暑假集训第三周周三赛 STL A - So easy 判断集合

A - So easy

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 5058

Description

Small W gets two files. There are n integers in each file. Small W wants to know whether these two files are same. So he invites you to write a program to check whether these two files are same. Small W thinks that two files are same when they have the same integer set. 
For example file A contains (5,3,7,7),and file B contains (7,5,3,3). They have the same integer set (3,5,7), so they are same. 
Another sample file C contains(2,5,2,5), and file D contains (2,5,2,3). 
The integer set of C is (2,5),but the integer set of D is (2,3,5),so they are not same. 
Now you are expected to write a program to compare two files with size of n.

 

Input

Multi test cases (about 100). Each case contain three lines. The first line contains one integer n represents the size of file. The second line contains n integers $a_1, a_2, a_3, \ldots, a_n$ - represents the content of the first file. The third line contains n integers $b_1, b_2, b_3, \ldots, b_n$ - represents the content of the second file. 
Process to the end of file. 
$1 \leq n \leq 100$ 
$1 \leq a_i , b_i \leq 1000000000$

 

Output

For each case, output "YES" (without quote) if these two files are same, otherwise output "NO" (without quote).

 

Sample Input

31 1 21 2 245 3 7 77 5 3 342 5 2 32 5 2 531 2 31 2 4 

Sample Output

YESYESNONO 

分析：

可能昨天就是心情不好吧，提都不想做，就这一道都是好不容易才做的。

这题就是找他们的集合是不是相同，所以要用set

它的作用主要就是去重，排序，如果经过处理后他们的集合一样，就输出YES

set只有一个自变量，所以没有first或second，只能用*p1！=*p2

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#include<stdio.h>#include<algorithm>#include<set>using namespace std;int main(){    int n,a,i,f;    set<int>s1,s2;    while(scanf("%d",&n)!=EOF)    {        s1.clear();        s2.clear();        for(i=0; i<n; i++)        {            scanf("%d",&a);            s1.insert(a);        }        for(i=0; i<n; i++)        {            scanf("%d",&a);            s2.insert(a);        }        if(s1.size()!=s2.size())            printf("NO\n");        else        {            f=0;            set<int>::iterator p1,p2;            for(p1=s1.begin(),p2=s2.begin(); p1!=s1.end()&&p2!=s2.end(); p1++,p2++)                if(*p1!=*p2)                {                    f=1;                    break;                }            if(f==0)                printf("YES\n");            else                printf("NO\n");        }    }    return 0;}