Question

分发奖章这是奖牌分发仪式。 10 ^ 6名警察，编号从1到10 ^ 6，排成一列。奖牌分发有N (1<=N<=1000)次迭代。迭代i (0 < = i < N), count[i] ( 1 < = count[i] < = 100)奖牌将颁发给from[i] to to[i] ( 1 < = from[i] < = to[i] < = 10^6 )

的所有官员

如果我们总结从副驾驶开始收到的奖牌数量，谁将是累积金额超过给定奖牌count THRESHOLD ( 1 < = THRESHOLD < = 10^9 )的副驾驶？

输入/输出规格输入格式：

您将获得5个输入：

input1 = N，迭代次数
input2 = count，每次迭代中奖牌计数的数组
input3 = from，每次迭代中的起始索引数组
input4 = to，每次迭代中的结束索引数组
input5 = THRESHOLD，奖牌数阈值

输出格式：

一个整数，表示副驾驶的号码，使得从第一军官开始到该军官的累积奖牌总数超过THRESHOLD。如果这样的官员不存在，输出应为-1。

Answer 1

它在c＃语言中工作

public static int DistributingMedals(int input1, int[] input2, int[] input3, int[] input4, int input5)
{
    int officerIndex = -1;
    if (InputsValid(input1, input2, input3, input4, input5))
    {
        int medalsCount = 0;
        for (int i = 0; i < input1; i++)
        {
            for (int o = input3[i]; o <= input4[i]; o++)
            {
                medalsCount += input2[i];
                if (medalsCount > input5)
                {
                    officerIndex = o;
                    break;
                }
            }
            if (medalsCount > input5)
                break;
        }
    }
    return officerIndex;
}

private static bool InputsValid(int input1, int[] input2, int[] input3, int[] input4, int input5)
{
    if (((1 <= input1) && (input1 <= 1000))
        && ((input2.Length == input1) && (input3.Length == input1) && (input4.Length == input1))
        && ((1 <= input5) && (input5 <= 1000000000)))
    {
        int ok = 0;
        for (int i = 0; i < input1; i++)
        {
            if ((1 <= input3[i] && input3[i] <= input4[i] && input4[i] <= 1000000)
                && (1 <= input2[i] && input2[i] <= 100))
                ok++;
        }
        if (ok == input1)
            return true;
    }
    return false;
}

Answer 2

In Java#
-----------------------

public class CandidateCode 
{ 
    public static void main(String[] args) {
        int answer = DistributingMedals(1,new int[]{1},new int[]{1},new int[]{10},2);
        System.out.println(answer);
    }

    /**
     * 
     * @param input1 = N, the number of iterations
     * @param input2 = count, the array of medal counts in each iteration
     * @param input3 = from, the array of starting indices in each iteration
     * @param input4 = to, the array of ending indices in each iteration
     * @param input5 = THRESHOLD, the medal count threshold

     * @return
     */

    public static int DistributingMedals(int input1,int[] input2,int[] input3,int[] input4,int input5)
    {

        int officerIndex = -1;
        if (InputsValid(input1, input2, input3, input4, input5))
        {
            int medalsCount = 0;
            for (int i = 0; i < input1; i++)
            {
                for (int o = input3[i]; o <= input4[i]; o++)
                {
                    medalsCount += input2[i];
                    if (medalsCount > input5)
                    {
                        officerIndex = o;
                        break;
                    }
                }
                if (medalsCount > input5)
                    break;
            }
        }
        return officerIndex;

    }   


    private static boolean InputsValid(int input1, int[] input2, int[] input3, int[] input4, int input5)
    {
        if (((1 <= input1) && (input1 <= 1000))
            && ((input2.length == input1) && (input3.length == input1) && (input4.length == input1))
            && ((1 <= input5) && (input5 <= 1000000000)))
        {
            int ok = 0;
            for (int i = 0; i < input1; i++)
            {
                if ((1 <= input3[i] && input3[i] <= input4[i] && input4[i] <= 1000000)
                    && (1 <= input2[i] && input2[i] <= 100))
                    ok++;
            }
            if (ok == input1)
                return true;
        }
        return false;
    }
}

用Java分发奖牌解决方案

2 个答案: