B

构造90倍数的最大数

最新推荐文章于 2025-04-12 14:14:32 发布

Polynomial Hierarchy

最新推荐文章于 2025-04-12 14:14:32 发布

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本文链接：https://blog.youkuaiyun.com/yuanhanchun/article/details/22582599

探讨了如何从一组仅包含数字0和5的卡片中构造出最大的且能被90整除的数字。该问题要求理解数学原理，并通过算法实现解决方案。

B - Jeff and Digits

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 352A

Description

Jeff's got n cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got?

Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards.

Input

The first line contains integer n(1 ≤ n ≤ 10³). The next line contains n integers a₁, a₂, ..., a_n (a_i = 0 or a_i = 5). Number a_irepresents the digit that is written on the i-th card.

Output

In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1.

Sample Input

Input

4
5 0 5 0

Output

Input

11
5 5 5 5 5 5 5 5 0 5 5

Output

5555555550

当9的倍数个5时，能被9整除，被90 整除，就必须有0；

结论性数论题

#include<stdio.h>
int main()
{
int n,a,b,p,d,i;
while(~scanf("%d",&n))
{
a=0;
b=0;
for( i=0;i<n;i++)
{
scanf("%d",&d);
if(d==5)a++;
if(d==0)b++;
}
p=a/9;
if(b==0)printf("-1\n");
else
{
if(p==0)printf("0");
else
{
for(i=0;i<p*9;i++)
printf("5");
for(i=0;i<b;i++)
printf("0");
}
printf("\n");

}
}
}