Color the fence
时间限制：1000 ms | 内存限制：65535 KB
难度：2
- 描述
-
Tom has fallen in love with Mary. Now Tom wants to show his love and write a number on the fence opposite to
Mary’s house. Tom thinks that the larger the numbers is, the more chance to win Mary’s heart he has.
Unfortunately, Tom could only get V liters paint. He did the math and concluded that digit i requires ai liters paint.
Besides,Tom heard that Mary doesn’t like zero.That’s why Tom won’t use them in his number.
Help Tom find the maximum number he can write on the fence.
- 输入
- There are multiple test cases.
Each case the first line contains a nonnegative integer V(0≤V≤10^6).
The second line contains nine positive integers a1,a2,……，a9(1≤ai≤10^5). - 输出
- Printf the maximum number Tom can write on the fence. If he has too little paint for any digit, print -1.
- 样例输入
-
5
5 4 3 2 1 2 3 4 5
2
9 11 1 12 5 8 9 10 6 - 样例输出
-
55555
33 思路：贪心思想，我们要让能得到数尽量大，那么就要保证让位数尽量长，在位数
尽量长的基础上，让高位数尽量大。除以最小的数，可以
让位数最长，然而最长可以不仅仅是由最小的数得到。如：V=5,array={2,3,24,32,31,14,15,7,9}，
虽然可以得到最长长度是2，且由除以2可得到结果11，但是最优的结果却是21。所以只要我们能
保证我们得到的高位数字比选择最小的数所得高位数字大并且位数相等即可。 AC代码：#include <stdio.h> int main(void)
{
int V;
while (~scanf("%d", &V))
{
int array[],i;
int min = ;
for (i = ; i <= ; i++)
{
scanf("%d", &array[i]);
if (array[i] <= min)
{
min = array[i];
}
} if (min > V)
{
printf("-1\n");
continue;
}
int bit = V / min;
while(bit--)
{
for (int j = ; j >= ; j--)
{
if (V >= array[j] && (V - array[j] )/ min == bit)
{
printf("%d", j);
V -= array[j];
break;
}
}
}
putchar('\n');
}
return ;
}