zoukankan      html  css  js  c++  java
  • Codeforces Round #204 (Div. 2)——A找规律——Jeff and Digits

    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 a1a2..., 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
    ");continue;}
            if(num1 == 0 && num2 != 0){ printf("0
    ");continue;}
            if(num1 == 0 && num2 == 0){printf("-1
    "); continue;}
            else {
            for(int i = 1; i <= num1; i++){
                printf("555555555");
            }
            for(int i = 1; i <= num2; i++)
             printf("0");
            }
            puts("");
        }
        return 0;
    }
    

      

  • 相关阅读:
    bzoj 1497 最小割模型
    bzoj 1024 暴力深搜
    POJ1163(简单的DP)
    POJ3287(BFS水题)
    N皇后问题(DFS)
    BFS求解迷宫的最短路径问题
    poj2386(简单的dfs/bfs)
    Fence Repair(poj3253)
    Best cow Line(POJ 3617)
    全排列
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4652373.html
Copyright © 2011-2022 走看看