cf509C Sums of Digits

C. Sums of Digits

time limit per test 2 seconds

memory limit per test 256 megabytes

input standard input

output standard output

Vasya had a strictly increasing sequence of positive integers a1, ..., an. Vasya used it to build a new sequence b1, ..., bn, where bi is the sum of digits of ai's decimal representation. Then sequence ai got lost and all that remained is sequence bi.

Vasya wonders what the numbers ai could be like. Of all the possible options he likes the one sequence with the minimum possible last number an. Help Vasya restore the initial sequence.

It is guaranteed that such a sequence always exists.

Input

The first line contains a single integer number n (1 ≤ n ≤ 300).

Next n lines contain integer numbers b1, ..., bn  — the required sums of digits. All bi belong to the range 1 ≤ bi ≤ 300.

Output

Print n integer numbers, one per line — the correct option for numbers ai, in order of following in sequence. The sequence should be strictly increasing. The sum of digits of the i-th number should be equal to bi.

If there are multiple sequences with least possible number an, print any of them. Print the numbers without leading zeroes.

Sample test(s)

input

3
1
2
3

output

1
2
3

input

3
3
2
1

output

3
11
100

题意是给你一个递增序列的每个数的各位数字之和，求还原这个数列
显然贪心，每次用比上一个数大的最小的那个就行了
但是这题模拟实现太蛋疼了
而且极限数据是一开始a[1]=300要凑出3后面33个9，然后a[2]到a[300]是299个1……这样只开300零几位的就死了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7fffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,len;
int a[500],s[500],t[500];
inline void calc(int x,int l)
{
    memset(s,0,sizeof(s));
    len=0;x--;s[l]=1;
    while (x>=9&&len<l)
    {
        s[++len]=9;
        x-=9;
    }
    if (x)s[++len]+=x;
    len=l;
}
inline bool getmn(int x)
{
    t[len+1]=0;
    for (int i=len;i>=1;i--)
        t[i]=t[i+1]+s[i];
    for (int i=1;i<=len;i++)
    {
        if (t[i]+1>x||s[i]==9)continue;
        s[i]++;
        while (x>9*(i-1)+t[i+1]+s[i]&&s[i]<9)s[i]++;
        if (x>9*(i-1)+t[i+1]+s[i])continue;
        x=x-(t[i+1]+s[i]);
        int now=1;
        while(now<i&&x>=9)
        {
            s[now++]=9;
            x-=9;
        }
        if (now!=i)s[now++]=x;
        for (int j=now;j<i;j++)s[j]=0;
        return 1;
    }
    return 0;
}
inline void put()
{
    for (int i=len;i>=1;i--)
        printf("%d",s[i]);
    printf("
");
}
int main()
{
    n=read();
    for (int i=1;i<=n;i++)a[i]=read();
    calc(a[1],a[1]/9+(a[1]%9!=0));
    put();
    
    for (int i=2;i<=n;i++)
    {
        if (a[i]>9*len)calc(a[i],a[i]/9+(a[i]%9!=0));
        if (!getmn(a[i]))calc(a[i],max(a[i]/9+(a[i]%9!=0),len+1));
        put();
    }
}