Codeforces-1250-H. Happy Birthday（思维+枚举）

H. Happy Birthday

You have a set of birthday cake candles. Each of such candles represents a digit between 0 and 9, inclusive.

Example of birthday cake candles.
Let’s denote the candle representing the digit d as d-candle.

Your set contains c0 instances of 0-candles, c1 instances of 1-candles and so on. So, the total number of candles is c0+c1+⋯+c9.

These digits are needed to wish your cat a happy birthday. For each birthday, starting with the first, you want to compose the age of the cat using the digits from the set.

Since you light candles for a very short time, candles don’t have time to burn out. For this reason you can reuse candles an arbitrary number of times (therefore your set of candles never changes).

For example, if you have one instance of each digit (i.e. c0=c1=⋯=c9=1), you can compose any number from 1 to 10 using this set, but you cannot compose 11.

You have to determine the first birthday, on which you cannot compose the age of the cat using the candles from your set. In other words, find the minimum number y such that all numbers from 1 to y−1 can be composed by digits from your set, but y cannot be composed.

Input

The first line contains an integer t (1≤t≤104) — the number of test cases in the input.

The only line of each test case contains ten integer numbers c0,c1,…,c9 (0≤ci≤105) — the number of 0-candles, 1-candles, 2-candles and so on.

It is guaranteed that the sum of all ci in the input does not exceed 106.

Output

For each test case, output one integer in single line — the minimum age which cannot be composed by candles from your set. Please note that the age can be quite large (it may exceed the standard 64-bit integer types in your programming language).

Example

input

4
1 1 1 1 1 1 1 1 1 1
0 0 1 1 2 2 3 3 4 4
1 2 1 2 1 3 1 0 0 0
0 1 2 1 4 3 1 1 2 1

output

11
1
7
10

解题思路：

题目给你0-9各n个数，问这些数字不能组成的最小数是多少。首先题目有一个隐含条件没有明说就是答案不能为0。那么考虑问题的解有哪些情况，倘若每一个数字都有一位，那么就可以组成所有的一位数，倘若每一个数字都有两位，那么就可以组成任意2位数。然而若其他数位都有n+个 而某个数位只有小于n个，那么答案就是这个数字输出n次，需要注意的事，答案会有诸如，11，22，3333，4444，但是，因为0不能组成0000，所以0可以比恒其他数位少一。所以0需要特判一下。那么枚举位数n，去找有没有小于n的数位就行了。

AC代码：

#include 
using namespace std;

const int N = 1e6+10;

int a[11];
int main()
{
    int n;
    cin>>n;
    for(int i = 0 ; i < n ; i ++)
    {
        for (int j = 0 ; j < 10 ; j ++)
            cin>>a[j];
        int flag = 0;
        for(int k = 0 ; k < 1000005;k ++)
        {
            if(a[0] == k-1)
            {
                cout<<1;
                for(int j = 0 ; j < k ; j ++)
                    cout<<0;
                cout<<endl;
                break;
            }
            for(int j = 1 ; j < 10 ; j ++)
            {
                if(a[j] == k)
                {
                    for(int l = 0 ; l <= k ; l ++)
                        cout<<j;
                    cout<<endl;
                    flag = 1;
                    break;
                }
            }
            if(flag) break;
        }

    }
}