【CodeForces - 653A】Bear and Three Balls（sort）

原创于 2017-08-03 00:02:15 发布 · 309 阅读

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本文探讨了一个关于选择三个特定大小球的问题，确保任意两个球的大小相差不超过2且不相同。通过输入不同数量和大小的球进行测试，判断是否能选出符合条件的三个球。

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题目

A. Bear and Three Balls
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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, 5 and 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 sizes 18, 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:

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

去重时要注意

#include<stdio.h>
#include<algorithm>
using namespace std;
#include<string.h>
//#include<iostream>
int a[51];
bool cmp(int a,int b)
{
    return a<b;
}
int main()
{
    int m,n,i,j,t,r;
    int flag=1;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    sort(a,a+n,cmp);
    //unique(a,a+n);去重复的数 
    for(i=0;i<n-1;i++)
    {
        if(a[i]==a[i+1])
        {
            for(j=i;j<n-1;j++)
            {
                a[j]=a[j+1];
            }
            n--;i--;
        }
    }

    for(i=0;i<n-2;i++)
    {
        if(a[i+2]-a[i]<3&&a[i]!=a[i+1]&&a[i]!=a[i+2]&&a[i+1]!=a[i+2])
        {
            flag=0;
            break;
        }
    }
    if(flag==0)
    printf("YES");
    else printf("NO");
return 0;
}