Problem:
Score : 800 points

Problem Statement

You are given an integer sequence x of length N.
Determine if there exists an integer sequence a that satisfies all of the following conditions, and if it exists, construct an instance of a.

a is N^2 in length, containing N copies of each of the integers 1, 2, ..., N.

For each 1 ≤ i ≤ N, the i-th occurrence of the integer i from the left in a is the x_i-th element of a from the left.

Constraints

1 ≤ N ≤ 500

1 ≤ x_i ≤ N^2

All x_i are distinct.

Input

The input is given from Standard Input in the following format:

N
x_1 x_2 ... x_N

Output

If there does not exist an integer sequence a that satisfies all the conditions, print No.
If there does exist such an sequence a, print Yes in the first line, then print an instance of a in the second line, with spaces inbetween.

Sample Input 1

3
1 5 9

Sample Output 1

Yes
1 1 1 2 2 2 3 3 3

For example, the second occurrence of the integer 2 from the left in a in the output is the fifth element of a from the left.
Similarly, the condition is satisfied for the integers 1 and 3.

Sample Input 2

2
4 1

Sample Output 2

No

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)


#include<stdio.h>
#define NIL -1

int main(){
    int n;
    scanf("%d",&n);
    int x[n+1],i;
    for(i=1;i<=n;i++)scanf("%d",&x[i]);
    int ans[n*n+1],first[n*n+1];
    for(i=1;i<=n*n;i++)ans[i]=NIL,first[i]=NIL;
    for(i=1;i<=n;i++)ans[x[i]]=i,first[x[i]]=i;
    
    int j=1,flag=1;  /*flag 1は真 0は偽*/
    
    while(j<=n*n){
        if(ans[j]!=NIL){
            int count=1,k=1;
            while(count<=ans[j]-1 && k<=j-1){
                if(ans[k]==NIL){
                    ans[k]=ans[j];
                    count++;
                }
                k++;
            }
            if(count!=ans[j])flag=0;
        }
        j++;
    }
    
    j=n*n;
    while(j>=1){
        if(first[j]!=NIL){
            int count=1,k=n*n;
            while(count<=n-ans[j] && k>=1){
                if(ans[k]==NIL){
                    ans[k]=ans[j];
                    count++;
                }
                k--;
            }
            if(count!=n-ans[j]+1)flag=2;
        }
        j--;
    }
    
    if(flag!=1)printf("No\n");
    else{
        printf("Yes\n");
        for(i=1;i<=n*n;i++){
            printf("%d",ans[i]);
            if(i!=n*n)printf(" ");
            else printf("\n");
        }
    }
    
    return 0;
}