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

55 4 3 2 1 2 3 4 5

output

55555

input

29 11 1 12 5 8 9 10 6

output

33

input

01 1 1 1 1 1 1 1 1

output

-1

题意：一个人用油漆在栅栏上写数字，每个数字i需要ai的油漆，给你v升油，问你能写出的最大数字，数字钟不包含零

解题思路：贪心，先找到花费最小且数字最大的数，如果他能恰好用完油漆则输出v/mi个这个数字就是最大数字，

否则就是找长度满足v/mi前几位为更大的数但满足所用油漆小于等于v，详见代码。

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 1000005#define LL long long#define inf 9999999int 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)//找到满足长度为xx，但前几位数大于x的数                k=i;            }            if(k)            {                v-=a[k];                xx--;                printf("%d",k);            }        }        printf("\n");    }    return 0;}