CodeForces 653 A. Bear and Three Balls——(IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2))

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CodeForces 653 A. Bear and Three Balls——(IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2))

angel_imp 2016-03-27 19:48:00 浏览1164
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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 45 and 3, or balls with sizes 9091 and 92. But he can't choose balls with sizes 55and 6 (two friends would get balls of the same size), and he can't choose balls with sizes 3031 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
题目大意:
就是给定 n 个数,让你求的就是是否存在三个连续的数。。。

解题思路:

这个题本来这么简单不想写的,但是突然发现一个函数还是比较不错的,unique(),就是这个函数,本来我是打算暴力的后来一寻思太麻烦了,就用集合写的,写到一半突然发现迭代器 it 不能加2,所以炸了,后来就用一个数组存的每一个*it 的数,有点小麻烦,具体看我第一个代码
My First AC Code:
<span style="font-size:18px;">#include <iostream>
#include <algorithm>
#include <set>
#include <cstdio>
#include <cstring>
using namespace std;
set <int> s;
set <int> ::iterator it;
int a[105];
int main()
{
    int n, x;
    while(cin>>n)
    {
        for(int i=0; i<n; i++)
        {
            cin>>x;
            s.insert(x);
        }
        memset(a, 0, sizeof(a));
        int cnt = 0;
        for(it=s.begin(); it!=s.end(); it++)
            a[cnt++] = *it;///每个都存一遍 实在是没招了 set学的不好呀。。。
        bool ok = 0;
        for(int i=1; i<cnt; i++)
        {
            if(a[i]==a[i-1]+1 && a[i]+1==a[i+1])
            {
                ok = 1;
                break;
            }
        }
        if(ok)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}</span>

接下来我要说的就是这篇博客的重点了(有没有想到高中老师呀...)介绍一个 STL 算法,unique(),这个函数是用来去重的,这还是我一队友告诉我的呢,我感觉自己好low啊,我详细的说一下:
unique的功能是去除相邻的重复元素(只保留一个),其实它并不是真正意思上的把重复的元素删除,只是把重复的元素移到后面去了,然后还是保存到了原数组中,然后 返回去重后最后一个元素的地址,因为unique去除的是相邻的重复元素,所以一般用之前都会要排一下序。
int tmp = unique(a,a+n)-a;我们只需要写上这么一句话,就行了,别忘了最后还得减a因为这个函数返回的就是地址,所以要减去,而tmp这个变量里存的就是去重之后的数组的长度,还是比较不错的吧,上我第二个代码:
My Second AC Code:

<span style="font-size:18px;">#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;

int a[105];
int main()
{
    int n, x;
    while(cin>>n)
    {
        for(int i=0; i<n; i++)
            cin>>a[i];
        sort(a,a+n);
        int tmp = unique(a,a+n)-a;///头文件 <algorithm>,还是挺好的,去重用的
        bool ok = 0;
        for(int i=1; i<tmp; i++)
        {
            if(a[i]==a[i-1]+1 && a[i]+1==a[i+1])
            {
                ok = 1;
                break;
            }
        }
        if(ok)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}
</span>


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