CF#202DIV2:B. Color the Fence

http://codeforces.com/contest/349/problem/B

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 digitd requiresa_{d 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 ≤ 10^{6). The second line contains nine positive integersa_{1, 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

思路：貪心即可

 

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int main(){    int a[10],sum,minn,k,i,j,cnt,r;    while(~scanf("%d",&sum))    {        minn = 1000000005;        k = 0;        for(i = 1; i<=9; i++)        {            scanf("%d",&a[i]);            if(a[i]<minn)//找出花費最小的顏料數            {                minn = a[i];                k = i;            }            else if(a[i] == minn && k<i)//花費與最少的相等，自然取數字大的                k = i;        }        if(sum<minn)//最少的花費都大於總顏料，自然不行        {            printf("-1\n");            continue;        }        cnt = sum/minn;//最小花費能得到的數字長度        r = sum%minn;        if(!r)//若總顏料能整出最小花費，全部輸出這個數一定是最大        {            for(i = 1; i<=cnt; i++)                printf("%d",k);            printf("\n");            continue;        }        while(sum>0)//總顏料還沒用完，找出與最小花費長度相等的最大數        {            int x = 0;            for(i = 1; i<=9; i++)            {                int s = sum-a[i];//減去這個數位花費                if(s<0)//這個數字會使顏料用完，換下一個顏料                    continue;                if(s/minn == cnt-1 && x<i)//減去該數位花費之後，剩下的除以最小的長度是原來長度-1，必定是最符合的，在所有最符合的狀況中找到最大的                    x = i;            }            if(x)            {                sum-=a[x];//放了一個數字，總花費減少                cnt--;//長度減一                printf("%d",x);//輸出這個數字            }        }        printf("\n");    }    return 0;}