Codeforces Round #289 C. Sums of Digits（构造）

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

Vasya wonders what the numbers a_i could be like. Of all the possible options he likes the one sequence with the minimum possible last number a_n. 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, ..., b_n  — the required sums of digits. All b_i belong to the range 1 ≤ b_i ≤ 300.

Output

Print n integer numbers, one per line — the correct option for numbers a_i, 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 b_i.

If there are multiple sequences with least possible number a_n, 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：

题目给出n个数，表示为每一个数的位数和，要求得到的数列是从大到小且都最小的。

思路：

根据上一个数来构造下一个数，先根据上一个数确定高位，使得此时的数比上一个数大，接着从最低位开始填数，在数位和的范围内尽可能的填。

分为两种情况：1‘上一个和大于下一个和，则从高位开始遍历，直到小于数位和的最低位，此位增加1（在不是9的情况下），接着从最低位开始赋值，尽可能大的赋值。2’上一个和小于下一个和，从最低位开始尽可能大的赋值。

CODE：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int maxn = 305;
int N, n[maxn];
int ans[maxn][1005], cnt[maxn];

void getnum(int n1, int i)
{
    int ai = 0;
    while(n1 > 0) {
        ans[i][ai++] = n1-9 >= 0? 9: n1;
        n1 -= 9;
    }
}

int main()
{  // freopen("in", "r", stdin);

    while(~scanf("%d", &N)) {
        memset(ans, 0, sizeof(ans));
        for(int i = 0; i < N; ++i) {
            scanf("%d", &n[i]);
        }
        int sum;
        getnum(n[0], 0);
         for(int i = 1; i <= N; ++i) {

            for(int k = 1000; k >= 0; --k) {
                if(ans[i-1][k]) {
                    cnt[i-1] = k;
                    if(i == N) break;
                    sum = 0;
                    for(int j = k; j >= 0; --j){

                        if(sum + ans[i-1][j] >= n[i]) {
                            j++;
                            while(ans[i-1][j] == 9) {
                                sum -= ans[i-1][j];
                                ans[i][j] = 0;
                                j++;
                            }

                            ans[i][j] = ans[i-1][j] + 1;

                            int s = n[i] - sum - 1;
                            //if(i == 45) printf("s = %d\n", s);
                            if(s > 0) {
                                getnum(s, i);
                            }
                            sum = sum + s + 1;
                            break;
                        }
                        else {
                            sum += ans[i-1][j];
                            ans[i][j] = ans[i-1][j];
                        }
                    }
                    break;
                }
            }

            if(n[i] - n[i-1]) {
                int x = n[i] - n[i-1];
                int ai = 0;
                while(x > 0) {
                    if(9 - ans[i][ai] <= x) {
                        x = x - (9 - ans[i][ai]);
                        ans[i][ai] = 9;
                    }
                    else {
                        ans[i][ai] += x;
                        x -= x;
                    }
                    ai++;
                }

            }
        }

        for(int i = 0; i < N; ++i) {
            for(int j = cnt[i]; j >= 0; --j) {
                printf("%d", ans[i][j]);
            }
            printf("\n");
        }

    }
    return 0;
}