Codeforces Round #202 (Div. 2)B. Color the Fence

B. Color the Fence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Igor has fallen in love with Tanya. Now Igor wants to show his feelings and write a number on the fence opposite to Tanya's house. Igor thinks that the larger the number is, the more chance to win Tanya's heart he has.

Unfortunately, Igor could only get v liters of paint. He did the math and concluded that digit d requires ad liters of paint. Besides, Igor heard that Tanya doesn't like zeroes. That's why Igor won't use them in his number.

Help Igor find the maximum number he can write on the fence.

Input

The first line contains a positive integer v (0 ≤ v ≤ 106). The second line contains nine positive integersa1, a2, ..., a9 (1 ≤ ai ≤ 105).

Output

Print the maximum number Igor can write on the fence. If he has too little paint for any digit (so, he cannot write anything), print -1.

Sample test(s)

input

5
5 4 3 2 1 2 3 4 5

output

55555

input

2
9 11 1 12 5 8 9 10 6

output

33

input

0
1 1 1 1 1 1 1 1 1

output

-1

題意：一個人用油漆在柵欄上寫數字，每個數字i需要ai的油漆，給你v升油，問你能寫出的最大數字，數字鐘不包含零

解題思路：貪心，先找到花費最小且數字最大的數，如果他能恰好用完油漆則輸出v/mi個這個數字就是最大數字，

否則就是找長度滿足v/mi前幾位爲更大的數但滿足所用油漆小於等於v，詳見代碼。

代碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 1000005
#define LL long long
#define inf 9999999
int a[15];
int main()
{
    int v,i,j;
    while(scanf("%d",&v)!=EOF)
    {
        int x;
        int mi=inf;
        for(i=1;i<=9;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]<=mi)
            {
                mi=a[i];
                x=i;
            }
        }
        int xx=v/mi;
        int fg=v%mi;
        if(mi>v)
        {
            printf("-1\n");
            continue;
        }
        if(fg==0)
        {
            for(i=1;i<=xx;i++)
            {
                printf("%d",x);
            }
            printf("\n");
            continue;
        }

        while(v>0&&xx>0)
        {
            int k=0;
            for(i=x;i<=9;i++)
            {
                int d=v-a[i];
                if(d>=0&&k<i&&d/mi==xx-1)//找到滿足長度爲xx，但前幾位數大於x的數
                k=i;
            }
            if(k)
            {
                v-=a[k];
                xx--;
                printf("%d",k);
            }
        }
        printf("\n");
    }
    return 0;
}