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 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.

原题链接 http://codeforces.com/contest/349/problem/B

(1) 题目大意 

给你一个maxmoney，代表能用的最大钱数，之后给你1-9每位需要花的钱数，问能组成最大的数是多少

（2）思路

我的思路就是先贪心位数最大

然后从首位开始从新贪心，用剩余的钱数去尽可能买数字大的，

贴代码了（通俗易懂

#include <algorithm>
#include <stack>
#include <istream>
#include <stdio.h>
#include <map>
#include <math.h>
#include <vector>
#include <iostream>
#include <queue>
#include <string.h>
#include <set>
#include <cstdio>
#define FR(i,n) for(int i=0;i<n;i++)
#define MAX 2005
#define mkp pair <int,int>
using namespace std;
const int maxn = 1e6+5;
typedef long long ll;
const int  inf = 0x3fffff;
void read(int  &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar());
    x *= 1 - 2 * flag;
}

int n,arr[1005];
int ans[maxn],top = 0;
int main() {
    read(n);
    int minn = 10000000;
    int id = 0;
    for(int i=1;i<=9;i++){
        read(arr[i]);
        if(minn>arr[i]){
            minn = arr[i];
            id = i;
        }
    }
    //cout<<1<<endl;
    if(n<minn){return 0*printf("-1
");}
    top = n/minn;
    for(int i=0;i<top;i++)ans[i]=id;
    int last = n-top*minn;
    for(int i=0;i<top;i++)
    {
        last  += arr[ans[i]];
        bool an= 1;
        for(int j=9;j>=1;j--){
            if(last>=arr[j]){
                ans[i]=j;
                an = 0;
                last-=arr[j];
                break;
            }
        }
        if(an)break;
       // last-=arr[ans[i]];
    }
    for(int i=0;i<top;i++){
        printf("%d",ans[i]);
    }
    return 0;
}