CodeForces 349B Color the Fence(完全背包）

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 integers a1, 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

题意：给出九个数，分别表示用写出1~9这九个数所需要的油漆量，求V升油漆能写出的最大数。

思路：1~9代表九种东西，所给的九个数代表东西的重量，每种东西的价值都为1.问题归根为完全背包问题。求出的最大价值即是所求数的位数

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <math.h>
#include <vector>
using namespace std;


int w[20], v[20], dp[20][1000009];
int main()
{
    int V;
    scanf("%d", &V);
    for(int i=0; i<9; i++)
    {
        scanf("%d", &w[i]);
    }
    for(int i=0; i<9; i++)
    {
        v[i]=1;
    }
    for(int i=0; i<9; i++)
    {
        for(int j=0; j<=V; j++)
        {
            if(j<w[i])
            {
                dp[i+1][j]=dp[i][j];
            }
            else
            {
                dp[i+1][j]=max(dp[i][j], dp[i+1][j-w[i]]+v[i]);
            }
        }
    }
    if(dp[9][V]==0)
        printf("-1\n");


    else
    {
        for(int i=9; i>0; i--)
        {
            int j=V;
            for(; j>=0; )
            {
                if(j-w[i-1]>=0 && dp[i][j]-dp[i][j-w[i-1]]==1)
                {
                    printf("%d", i);
                    j-=w[i-1];
                }
                else
                {
                    V=j;
                    break;
                }
            }
            if(dp[i][j]==0)
                break;
        }
        printf("\n");
    }
    return 0;
}