HDU 5058 So easy(STL set运用)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5058

Problem 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

 

Source

BestCoder Round #12

代码如下：

#include <cstdio>#include <algorithm>#include <set>using namespace std;int main(){    set<int>s1,s2;    int n;    while(~scanf("%d",&n))    {        s1.clear();        s2.clear();        int tt;        for(int i = 0; i < n; i++)        {            scanf("%d",&tt);            s1.insert(tt);        }        for(int i = 0; i < n; i++)        {            scanf("%d",&tt);            s2.insert(tt);        }        if(s1.size() != s2.size())        {            printf("NO\n");            continue;        }        int flag = 0;        set<int>::iterator it1, it2;        for(it1 = s1.begin(),it2 = s2.begin(); it1!=s1.end()&&it2!=s2.end(); it1++,it2++)        {            if(*it1 != *it2)            {                flag = 1;                break;            }        }        if(flag)        {            printf("NO\n");        }        else            printf("YES\n");    }    return 0;}