(x>i) {cout<<-1<<endl;return 0;}
v.insert(v.begin()+x-1,x);
}
for(auto it:v)cout<<it<<endl;
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(){
int n,x,i;cin>>n;
vector<int>v;
for(i=1;i<=n;i++){
cin>>x;
if(x>i) {cout<<-1<<endl;return 0;}
v.insert(v.begin()+x-1,x);
}
for(auto it:v)cout<<it<<endl;
return 0;
}
``` |
#include <iostream>
using namespace std;
int b[100],a[100];
int main(){
int i,N,j,k,maxj;
cin >> N;
for (i=0;i<N;++i) {
cin >> b[i];
if (b[i]>i+1) {
cout << -1 << endl;
return 0;
}
}
for (i=0;i<N;++i) {
maxj=0;
for (j=0;j<N-i;j++) {
if (b[j] == j+1) {
maxj=j+1;
}
}
a[N-1-i] = maxj;
for (k=maxj;k<N-i;++k) {
b[k-1] = b[k];
}
}
for (i=0;i<N;++i) {
cout << a[i] << endl;
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <iostream>
using namespace std;
int b[100],a[100];
int main(){
int i,N,j,k,maxj;
cin >> N;
for (i=0;i<N;++i) {
cin >> b[i];
if (b[i]>i+1) {
cout << -1 << endl;
return 0;
}
}
for (i=0;i<N;++i) {
maxj=0;
for (j=0;j<N-i;j++) {
if (b[j] == j+1) {
maxj=j+1;
}
}
a[N-1-i] = maxj;
for (k=maxj;k<N-i;++k) {
b[k-1] = b[k];
}
}
for (i=0;i<N;++i) {
cout << a[i] << endl;
}
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main(){
int N;cin>>N;
vector<int> B(N);bool flag=true;
for(int i=0;i<N;i++){
cin>>B[i];
if(B[i]>i+1){
flag=false;
}
}if(!flag){
cout<<-1<<endl;
return 0;
}vector<int> ans(N);
for(int i=N-1;i>=0;i--){
for(int j=i;j>=0;j--){
if(B[j]==j+1){
B.erase(B.begin()+j);
ans[i]=j+1;
break;
}
}
}for(int i=0;i<N;i++){
cout<<ans[i]<<endl;
}
} | ### Prompt
Develop a solution in CPP to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int N;cin>>N;
vector<int> B(N);bool flag=true;
for(int i=0;i<N;i++){
cin>>B[i];
if(B[i]>i+1){
flag=false;
}
}if(!flag){
cout<<-1<<endl;
return 0;
}vector<int> ans(N);
for(int i=N-1;i>=0;i--){
for(int j=i;j>=0;j--){
if(B[j]==j+1){
B.erase(B.begin()+j);
ans[i]=j+1;
break;
}
}
}for(int i=0;i<N;i++){
cout<<ans[i]<<endl;
}
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
vector<int> b(n);
for(int i=0;i<n;i++)
cin>>b.at(i);
vector<int> a;
for(int s=n;s>0;s--){
int p=s;
for(int i=0;i<s;i++)
p=(b.at(i)==i+1?i:p);
if(p==s){
cout<<-1<<endl;
return 0;
}
a.push_back(p+1);
for(int i=p;i+1<s;i++)
b.at(i)=b.at(i+1);
}
for(int i=n-1;i>=0;i--)
cout<<a.at(i)<<endl;
} | ### Prompt
Generate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
vector<int> b(n);
for(int i=0;i<n;i++)
cin>>b.at(i);
vector<int> a;
for(int s=n;s>0;s--){
int p=s;
for(int i=0;i<s;i++)
p=(b.at(i)==i+1?i:p);
if(p==s){
cout<<-1<<endl;
return 0;
}
a.push_back(p+1);
for(int i=p;i+1<s;i++)
b.at(i)=b.at(i+1);
}
for(int i=n-1;i>=0;i--)
cout<<a.at(i)<<endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int a[N];
int b[N];
void del(int id) {
for (int i = id; i < N - 1; ++i) {
b[i] = b[i + 1];
}
}
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> b[i];
}
for (int i = n; i > 0; --i) {
int id = -1;
for (int j = 1; j <= i; ++j) {
if (b[j] == j) {
id = j;
}
}
if (id == -1) {
puts("-1");
return 0;
}
del(id);
a[i] = id;
}
for (int i = 1; i <= n; ++i) {
printf("%d\n", a[i]);
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int a[N];
int b[N];
void del(int id) {
for (int i = id; i < N - 1; ++i) {
b[i] = b[i + 1];
}
}
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> b[i];
}
for (int i = n; i > 0; --i) {
int id = -1;
for (int j = 1; j <= i; ++j) {
if (b[j] == j) {
id = j;
}
}
if (id == -1) {
puts("-1");
return 0;
}
del(id);
a[i] = id;
}
for (int i = 1; i <= n; ++i) {
printf("%d\n", a[i]);
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(void)
{
int n;
cin >> n;
vector<int> b(n+1);
for (int i = 1; i <= n; i++) {
cin >> b[i];
}
deque<int> a;
while (b.size() != 1) {
for (int i = b.size()-1; i >= 1; i--) {
if (b[i] == i) {
a.push_front(b[i]);
b.erase(b.begin() + i);
break;
}
if (i == 1) {
cout << -1 << endl;
return 0;
}
}
}
for (int i = 0; i < n; i++) {
cout << a.front() << endl;
a.pop_front();
}
return 0;
} | ### Prompt
Please provide a cpp coded solution to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(void)
{
int n;
cin >> n;
vector<int> b(n+1);
for (int i = 1; i <= n; i++) {
cin >> b[i];
}
deque<int> a;
while (b.size() != 1) {
for (int i = b.size()-1; i >= 1; i--) {
if (b[i] == i) {
a.push_front(b[i]);
b.erase(b.begin() + i);
break;
}
if (i == 1) {
cout << -1 << endl;
return 0;
}
}
}
for (int i = 0; i < n; i++) {
cout << a.front() << endl;
a.pop_front();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(void) {
int N;
cin >> N;
vector<int> b;
for(int i = 1; i <= N; ++i) {
int t;
cin >> t;
if(t > i) {
cout << "-1\n";
return 0;
}
b.push_back(t);
}
vector<int> ans;
while(b.size() > 0) {
for(int i = b.size(); i >= 1; --i) {
if(b[i - 1] == i) {
ans.push_back(i);
b.erase(b.begin() + (i - 1));
break;
}
}
}
reverse(ans.begin(), ans.end());
for(int i = 0; i < N; ++i) {
cout << ans[i] << '\n';
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(void) {
int N;
cin >> N;
vector<int> b;
for(int i = 1; i <= N; ++i) {
int t;
cin >> t;
if(t > i) {
cout << "-1\n";
return 0;
}
b.push_back(t);
}
vector<int> ans;
while(b.size() > 0) {
for(int i = b.size(); i >= 1; --i) {
if(b[i - 1] == i) {
ans.push_back(i);
b.erase(b.begin() + (i - 1));
break;
}
}
}
reverse(ans.begin(), ans.end());
for(int i = 0; i < N; ++i) {
cout << ans[i] << '\n';
}
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
vector<int>ans;
for(int i=1;i<=n;i++){
int a;
cin>>a;
if(a>i){cout<<-1;return 0;}
else{ans.insert(ans.begin()+a-1,a);}
}
for(int i=0;i<n;i++){
cout<<ans[i]<<endl;
}
} | ### Prompt
Your task is to create a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
vector<int>ans;
for(int i=1;i<=n;i++){
int a;
cin>>a;
if(a>i){cout<<-1;return 0;}
else{ans.insert(ans.begin()+a-1,a);}
}
for(int i=0;i<n;i++){
cout<<ans[i]<<endl;
}
}
``` |
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(long long int i=0;i<n;++i)
typedef long long int ll;
int main(){
int n;
cin >> n;
vector<int> b(n);
rep(i,n)cin >> b[i];
vector<int> a;
for(int i=0;i<n;i++){
if(b[i]<=i+1){
a.insert(a.begin()+b[i]-1,b[i]);
}else{
cout << -1 << endl;
return 0;
}
}
rep(i,n){
cout << a[i] << " ";
}cout << endl;
return 0;
} | ### Prompt
Your challenge is to write a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(long long int i=0;i<n;++i)
typedef long long int ll;
int main(){
int n;
cin >> n;
vector<int> b(n);
rep(i,n)cin >> b[i];
vector<int> a;
for(int i=0;i<n;i++){
if(b[i]<=i+1){
a.insert(a.begin()+b[i]-1,b[i]);
}else{
cout << -1 << endl;
return 0;
}
}
rep(i,n){
cout << a[i] << " ";
}cout << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int b[1000],ans[1000];
bool check(int n){
if(!n)return 1;
for(int i=n;i;--i)
if(b[i]==i){
ans[n]=i;
for(int j=i;j<n;++j)b[j]=b[j+1];
return check(n-1);
}
return 0;
}
int main(){
int n;
scanf("%d",&n);
for(int i=1;i<=n;++i)scanf("%d",&b[i]);
if(!check(n))puts("-1");
else for(int i=1;i<=n;++i)printf("%d\n",ans[i]);
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int b[1000],ans[1000];
bool check(int n){
if(!n)return 1;
for(int i=n;i;--i)
if(b[i]==i){
ans[n]=i;
for(int j=i;j<n;++j)b[j]=b[j+1];
return check(n-1);
}
return 0;
}
int main(){
int n;
scanf("%d",&n);
for(int i=1;i<=n;++i)scanf("%d",&b[i]);
if(!check(n))puts("-1");
else for(int i=1;i<=n;++i)printf("%d\n",ans[i]);
}
``` |
#include<iostream>
using namespace std;
int main(){
int a[100];
int ans[100];
int n;
cin>>n;
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=n-1;i>=0;i--){
int add=-1;
for(int j=0;j<=i;j++){
if(a[j]==j+1){add=j;}}
if(add==-1){cout<<-1<<endl;return 0;}
else{
ans[i]=add+1;
for(int k=add;k<=i;k++){
a[k]=a[k+1];
}
}
}
for(int i=0;i<n;i++){
cout<<ans[i]<<endl;
}
}
| ### Prompt
In cpp, your task is to solve the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<iostream>
using namespace std;
int main(){
int a[100];
int ans[100];
int n;
cin>>n;
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=n-1;i>=0;i--){
int add=-1;
for(int j=0;j<=i;j++){
if(a[j]==j+1){add=j;}}
if(add==-1){cout<<-1<<endl;return 0;}
else{
ans[i]=add+1;
for(int k=add;k<=i;k++){
a[k]=a[k+1];
}
}
}
for(int i=0;i<n;i++){
cout<<ans[i]<<endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main() {
int N,b;
cin >> N;
vector<int> ans;
for(int i=0; i<N; i++){
cin >> b;
if(i+1<b){cout << "-1" << endl; return 0;}
ans.insert(ans.begin()+b-1,b);
}
for(int i=0; i<N; i++){
cout << ans[i] << endl;
}
return 0;
} | ### Prompt
In CPP, your task is to solve the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main() {
int N,b;
cin >> N;
vector<int> ans;
for(int i=0; i<N; i++){
cin >> b;
if(i+1<b){cout << "-1" << endl; return 0;}
ans.insert(ans.begin()+b-1,b);
}
for(int i=0; i<N; i++){
cout << ans[i] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(){
int N;
cin >> N;
vector<int> b(N),ans(N),index(N);
for(int i=0;i<N;i++){
cin >> b.at(i);
index.at(i)=i;
}
for(int i=N-1;i>=0;i--){
if(index.size()<b.at(i)){
cout << -1 << endl;
return 0;
}
ans.at(index.at(b.at(i)-1))=b.at(i);
index.erase(index.begin()+b.at(i)-1);
}
for(int i=0;i<N;i++){
cout << ans.at(i) << endl;
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(){
int N;
cin >> N;
vector<int> b(N),ans(N),index(N);
for(int i=0;i<N;i++){
cin >> b.at(i);
index.at(i)=i;
}
for(int i=N-1;i>=0;i--){
if(index.size()<b.at(i)){
cout << -1 << endl;
return 0;
}
ans.at(index.at(b.at(i)-1))=b.at(i);
index.erase(index.begin()+b.at(i)-1);
}
for(int i=0;i<N;i++){
cout << ans.at(i) << endl;
}
return 0;
}
``` |
#include<stdio.h>
int a[110],s[110];
int main(){
int n,i,j;
scanf("%d",&n);
for(i=1;i<=n;i++)scanf("%d",a+i);
for(i=n;i>0;i--){
for(j=i;j>0;j--){
if(a[j]==j)break;
}
if(!j){
printf("-1");
return 0;
}
s[i]=j;
for(;j<i;j++)a[j]=a[j+1];
}
for(i=1;i<=n;i++)printf("%d ",s[i]);
} | ### Prompt
In cpp, your task is to solve the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<stdio.h>
int a[110],s[110];
int main(){
int n,i,j;
scanf("%d",&n);
for(i=1;i<=n;i++)scanf("%d",a+i);
for(i=n;i>0;i--){
for(j=i;j>0;j--){
if(a[j]==j)break;
}
if(!j){
printf("-1");
return 0;
}
s[i]=j;
for(;j<i;j++)a[j]=a[j+1];
}
for(i=1;i<=n;i++)printf("%d ",s[i]);
}
``` |
#include <bits/stdc++.h>
#define int long long
using namespace std;
int a[10010];
int32_t main()
{
int n;
cin >> n;
vector <int> ans;
for(int i=1;i<=n;i++)
{
cin >> a[i];
if(ans.size() < a[i]-1)
{
cout << -1 << endl;
return 0;
}
ans.insert(ans.begin()+a[i]-1,a[i]);
}
for(int i:ans)
cout << i << endl;
} | ### Prompt
Create a solution in Cpp for the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
#define int long long
using namespace std;
int a[10010];
int32_t main()
{
int n;
cin >> n;
vector <int> ans;
for(int i=1;i<=n;i++)
{
cin >> a[i];
if(ans.size() < a[i]-1)
{
cout << -1 << endl;
return 0;
}
ans.insert(ans.begin()+a[i]-1,a[i]);
}
for(int i:ans)
cout << i << endl;
}
``` |
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
int main()
{
int n; cin >> n;
vector<int> a(n), b;
for(int i = 0; i < n; i++)cin >> a[i];
for(int i = 0; i < n; i++) {
if(a[i] - 1 > (int)b.size()) {
cout << -1 << endl;
return 0;
}
b.insert(b.begin() + a[i] - 1, a[i]);
}
for(auto i : b)cout << i << endl;
} | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
int main()
{
int n; cin >> n;
vector<int> a(n), b;
for(int i = 0; i < n; i++)cin >> a[i];
for(int i = 0; i < n; i++) {
if(a[i] - 1 > (int)b.size()) {
cout << -1 << endl;
return 0;
}
b.insert(b.begin() + a[i] - 1, a[i]);
}
for(auto i : b)cout << i << endl;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >> n;
vector<int> b(n);
for(int i = 0; i < n; i++) cin >> b[i];
vector<int> ans;
for(int i = n - 1; i >= 0; i--){
for(int j = i; j >= 0; j--){
if(b[j] == j + 1){
ans.push_back(j + 1);
b.erase(b.begin() + j);
break;
}
}
}
if(ans.size() == n){
for(int i = n - 1; i >= 0; i--){
cout << ans[i] << endl;
}
}else{
cout << -1 << endl;
}
return 0;
} | ### Prompt
Please formulate a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >> n;
vector<int> b(n);
for(int i = 0; i < n; i++) cin >> b[i];
vector<int> ans;
for(int i = n - 1; i >= 0; i--){
for(int j = i; j >= 0; j--){
if(b[j] == j + 1){
ans.push_back(j + 1);
b.erase(b.begin() + j);
break;
}
}
}
if(ans.size() == n){
for(int i = n - 1; i >= 0; i--){
cout << ans[i] << endl;
}
}else{
cout << -1 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
ll n;
cin >> n;
vector <ll> arr(n);
for (int z=0;z<n;z++){
cin >> arr[z];
}
vector <ll> aaa;
ll count = n;
while (arr.size() > 0){
if (count < 0 ){
cout << -1 << endl;
return 0;
}
for (int z=n-1;z>-1;z--){
if (arr[z] == z+1 && arr[z] <= count){
aaa.push_back(z+1);
arr.erase(arr.begin()+z);
break;
}
}
count--;
}
for (int z=n-1;z>-1;z--){
cout << aaa[z] << endl;
}
cin >> n;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
ll n;
cin >> n;
vector <ll> arr(n);
for (int z=0;z<n;z++){
cin >> arr[z];
}
vector <ll> aaa;
ll count = n;
while (arr.size() > 0){
if (count < 0 ){
cout << -1 << endl;
return 0;
}
for (int z=n-1;z>-1;z--){
if (arr[z] == z+1 && arr[z] <= count){
aaa.push_back(z+1);
arr.erase(arr.begin()+z);
break;
}
}
count--;
}
for (int z=n-1;z>-1;z--){
cout << aaa[z] << endl;
}
cin >> n;
}
``` |
#include<cstdio>
#include<list>
using namespace std;
int main(){
int N;
scanf("%d", &N);
list<int> b;
for (int i = 0; i < N; ++i){
int tmp;
scanf("%d", &tmp);
b.push_front(tmp);
}
int a[N];
for (int i = N - 1; i >= 0; --i){
list<int>::iterator it = b.begin();
int j = i + 1;
while(*it < j){
j--;
it++;
}
if(*it > j){
printf("-1\n");
return 0;
}else{
a[i] = *it;
b.erase(it);
}
}
for (int i = 0; i < N; ++i) printf("%d\n", a[i]);
return 0;
} | ### Prompt
Please provide a cpp coded solution to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<cstdio>
#include<list>
using namespace std;
int main(){
int N;
scanf("%d", &N);
list<int> b;
for (int i = 0; i < N; ++i){
int tmp;
scanf("%d", &tmp);
b.push_front(tmp);
}
int a[N];
for (int i = N - 1; i >= 0; --i){
list<int>::iterator it = b.begin();
int j = i + 1;
while(*it < j){
j--;
it++;
}
if(*it > j){
printf("-1\n");
return 0;
}else{
a[i] = *it;
b.erase(it);
}
}
for (int i = 0; i < N; ++i) printf("%d\n", a[i]);
return 0;
}
``` |
#include <cstdio>
#include <cstring>
#include <algorithm>
int n;
int A[111];
int ans[111];
bool vis[111];
int main(){
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d", &A[i]);
for(int i = 1; i <= n; i++){
int k = -1;
for(int j = 1, mx = n - i + 1; j <= mx; j++)
if(A[j] == j) k = j;
if(k == -1) return !printf("-1\n");
ans[i] = k;
for(int j = k, mx = n - i + 1; j < mx; j++) A[j] = A[j + 1];
}
for(int i = n; i; i--) printf("%d\n", ans[i]);
return 0;
} | ### Prompt
Generate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <cstdio>
#include <cstring>
#include <algorithm>
int n;
int A[111];
int ans[111];
bool vis[111];
int main(){
scanf("%d", &n);
for(int i = 1; i <= n; i++) scanf("%d", &A[i]);
for(int i = 1; i <= n; i++){
int k = -1;
for(int j = 1, mx = n - i + 1; j <= mx; j++)
if(A[j] == j) k = j;
if(k == -1) return !printf("-1\n");
ans[i] = k;
for(int j = k, mx = n - i + 1; j < mx; j++) A[j] = A[j + 1];
}
for(int i = n; i; i--) printf("%d\n", ans[i]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const ll MOD=(ll)1e9+7;
int main() {
int N; cin>>N;
vector<int> a,b(N);
for (int i = 0; i < N; ++i) {
cin>>b[i];
}
for (int i = 0; i < N; ++i) {
if(i+1<b[i]){
cout<<-1<<endl;
return 0;
}
a.insert(a.begin()+b[i]-1,b[i]);
}
for (auto i:a) {
cout<<i<<endl;
}
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const ll MOD=(ll)1e9+7;
int main() {
int N; cin>>N;
vector<int> a,b(N);
for (int i = 0; i < N; ++i) {
cin>>b[i];
}
for (int i = 0; i < N; ++i) {
if(i+1<b[i]){
cout<<-1<<endl;
return 0;
}
a.insert(a.begin()+b[i]-1,b[i]);
}
for (auto i:a) {
cout<<i<<endl;
}
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
stack<int> s;
int num[110];
int n;
void chg(int st,int ed)
{
for(int i=st;i<ed;i++)
{
num[i]=num[i+1];
}
}
int main()
{
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&num[i]);
}
for(int i=0;i<n;i++)
{
int dd=0;
for(int j=n-1-i;j>=0;j--)
{
if(num[j]==j+1)
{
dd=1;
s.push(j+1);
chg(j,n-i-1);
break;
}
}
if(dd==0)
{
printf("-1");
return 0;
}
}
while(!s.empty())
{
printf("%d\n",s.top());
s.pop();
}
} | ### Prompt
Create a solution in CPP for the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
stack<int> s;
int num[110];
int n;
void chg(int st,int ed)
{
for(int i=st;i<ed;i++)
{
num[i]=num[i+1];
}
}
int main()
{
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&num[i]);
}
for(int i=0;i<n;i++)
{
int dd=0;
for(int j=n-1-i;j>=0;j--)
{
if(num[j]==j+1)
{
dd=1;
s.push(j+1);
chg(j,n-i-1);
break;
}
}
if(dd==0)
{
printf("-1");
return 0;
}
}
while(!s.empty())
{
printf("%d\n",s.top());
s.pop();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >> n;
vector<int> b(n);
for(int i = 0; i < n; i++) cin >> b.at(i);
vector<int> a(n);
int now;
for(int i = 0; i < n; i++){
now = -1;
for(int j = 0; j < n-i; j++){
if(j+1 == b.at(j)) now = j+1;
}
a.at(i) = now;
if(now == -1) break;
b.erase(b.begin()+now-1);
}
if(now == -1) cout << now << endl;
else{
for(int i = n-1; i >= 0; i--){
cout << a.at(i) << endl;
}
}
} | ### Prompt
Please formulate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >> n;
vector<int> b(n);
for(int i = 0; i < n; i++) cin >> b.at(i);
vector<int> a(n);
int now;
for(int i = 0; i < n; i++){
now = -1;
for(int j = 0; j < n-i; j++){
if(j+1 == b.at(j)) now = j+1;
}
a.at(i) = now;
if(now == -1) break;
b.erase(b.begin()+now-1);
}
if(now == -1) cout << now << endl;
else{
for(int i = n-1; i >= 0; i--){
cout << a.at(i) << endl;
}
}
}
``` |
#include<iostream>
#include<vector>
using namespace std;
int N_MAX = 100;
int main()
{
int n,m,b,i;
int res[N_MAX];
vector<int> v;
cin >> n;
for(i=0;i<n;i++){
cin >> b;
v.push_back(b);
}
for(m=n;m>0;m--){
for(i=m-1;i>=0;i--)
if(v[i]==i+1){
res[m-1]=i+1;
v.erase(v.begin()+i);
break;
}
if(i==-1){
cout << "-1" << endl;
return 0;
}
}
for(i=0;i<n;i++)
cout << res[i] << endl;
return 0;
} | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<iostream>
#include<vector>
using namespace std;
int N_MAX = 100;
int main()
{
int n,m,b,i;
int res[N_MAX];
vector<int> v;
cin >> n;
for(i=0;i<n;i++){
cin >> b;
v.push_back(b);
}
for(m=n;m>0;m--){
for(i=m-1;i>=0;i--)
if(v[i]==i+1){
res[m-1]=i+1;
v.erase(v.begin()+i);
break;
}
if(i==-1){
cout << "-1" << endl;
return 0;
}
}
for(i=0;i<n;i++)
cout << res[i] << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main()
{
int N;
cin >> N;
vector<int> b(N);
for (int i = 0; i < N; i++) {
cin >> b[i];
}
vector<int> ans(N, 0);
for (int i = 0; i < N; i++) {
int p = -1;
for (int j = 0; j < b.size(); j++) {
if (b[j] == j + 1) p = j;
}
if (p == -1) {
cout << -1 << endl;
return 0;
}
b.erase(b.begin() + p);
ans[N - i - 1] = p+1;
}
for (int i = 0; i < N; i++) {
cout << ans[i] << endl;
}
} | ### Prompt
In Cpp, your task is to solve the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main()
{
int N;
cin >> N;
vector<int> b(N);
for (int i = 0; i < N; i++) {
cin >> b[i];
}
vector<int> ans(N, 0);
for (int i = 0; i < N; i++) {
int p = -1;
for (int j = 0; j < b.size(); j++) {
if (b[j] == j + 1) p = j;
}
if (p == -1) {
cout << -1 << endl;
return 0;
}
b.erase(b.begin() + p);
ans[N - i - 1] = p+1;
}
for (int i = 0; i < N; i++) {
cout << ans[i] << endl;
}
}
``` |
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;
const int N=1010;
int n,a[N],res[N];
int main(){
scanf("%d",&n);
rep(i,1,n) scanf("%d",&a[i]);
rep(i,1,n){
int k=0;
for (int j=n-i+1; j; j--) if (a[j]==j) { k=j; break; }
if (!k){ puts("-1"); return 0; }
res[i]=k;
rep(j,k,n-i) a[j]=a[j+1];
}
for (int i=n; i; i--) printf("%d\n",res[i]);
return 0;
} | ### Prompt
In Cpp, your task is to solve the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;
const int N=1010;
int n,a[N],res[N];
int main(){
scanf("%d",&n);
rep(i,1,n) scanf("%d",&a[i]);
rep(i,1,n){
int k=0;
for (int j=n-i+1; j; j--) if (a[j]==j) { k=j; break; }
if (!k){ puts("-1"); return 0; }
res[i]=k;
rep(j,k,n-i) a[j]=a[j+1];
}
for (int i=n; i; i--) printf("%d\n",res[i]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
vector<int> A(N);
for (int i = 0; i < N; i++) cin >> A.at(i);
vector<int> ans;
while (A.size()) {
bool b = 0;
for (int i = A.size() - 1; i >= 0; i--) {
if (A.at(i) == i + 1) {
ans.push_back(A.at(i));
A.erase(A.begin() + i);
b = 1;
break;
}
}
if (!b) return cout << -1 << "\n", 0;
}
reverse(ans.begin(), ans.end());
for (auto a : ans) cout << a << "\n";
} | ### Prompt
Please formulate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
vector<int> A(N);
for (int i = 0; i < N; i++) cin >> A.at(i);
vector<int> ans;
while (A.size()) {
bool b = 0;
for (int i = A.size() - 1; i >= 0; i--) {
if (A.at(i) == i + 1) {
ans.push_back(A.at(i));
A.erase(A.begin() + i);
b = 1;
break;
}
}
if (!b) return cout << -1 << "\n", 0;
}
reverse(ans.begin(), ans.end());
for (auto a : ans) cout << a << "\n";
}
``` |
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<memory.h>
#include<map>
#include<set>
#include<queue>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
int n,a[105],ans[105];
int main()
{
cin>>n;
rep(i,1,n)cin>>a[i];
per(j,n,1)
{
int t=0;
per(i,j,1)if(a[i]==i){t=i;break;}
if(!t){puts("-1");return 0;}
ans[j]=t;
rep(i,t,j-1)a[i]=a[i+1];
}
rep(i,1,n)cout<<ans[i]<<endl;
}
| ### Prompt
Generate a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<memory.h>
#include<map>
#include<set>
#include<queue>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
int n,a[105],ans[105];
int main()
{
cin>>n;
rep(i,1,n)cin>>a[i];
per(j,n,1)
{
int t=0;
per(i,j,1)if(a[i]==i){t=i;break;}
if(!t){puts("-1");return 0;}
ans[j]=t;
rep(i,t,j-1)a[i]=a[i+1];
}
rep(i,1,n)cout<<ans[i]<<endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,x,i;cin>>n;vector<int>v;
for(i=1;i<=n;i++){
cin>>x;
if(x>i) {cout<<-1<<endl;return 0;}
v.insert(v.begin()+x-1,x);
}
for(auto it:v)cout<<it<<endl;
} | ### Prompt
Please formulate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,x,i;cin>>n;vector<int>v;
for(i=1;i<=n;i++){
cin>>x;
if(x>i) {cout<<-1<<endl;return 0;}
v.insert(v.begin()+x-1,x);
}
for(auto it:v)cout<<it<<endl;
}
``` |
#include <stdio.h>
int main()
{
int N, A[101], B[101];
scanf ("%d", &N);
for (int i = 1; i <= N; i++) scanf ("%d", &A[i]);
for (int k = N; k >= 1; k--){
bool f = 0;
for (int i = k; i >= 1; i--) if (A[i] == i){
for (int j = i + 1; j <= k; j++) A[j - 1] = A[j];
B[k] = i;
f = 1;
break;
}
if (!f){
puts("-1");
return 0;
}
}
for (int i = 1; i <= N; i++) printf ("%d\n", B[i]);
return 0;
} | ### Prompt
Please provide a Cpp coded solution to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <stdio.h>
int main()
{
int N, A[101], B[101];
scanf ("%d", &N);
for (int i = 1; i <= N; i++) scanf ("%d", &A[i]);
for (int k = N; k >= 1; k--){
bool f = 0;
for (int i = k; i >= 1; i--) if (A[i] == i){
for (int j = i + 1; j <= k; j++) A[j - 1] = A[j];
B[k] = i;
f = 1;
break;
}
if (!f){
puts("-1");
return 0;
}
}
for (int i = 1; i <= N; i++) printf ("%d\n", B[i]);
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >>n;
vector<int> b(n),a(n);
for(int i=0;i<n;i++)cin>>b[i];
for(int j=0;j<n;j++){
for(int i=b.size()-1;i>=0;i--){
if(i+1==b[i]){
a[j]=b[i];
b.erase(b.begin()+i);
break;
}
}
}
if(a[n-1]!=1)cout<<-1<<endl;
else {for(int i=n-1;i>=0;i--)cout << a[i]<<endl;}
} | ### Prompt
Create a solution in CPP for the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >>n;
vector<int> b(n),a(n);
for(int i=0;i<n;i++)cin>>b[i];
for(int j=0;j<n;j++){
for(int i=b.size()-1;i>=0;i--){
if(i+1==b[i]){
a[j]=b[i];
b.erase(b.begin()+i);
break;
}
}
}
if(a[n-1]!=1)cout<<-1<<endl;
else {for(int i=n-1;i>=0;i--)cout << a[i]<<endl;}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a[100];
int n=0;
int b[100];
int N;
bool check(void){
for(int i=N-1;i>=0;i--){
if(b[i]==i+1){
a[n]=i+1;
n+=1;
for(int j=i;j<N-1;j++) b[j]=b[j+1];
N-=1;
if(N>0){
if(check()==false) return false;
}
return true;
}
}
return false;
}
int main(void){
cin>>N;
for(int i=0;i<N;i++) cin>>b[i];
if(check())
for(int i=n-1;i>=0;i--) cout<<a[i]<<endl;
else cout<<-1<<endl;
return 0;
} | ### Prompt
Create a solution in cpp for the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[100];
int n=0;
int b[100];
int N;
bool check(void){
for(int i=N-1;i>=0;i--){
if(b[i]==i+1){
a[n]=i+1;
n+=1;
for(int j=i;j<N-1;j++) b[j]=b[j+1];
N-=1;
if(N>0){
if(check()==false) return false;
}
return true;
}
}
return false;
}
int main(void){
cin>>N;
for(int i=0;i<N;i++) cin>>b[i];
if(check())
for(int i=n-1;i>=0;i--) cout<<a[i]<<endl;
else cout<<-1<<endl;
return 0;
}
``` |
#include <iostream>
#include <list>
int main() {
size_t N;
std::cin >> N;
std::list<size_t> b;
for (size_t i=0; i<N; i++) {
size_t a;
std::cin >> a;
if (a > i+1) {
std::cout << -1 << std::endl;
return 0;
}
auto it = std::next(b.begin(), a-1);
b.insert(it, a);
}
for (auto it = b.begin(); it != b.end(); it++) {
std::cout << *it << std::endl;
}
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <iostream>
#include <list>
int main() {
size_t N;
std::cin >> N;
std::list<size_t> b;
for (size_t i=0; i<N; i++) {
size_t a;
std::cin >> a;
if (a > i+1) {
std::cout << -1 << std::endl;
return 0;
}
auto it = std::next(b.begin(), a-1);
b.insert(it, a);
}
for (auto it = b.begin(); it != b.end(); it++) {
std::cout << *it << std::endl;
}
return 0;
}
``` |
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {
int N; cin >> N;
vector<int> b(N+1);
for(int i = 1; i <= N; ++i) cin >> b[i];
vector<int> a;
for(int i = N; i >= 1; --i) {
int r = -1;
for(int j = 1; j <= i; ++j) if(b[j] == j) r = j;
if(r == -1) {
cout << -1 << endl;
return 0;
}
a.push_back(r);
for(int j = r; j < i; ++j) b[j] = b[j+1];
}
for(int i = N-1; i >= 0; --i) {
std::cout << a[i] << std::endl;
}
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {
int N; cin >> N;
vector<int> b(N+1);
for(int i = 1; i <= N; ++i) cin >> b[i];
vector<int> a;
for(int i = N; i >= 1; --i) {
int r = -1;
for(int j = 1; j <= i; ++j) if(b[j] == j) r = j;
if(r == -1) {
cout << -1 << endl;
return 0;
}
a.push_back(r);
for(int j = r; j < i; ++j) b[j] = b[j+1];
}
for(int i = N-1; i >= 0; --i) {
std::cout << a[i] << std::endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
vector<int> v(N);
stack<int> ans;
for (int &e : v) cin >> e;
while (v.size() > 0) {
int p = -1;
for (int i = 0; i < N; i++) {
if (v[i] == i+1) p = i;
}
if (p < 0) {
cout << -1 << endl;
return 0;
}
ans.push(v[p]);
v.erase(v.begin()+p);
N--;
}
while (ans.size() > 0) {
cout << ans.top() << endl;
ans.pop();
}
} | ### Prompt
Generate a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int N;
cin >> N;
vector<int> v(N);
stack<int> ans;
for (int &e : v) cin >> e;
while (v.size() > 0) {
int p = -1;
for (int i = 0; i < N; i++) {
if (v[i] == i+1) p = i;
}
if (p < 0) {
cout << -1 << endl;
return 0;
}
ans.push(v[p]);
v.erase(v.begin()+p);
N--;
}
while (ans.size() > 0) {
cout << ans.top() << endl;
ans.pop();
}
}
``` |
#include<cstdio>
#include<algorithm>
using namespace std;
int n, w[110];
int vis[110], R[110];
int main() {
int i, j;
scanf("%d", &n);
for (i = 1; i <= n; i++)scanf("%d", &w[i]);
for (i = 1; i <= n; i++) {
int cur = 1, pv = -1;
for (j = 1; j <= n; j++) {
if (vis[j])cur++;
else {
if (w[j] == cur && cur <= i) {
pv = j;
}
}
}
if (pv == -1) {
puts("-1");
return 0;
}
R[i] = w[pv];
vis[pv] = 1;
}
for (i = 1; i <= n; i++)printf("%d\n", R[i]);
} | ### Prompt
Generate a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<cstdio>
#include<algorithm>
using namespace std;
int n, w[110];
int vis[110], R[110];
int main() {
int i, j;
scanf("%d", &n);
for (i = 1; i <= n; i++)scanf("%d", &w[i]);
for (i = 1; i <= n; i++) {
int cur = 1, pv = -1;
for (j = 1; j <= n; j++) {
if (vis[j])cur++;
else {
if (w[j] == cur && cur <= i) {
pv = j;
}
}
}
if (pv == -1) {
puts("-1");
return 0;
}
R[i] = w[pv];
vis[pv] = 1;
}
for (i = 1; i <= n; i++)printf("%d\n", R[i]);
}
``` |
// failed to generate code
#include <bits/stdc++.h>
using namespace std;
#define PB push_back
#define MP make_pair
int main(){
ios::sync_with_stdio(false);
int n; cin>>n; vector<int> b(n);
for (int i=0;i<n;i++) cin>>b[i];
vector<int> vans; vans.PB(-1);
for (int i=0;i<n;i++) {
int x=b[i];
if (x>vans.size()) {
cout<<-1<<endl; return 0;
}
vans.insert(vans.begin()+x,x);
//cout<<x<<":"<<vans.size()<<endl;
}
for (int i=1;i<=n;i++) cout<<vans[i]<<endl;
}
| ### Prompt
Please create a solution in cpp to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
// failed to generate code
#include <bits/stdc++.h>
using namespace std;
#define PB push_back
#define MP make_pair
int main(){
ios::sync_with_stdio(false);
int n; cin>>n; vector<int> b(n);
for (int i=0;i<n;i++) cin>>b[i];
vector<int> vans; vans.PB(-1);
for (int i=0;i<n;i++) {
int x=b[i];
if (x>vans.size()) {
cout<<-1<<endl; return 0;
}
vans.insert(vans.begin()+x,x);
//cout<<x<<":"<<vans.size()<<endl;
}
for (int i=1;i<=n;i++) cout<<vans[i]<<endl;
}
``` |
#include<iostream>
#include<vector>
using namespace std;
int main(){
int n,i,j;
cin>>n;
vector<int> a(n);
for(i=0;i<n;i++){
cin>>a[i];
}
int buf[n];
for(i=0;i<n;i++){
for(j=n-i;j>=1;j--){
if(a[j-1]==j){
buf[i]=j;
a.erase(a.begin()+j-1);
break;
}
}
if(j==0){
cout<<-1<<endl;
return 0;
}
}
for(i=0;i<n;i++) cout<<buf[n-i-1]<<endl;
return 0;
} | ### Prompt
Generate a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<iostream>
#include<vector>
using namespace std;
int main(){
int n,i,j;
cin>>n;
vector<int> a(n);
for(i=0;i<n;i++){
cin>>a[i];
}
int buf[n];
for(i=0;i<n;i++){
for(j=n-i;j>=1;j--){
if(a[j-1]==j){
buf[i]=j;
a.erase(a.begin()+j-1);
break;
}
}
if(j==0){
cout<<-1<<endl;
return 0;
}
}
for(i=0;i<n;i++) cout<<buf[n-i-1]<<endl;
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int n,a;
vector<int> v;
int main()
{
cin>>n;
for(int i=1;i<=n;i++)cin>>a,v.push_back(a-1);
vector<int> ans;
while(!v.empty())
{
bool f=1;
for(int i=v.size()-1;i>=0;i--)
if(v[i]==i)
{
ans.push_back(i+1);
v.erase(v.begin()+i);
f=0;
break;
}
if(f)cout<<"-1\n",exit(0);
}
reverse(ans.begin(),ans.end());
for(int i:ans)cout<<i<<'\n';
}
| ### Prompt
Develop a solution in cpp to the problem described below:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int n,a;
vector<int> v;
int main()
{
cin>>n;
for(int i=1;i<=n;i++)cin>>a,v.push_back(a-1);
vector<int> ans;
while(!v.empty())
{
bool f=1;
for(int i=v.size()-1;i>=0;i--)
if(v[i]==i)
{
ans.push_back(i+1);
v.erase(v.begin()+i);
f=0;
break;
}
if(f)cout<<"-1\n",exit(0);
}
reverse(ans.begin(),ans.end());
for(int i:ans)cout<<i<<'\n';
}
``` |
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(void){
int n,l;
vector<int> b;
vector<int> a;
cin >> n;
l = n;
for(int i=0;i<n;++i){
int k;
cin >> k;
b.push_back(k);
}
for(int i=0;i<n;++i){
for(int j=l-1;j>=0;--j){
if(b[j] == j+1){
a.push_back(j+1);
b.erase(b.begin() + j);
l--;
break;
}
}
}
reverse(a.begin(),a.end());
if(a.size() == n){
for(int i=0;i<n;++i)cout << a[i] << endl;
}else{
cout << -1 << endl;
}
} | ### Prompt
Your task is to create a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(void){
int n,l;
vector<int> b;
vector<int> a;
cin >> n;
l = n;
for(int i=0;i<n;++i){
int k;
cin >> k;
b.push_back(k);
}
for(int i=0;i<n;++i){
for(int j=l-1;j>=0;--j){
if(b[j] == j+1){
a.push_back(j+1);
b.erase(b.begin() + j);
l--;
break;
}
}
}
reverse(a.begin(),a.end());
if(a.size() == n){
for(int i=0;i<n;++i)cout << a[i] << endl;
}else{
cout << -1 << endl;
}
}
``` |
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
int N; cin >> N;
vector<int> b(N);
for(int i = 0; i < N; i++){
cin >> b[i];
b[i]--;
}
vector<int> a;
for(int i = 0; i < N; i++){
for(int j = b.size() - 1; j >= 0; j--){
if(b[j] == j) {
a.emplace_back(b[j] + 1);
b.erase(b.begin() + j);
break;
}
}
}
if(b.empty()) {
for(int i = N - 1; i >= 0; i--){
cout << a[i] << endl;
}
}else {
cout << -1;
}
} | ### Prompt
Your task is to create a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
int N; cin >> N;
vector<int> b(N);
for(int i = 0; i < N; i++){
cin >> b[i];
b[i]--;
}
vector<int> a;
for(int i = 0; i < N; i++){
for(int j = b.size() - 1; j >= 0; j--){
if(b[j] == j) {
a.emplace_back(b[j] + 1);
b.erase(b.begin() + j);
break;
}
}
}
if(b.empty()) {
for(int i = N - 1; i >= 0; i--){
cout << a[i] << endl;
}
}else {
cout << -1;
}
}
``` |
#include<bits/stdc++.h>
using namespace std;
int n,a[101],ans[101],an,s[101];
bool bo[101];
int main()
{
scanf("%d",&n);
for (int i=1; i<=n; i++) scanf("%d",&a[i]);
memset(bo,0,sizeof(bo));
memset(s,0,sizeof(s));
for (int T=1; T<=n; T++)
{
bool fl=0;
for (int i=n; i; i--)
if (!bo[i]&&s[i]==a[i]-1)
{
ans[++an]=a[i];
for (int j=i; j<=n; j++) s[j]++;
bo[i]=1;
fl=1; break;
}
if (!fl) return puts("-1"),0;
}
for (int i=1; i<=an; i++) printf("%d\n",ans[i]);
return 0;
} | ### Prompt
Construct a cpp code solution to the problem outlined:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int n,a[101],ans[101],an,s[101];
bool bo[101];
int main()
{
scanf("%d",&n);
for (int i=1; i<=n; i++) scanf("%d",&a[i]);
memset(bo,0,sizeof(bo));
memset(s,0,sizeof(s));
for (int T=1; T<=n; T++)
{
bool fl=0;
for (int i=n; i; i--)
if (!bo[i]&&s[i]==a[i]-1)
{
ans[++an]=a[i];
for (int j=i; j<=n; j++) s[j]++;
bo[i]=1;
fl=1; break;
}
if (!fl) return puts("-1"),0;
}
for (int i=1; i<=an; i++) printf("%d\n",ans[i]);
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main(){
int N;
cin >> N;
vector<int> b(N);
for(int i=1;i<=N;i++)cin >> b[i];
for(int i=1;i<=N;i++){
if(b[i]>i){cout << -1 << endl; return 0;}
}
vector<int> a;
for(int i=1;i<=N;i++){
if(b[i]==1){a.push_back(1);}
else{a.insert(a.begin()+i-b[i],b[i]);}
}
reverse(a.begin(),a.end());
for(int i=0;i<N;i++){cout << a[i] << endl;}
} | ### Prompt
In CPP, your task is to solve the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int N;
cin >> N;
vector<int> b(N);
for(int i=1;i<=N;i++)cin >> b[i];
for(int i=1;i<=N;i++){
if(b[i]>i){cout << -1 << endl; return 0;}
}
vector<int> a;
for(int i=1;i<=N;i++){
if(b[i]==1){a.push_back(1);}
else{a.insert(a.begin()+i-b[i],b[i]);}
}
reverse(a.begin(),a.end());
for(int i=0;i<N;i++){cout << a[i] << endl;}
}
``` |
#include <cstdio>
#include <vector>
std::vector<int> bs;
int seq[105];
int main(){
int N;
scanf("%d",&N);
bs.resize(N+1);
for(int i=1;i<=N;i++){
scanf("%d",&bs[i]);
}
for(int t=N;t>0;t--){
for(int j=t;;j--){
if(j==0){
printf("-1\n");
return 0;
}
if(bs[j]==j){
seq[t]=j;
bs.erase(bs.begin()+j);
break;
}
}
}
for(int i=1;i<=N;i++){
printf("%d\n",seq[i]);
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <cstdio>
#include <vector>
std::vector<int> bs;
int seq[105];
int main(){
int N;
scanf("%d",&N);
bs.resize(N+1);
for(int i=1;i<=N;i++){
scanf("%d",&bs[i]);
}
for(int t=N;t>0;t--){
for(int j=t;;j--){
if(j==0){
printf("-1\n");
return 0;
}
if(bs[j]==j){
seq[t]=j;
bs.erase(bs.begin()+j);
break;
}
}
}
for(int i=1;i<=N;i++){
printf("%d\n",seq[i]);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 200;
int n, len;
int a[N], ans[N]; bool flag;
int main() {
scanf("%d", &n), len = n;
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = n; i; i--) {
flag = 1;
for(int j = len; j; j--) {
if(a[j] != j) continue;
for(int k = j; k < len; k++) a[k] = a[k + 1];
ans[i] = j, --len, flag = 0; break;
}
if(flag) { puts("-1"); return 0; }
}
for(int i = 1; i <= n; i++) printf("%d\n", ans[i]);
return 0;
} | ### Prompt
Construct a CPP code solution to the problem outlined:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 200;
int n, len;
int a[N], ans[N]; bool flag;
int main() {
scanf("%d", &n), len = n;
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = n; i; i--) {
flag = 1;
for(int j = len; j; j--) {
if(a[j] != j) continue;
for(int k = j; k < len; k++) a[k] = a[k + 1];
ans[i] = j, --len, flag = 0; break;
}
if(flag) { puts("-1"); return 0; }
}
for(int i = 1; i <= n; i++) printf("%d\n", ans[i]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)
int main(){
int N; cin >> N;
vector<int> a(N),b(N);
repp(i,0,N) cin >> b[i];
repm(i,N-1,-1){
repm(k,i,-1){
if(k+1 == b[k]){
a[i] = k+1;
break;
}
if(k == 0) return cout << -1 << endl, 0;
}
repp(k,a[i],i+1) b[k-1] = b[k];
}
repp(i,0,N) cout << a[i] << endl;
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)
int main(){
int N; cin >> N;
vector<int> a(N),b(N);
repp(i,0,N) cin >> b[i];
repm(i,N-1,-1){
repm(k,i,-1){
if(k+1 == b[k]){
a[i] = k+1;
break;
}
if(k == 0) return cout << -1 << endl, 0;
}
repp(k,a[i],i+1) b[k-1] = b[k];
}
repp(i,0,N) cout << a[i] << endl;
return 0;
}
``` |
#include<bits/stdc++.h>
using namespace std;
const int N=110;
int n,b[N],x[N],y[N],s[N];
int main(){
scanf("%d",&n);
for(int i=1; i<=n; ++i)
scanf("%d",&b[i]);
for(int i=1; i<=n; ++i)x[i]=i,y[i]=1;
for(int i=1; i<=n; ++i){
int k=0;
for(int j=1; j<=n; ++j)if(b[j]==x[j]&&y[j]==1)k=j;
if(k==0){puts("-1"); return 0;}
for(int j=k; j<=n; ++j)--x[j]; y[k]=0; s[++s[0]]=b[k];
}
for(int i=n; i>=1; --i)printf("%d\n",s[i]);
return 0;
} | ### Prompt
Create a solution in Cpp for the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
const int N=110;
int n,b[N],x[N],y[N],s[N];
int main(){
scanf("%d",&n);
for(int i=1; i<=n; ++i)
scanf("%d",&b[i]);
for(int i=1; i<=n; ++i)x[i]=i,y[i]=1;
for(int i=1; i<=n; ++i){
int k=0;
for(int j=1; j<=n; ++j)if(b[j]==x[j]&&y[j]==1)k=j;
if(k==0){puts("-1"); return 0;}
for(int j=k; j<=n; ++j)--x[j]; y[k]=0; s[++s[0]]=b[k];
}
for(int i=n; i>=1; --i)printf("%d\n",s[i]);
return 0;
}
``` |
#include <vector>
#include <iostream>
using namespace std;
int main() {
int n; cin >> n;
vector<int> b(n); for (int i = 0; i < n; i++) cin >> b[i];
vector<int> ans;
for (int i = 0; i < n; i++) {
int k = -1;
for (int j = 0; j < b.size(); j++) {
if (j == b[j]-1) k = j;
}
if (k == -1) { cout << "-1\n"; return 0; }
ans.push_back(k);
b.erase(b.begin() + k);
}
for (auto it = ans.rbegin(); it != ans.rend(); it++) cout << 1 + *it << endl;
}
| ### Prompt
Create a solution in cpp for the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <vector>
#include <iostream>
using namespace std;
int main() {
int n; cin >> n;
vector<int> b(n); for (int i = 0; i < n; i++) cin >> b[i];
vector<int> ans;
for (int i = 0; i < n; i++) {
int k = -1;
for (int j = 0; j < b.size(); j++) {
if (j == b[j]-1) k = j;
}
if (k == -1) { cout << "-1\n"; return 0; }
ans.push_back(k);
b.erase(b.begin() + k);
}
for (auto it = ans.rbegin(); it != ans.rend(); it++) cout << 1 + *it << endl;
}
``` |
#include<bits/stdc++.h>
using namespace std;
int main() {
int n; cin>>n;
vector<int> b(n);
for (int i=0; i<n; i++) cin>>b[i], b[i]--;
vector<int> ans;
while (b.size() > 0) {
int z = -1;
for (int i = 0; i < b.size(); i++) {
if(b[i] == i) z = i;
}
if( z==-1) {
cout << -1 << endl;
return 0;
}
ans.push_back(z);
b.erase(b.begin() + z);
}
reverse(ans.begin(), ans.end());
for (int i=0; i<n; i++) cout<<1+ans[i]<<endl;
return 0;
} | ### Prompt
Your task is to create a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
using namespace std;
int main() {
int n; cin>>n;
vector<int> b(n);
for (int i=0; i<n; i++) cin>>b[i], b[i]--;
vector<int> ans;
while (b.size() > 0) {
int z = -1;
for (int i = 0; i < b.size(); i++) {
if(b[i] == i) z = i;
}
if( z==-1) {
cout << -1 << endl;
return 0;
}
ans.push_back(z);
b.erase(b.begin() + z);
}
reverse(ans.begin(), ans.end());
for (int i=0; i<n; i++) cout<<1+ans[i]<<endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int n, b[N], a[N];
int main() {
scanf("%d",&n);
for(int i=1;i<=n;++i) scanf("%d",b+i);
for(int i=1;i<=n;++i) {
for(int j=n-i+1;j>=1;--j)
if(b[j]==j) {
a[i]=j; break;
}
if(!a[i]) return puts("-1"),0;
for(int j=a[i];j<=n-i;++j) b[j]=b[j+1];
}
for(int i=n;i>=1;--i) printf("%d\n",a[i]);
return 0;
} | ### Prompt
Your challenge is to write a CPP solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int n, b[N], a[N];
int main() {
scanf("%d",&n);
for(int i=1;i<=n;++i) scanf("%d",b+i);
for(int i=1;i<=n;++i) {
for(int j=n-i+1;j>=1;--j)
if(b[j]==j) {
a[i]=j; break;
}
if(!a[i]) return puts("-1"),0;
for(int j=a[i];j<=n-i;++j) b[j]=b[j+1];
}
for(int i=n;i>=1;--i) printf("%d\n",a[i]);
return 0;
}
``` |
#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
using namespace std;
const int N=111;
int n,a[N],p[N],del[N],ans[N];
int main(){
scanf("%d",&n);
rep (i,1,n) scanf("%d",&a[i]),p[i]=i,del[i]=0;
rep (i,1,n){
int cnt=0,x;
rep (j,1,n) if (!del[j]&&p[j]==a[j]) cnt++,x=j;
if (!cnt) return puts("-1"),0;
ans[n-i+1]=a[x],del[x]=1;
rep (j,x+1,n) if (!del[j]) p[j]--;
}
rep (i,1,n) printf("%d\n",ans[i]);
return 0;
} | ### Prompt
Please create a solution in Cpp to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
using namespace std;
const int N=111;
int n,a[N],p[N],del[N],ans[N];
int main(){
scanf("%d",&n);
rep (i,1,n) scanf("%d",&a[i]),p[i]=i,del[i]=0;
rep (i,1,n){
int cnt=0,x;
rep (j,1,n) if (!del[j]&&p[j]==a[j]) cnt++,x=j;
if (!cnt) return puts("-1"),0;
ans[n-i+1]=a[x],del[x]=1;
rep (j,x+1,n) if (!del[j]) p[j]--;
}
rep (i,1,n) printf("%d\n",ans[i]);
return 0;
}
``` |
#include <iostream>
#include <vector>
#include <list>
using namespace std;
int main() {
int n; cin >> n;
list<int> b(n);
vector<int> ans;
for (auto& e : b) cin >> e;
while (!b.empty()) {
int t = b.size();
for (auto it = b.rbegin(); it != b.rend(); ++it) {
if (*it == t) {
b.erase(--it.base());
ans.push_back(t);
break;
}
--t;
}
if (!t) break;
}
if (ans.size() != n) cout << -1 << endl;
else {
for (int i = n - 1; i >= 0; --i) cout << ans[i] << endl;
}
return 0;
} | ### Prompt
Please create a solution in Cpp to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <iostream>
#include <vector>
#include <list>
using namespace std;
int main() {
int n; cin >> n;
list<int> b(n);
vector<int> ans;
for (auto& e : b) cin >> e;
while (!b.empty()) {
int t = b.size();
for (auto it = b.rbegin(); it != b.rend(); ++it) {
if (*it == t) {
b.erase(--it.base());
ans.push_back(t);
break;
}
--t;
}
if (!t) break;
}
if (ans.size() != n) cout << -1 << endl;
else {
for (int i = n - 1; i >= 0; --i) cout << ans[i] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int c[110];
int main() {
int n;
cin>>n;
int b[110];
for (int i=0; i<n; i++) {
scanf("%d",b+i);
if (b[i]>i+1) {
cout<<-1<<endl;
return 0;
}
}
for (int i=n-1; i>=0; i--) {
if (b[i]==c[i]+1) {
printf("%d\n",b[i]);
for (int j=i; j<n; j++) {
c[j]++;
}
i=n;
}
}
return 0;
} | ### Prompt
Your task is to create a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int c[110];
int main() {
int n;
cin>>n;
int b[110];
for (int i=0; i<n; i++) {
scanf("%d",b+i);
if (b[i]>i+1) {
cout<<-1<<endl;
return 0;
}
}
for (int i=n-1; i>=0; i--) {
if (b[i]==c[i]+1) {
printf("%d\n",b[i]);
for (int j=i; j<n; j++) {
c[j]++;
}
i=n;
}
}
return 0;
}
``` |
#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
int n,a[105],vis[105],ans[105];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
{
for(int j=n;j>=1;j--)
{
if(vis[j]) continue;
int cc=0;
for(int k=1;k<j;k++) cc+=(!vis[k]);
if(cc==a[j]-1)
{
vis[j]=1,ans[i]=a[j];
break;
}
}
if(!ans[i])
{
puts("-1");
return 0;
}
}
for(int i=n;i>=1;i--) printf("%d\n",ans[i]);
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
int n,a[105],vis[105],ans[105];
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
{
for(int j=n;j>=1;j--)
{
if(vis[j]) continue;
int cc=0;
for(int k=1;k<j;k++) cc+=(!vis[k]);
if(cc==a[j]-1)
{
vis[j]=1,ans[i]=a[j];
break;
}
}
if(!ans[i])
{
puts("-1");
return 0;
}
}
for(int i=n;i>=1;i--) printf("%d\n",ans[i]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
int n,x,i;cin>>n;
vector<int>v;
for(i=1;i<=n;i++){
cin>>x;
if(x>i) {cout<<-1<<endl;return 0;}
v.insert(v.begin()+x-1,x);
}
for(auto it:v)cout<<it<<endl;
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
int n,x,i;cin>>n;
vector<int>v;
for(i=1;i<=n;i++){
cin>>x;
if(x>i) {cout<<-1<<endl;return 0;}
v.insert(v.begin()+x-1,x);
}
for(auto it:v)cout<<it<<endl;
return 0;
}
``` |
#include <iostream>
#include <vector>
using namespace std;
vector<int>v;
int main(void){
int n,b;
cin>>n;
bool flag=0;
for(int i=1;i<=n;i++){
cin>>b;
if(b>i){
flag=1;
break;
}
else v.insert(v.begin()+b-1,b);
}
if(flag) cout<<-1<<endl;
else for(int i=0;i<n;i++) cout<<v[i]<<endl;
} | ### Prompt
Construct a cpp code solution to the problem outlined:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <iostream>
#include <vector>
using namespace std;
vector<int>v;
int main(void){
int n,b;
cin>>n;
bool flag=0;
for(int i=1;i<=n;i++){
cin>>b;
if(b>i){
flag=1;
break;
}
else v.insert(v.begin()+b-1,b);
}
if(flag) cout<<-1<<endl;
else for(int i=0;i<n;i++) cout<<v[i]<<endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n;
vector <int> b, rlt;
int main()
{
scanf("%d", &n);
b.resize(n);
for (int i = 0; i < n; i ++) scanf("%d", &b[i]);
while (n --)
{
int k = n;
while (k >= 0 && b[k] != k + 1) k --;
if (k < 0) return printf("-1"), 0;
rlt.push_back(k + 1), b.erase(b.begin() + k);
}
reverse(rlt.begin(), rlt.end());
for (int i = 0; i < rlt.size(); i ++) printf("%d\n", rlt[i]);
}
| ### Prompt
In cpp, your task is to solve the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
vector <int> b, rlt;
int main()
{
scanf("%d", &n);
b.resize(n);
for (int i = 0; i < n; i ++) scanf("%d", &b[i]);
while (n --)
{
int k = n;
while (k >= 0 && b[k] != k + 1) k --;
if (k < 0) return printf("-1"), 0;
rlt.push_back(k + 1), b.erase(b.begin() + k);
}
reverse(rlt.begin(), rlt.end());
for (int i = 0; i < rlt.size(); i ++) printf("%d\n", rlt[i]);
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(){
int N;
cin >> N;
vector<int> b(N);
vector<int> ans(N);
for(auto &k:b)cin >> k;
for(int k=0;k<N;k++){
for(int i=b.size()-1;i>=0;i--){
if(b[i]>i+1){
cout << "-1" << endl;
return 0;
}
if(b[i]==i+1){
ans[N-k-1]=i+1;
b.erase(b.begin()+i);
break;
}
}
}
for(auto a:ans)cout << a << endl;
return 0;
} | ### Prompt
Generate a cpp solution to the following problem:
Snuke has an empty sequence a.
He will perform N operations on this sequence.
In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).
You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq b_i \leq N
Input
Input is given from Standard Input in the following format:
N
b_1 \dots b_N
Output
If there is no sequence of N operations after which a would be equal to b, print `-1`. If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.
Examples
Input
3
1 2 1
Output
1
1
2
Input
2
2 2
Output
-1
Input
9
1 1 1 2 2 1 2 3 2
Output
1
2
2
3
1
2
2
1
1
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(){
int N;
cin >> N;
vector<int> b(N);
vector<int> ans(N);
for(auto &k:b)cin >> k;
for(int k=0;k<N;k++){
for(int i=b.size()-1;i>=0;i--){
if(b[i]>i+1){
cout << "-1" << endl;
return 0;
}
if(b[i]==i+1){
ans[N-k-1]=i+1;
b.erase(b.begin()+i);
break;
}
}
}
for(auto a:ans)cout << a << endl;
return 0;
}
``` |
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