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  • codeforces 653A A. Bear and Three Balls(水题)

    题目链接:

    A. Bear and Three Balls

     

    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Limak is a little polar bear. He has n balls, the i-th ball has size ti.

    Limak wants to give one ball to each of his three friends. Giving gifts isn't easy — there are two rules Limak must obey to make friends happy:

    • No two friends can get balls of the same size.
    • No two friends can get balls of sizes that differ by more than 2.

    For example, Limak can choose balls with sizes 4, 5 and 3, or balls with sizes 90, 91 and 92. But he can't choose balls with sizes 5, 5and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 30, 31 and 33 (because sizes 30 and 33 differ by more than 2).

    Your task is to check whether Limak can choose three balls that satisfy conditions above.

    Input

    The first line of the input contains one integer n (3 ≤ n ≤ 50) — the number of balls Limak has.

    The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 1000) where ti denotes the size of the i-th ball.

    Output

    Print "YES" (without quotes) if Limak can choose three balls of distinct sizes, such that any two of them differ by no more than 2. Otherwise, print "NO" (without quotes).

    Examples
    input
    4
    18 55 16 17
    output
    YES
    input
    6
    40 41 43 44 44 44
    output
    NO
    input
    8
    5 972 3 4 1 4 970 971
    output
    YES
    Note

    In the first sample, there are 4 balls and Limak is able to choose three of them to satisfy the rules. He must must choose balls with sizes18, 16 and 17.

    In the second sample, there is no way to give gifts to three friends without breaking the rules.

    In the third sample, there is even more than one way to choose balls:

    1. Choose balls with sizes 3, 4 and 5.
    2. Choose balls with sizes 972, 970, 971.

    题意:就是问你是否有连续的三个数;

    AC代码:

    #include <bits/stdc++.h>
    using namespace std;
    int a[1000],num[1005];
    int main()
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            num[a[i]]++;
        }
        for(int i=1;i<=1000;i++)
        {
            if(num[i]&&num[i+1]&&num[i+2])
            {
                cout<<"YES";
                return 0;
            }
        }
        cout<<"NO";
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5296636.html
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