CF349 B. Color the Fence

题目传送门：https://codeforces.com/problemset/problem/349/B

题目大意：
给你总颜料数(v)，再给你9个数(a_{1...9})，(a_i)表示画(i)的颜料消耗，求所能画出的最大的数

---

要考虑画的数最大，首先得看位数，位数 (len) 可以用 $lfloor frac{v}{Min} floor $确定，而 (Min=min{a_i})

得到位数后，我们从高位，从大数开始

如果 (lfloorfrac{v-a_i}{Min} floor=len-1) ，则说明该位可填，否则不行

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=10;
int A[N+10];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int v=read(0),Min=int_inf;
	for (int i=1;i<10;i++)	A[i]=read(0),Min=min(Min,A[i]);
	if (v<Min){
		printf("-1
");
		return 0;
	}
	int Cnt=v/Min;
	while (Cnt--){
		for (int i=9;i;i--){
			if (v>=A[i]&&(v-A[i])/Min==Cnt){
				printf("%d",i);
				v-=A[i];
				break;
			}
		}
	}
	putchar('
');
	return 0;
}