Codeforces Round #204 (Div. 2)——A找规律——Jeff and Digits
2015-07-16 20:10
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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, ..., an (ai = 0 or ai = 5). Number airepresents 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
Output
Input
Output
Hint
In the first test you can make only one number that is a multiple of 90 — 0.
In the second test you can make number 5555555550, it is a multiple of 90.
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, ..., an (ai = 0 or ai = 5). Number airepresents 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
0
Input
11 5 5 5 5 5 5 5 5 0 5 5
Output
5555555550
Hint
In the first test you can make only one number that is a multiple of 90 — 0.
In the second test you can make number 5555555550, it is a multiple of 90.
/* 找规律 9个5能被9整除 把0全部放到最后,如果没有0就失败 */ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { int n; int a[1100]; while(~scanf("%d", &n)){ for(int i = 1; i <= n ;i++) scanf("%d", &a[i]); int num1 = 0, num2 = 0; for(int i = 1; i <= n ;i++){ if(a[i] == 5) num1++; if(a[i] == 0) num2++; } num1/= 9; if(num2 == 0 ){ printf("-1\n");continue;} if(num1 == 0 && num2 != 0){ printf("0\n");continue;} if(num1 == 0 && num2 == 0){printf("-1\n"); continue;} else { for(int i = 1; i <= num1; i++){ printf("555555555"); } for(int i = 1; i <= num2; i++) printf("0"); } puts(""); } return 0; }
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