[HDU](5058)So easy --去重

So easy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1268    Accepted Submission(s): 684

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 a1,a2,a3,…,an - represents the content of the first file. The third line contains n integers b1,b2,b3,…,bn - represents the content of the second file.
Process to the end of file.
1≤n≤100
1≤ai,bi≤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

自己曾经在参加一个学院小竞赛时，遇到过去重的问题，当时困扰了好久，然后通过学习，在csdn论坛上看到有前辈写了一个去重的算法，十分好理解，觉得不错，所以在做杭电这道题是，用到了，反馈还不错。听说这道题还可以用C++ 的STL做，自己能力还不到，还要努力学习~o(〃'▽'〃)o

题意：给你两个数字序列，去重后，比较相不相等。

思路：先进行排序，然后去重。 

#include<iostream>#include<algorithm>using namespace std;int main(){    int n;    int a[100];    int b[100];    int i;    int j;    while(cin>>n){    int c[100];    int d[100];    int nc=0;    int nd=0;    int count=0;    for(i=0;i<n;i++)        cin>>a[i];    for(i=0;i<n;i++)        cin>>b[i];    sort(a,a+n);    sort(b,b+n);    for(i=0;i<n;i++)    {        for(j=i+1;j<n;j++)        {            if(a[i]==a[j])                a[j]=-1;        }    }    for(i=0;i<n;i++)    {        for(j=i+1;j<n;j++)        {            if(b[i]==b[j])                b[j]=-1;        }    }    for(i=0;i<n;i++)    {        if(a[i]!=-1)        {            c[nc]=a[i];            nc++;        }    }    for(i=0;i<n;i++)    {        if(b[i]!=-1)        {            d[nd]=b[i];            nd++;        }    }    for(i=0;i<nd;i++)    {        if(c[i]==d[i])            count++;    }    if(count==nc && count==nd)        cout<<"YES"<<endl;    else        cout<<"NO"<<endl;    }    return 0;}